ISI MSQE ME(I) PREVIOUS YEAR’SANSWER KEY
ISI MSQE 2004-15 ME-I SAMPLE PAPER SOLUTION
ISI: Year 2004
1.(a) 2. (b)
3. (c)
4. (a)
5. (c) 6. (d) 7. (b)
8. (c) 9. (c) 10. (a) 11. (d) 12. (c) 13. (c) 14.(d)
15. (c) 16. (a) 17. (b) 18. (c) 19. (b) 20. (a) 21. (b)
22. (d) 23. (d) 24. (a) 25. (c) 26. (c) 27. (a)
28. (d) 29. (d) 30. (b)
ISI: Year 2005
1. (a) 2. (c) 3. (a) 4. (c) 5. (a) 6. (c)
7. (a) 8. (c) 9. (d) 10. (d) 11. (c) 12. (d)
ISI: Year 2006
1. (a) 2. (a) 3. (d) 4. (c) 5. (d)
6. (a)
7. (c) 8. (d)
9. (b) 10. (d) 11. (c) 12. (c) 13. (a) 14. (a) 15. (d)
ISI: Year 2007
1. (a)
8. (a)
15. (c)
22. (d)
2. (b) 3. (a) 4. (d) 5. (d)
9. (c) 10. (d) 11. (b) 12. (b)
16. (a) 17. (b) 18. (d) 19. (a)
23. (c) 24. (d) 25. (b) 26. (d)
29. (d) 30. (a)
6. (c)
13. (d)
20. (b)
27. (c)
7. (b)
14. (a)
21. (b)
28. (d)
ISI: Year 2008
1. (d) 2. (b) 3. (c) 4. (b) 5. (c)
6. (c)
8. (d) 9. (d) 10. (c) 11. (c) 12. (b) 13. (b)
15. (b) 16. (c) 17. (a) 18. (b) 19. (a) 20. (d)
22. (c) 23. (a) 24. (a) 25. (b) 26. (a) 27. (c)
29. (d) 30. (c)
1
7. (b)
14. (c)
21. (d)
28. (d)
ISI MSQE ME(I) PREVIOUS YEAR’SANSWER KEY
ISI: Year 2009
1. (b)
9. (c)
17. (b)
2. (d) 3. (b)
4. (a)
10. (c) 11. (b) 12. (b)
18. (a) 19. (c) 20. (a)
25. (b) 26. (a) 27. (b)
5. (d)
13. (c)
21. (c)
28. (b)
6. (c)
14. (b)
22. (a)
29. (a)
7. (c)
15. (d)
23. (b)
30. (d)
8. (c)
16. (b)
24. (d)
ISI: Year 2010
1. (a) 2. (a) 3. (a) 4. (d) 5. (a) 6. (c) 7. (b) 8. (b) 9. (c) 10. (a)
11. (c) 12. (c) 13. (b) 14. (b) 15. (c) 16. (c) 17. (b) 18. (c) 19. (b) 20. (d)
21. (d) 22. (b) 23. (d) 24. (c) 25. (a) 26. (b) 27. (c) 28. (d) 29. (b) 30. (c)
ISI: Year 2011
1. (d) 2. (d) 3. (b) 4. (d) 5. (c) 6. (c) 7. (d) 8. (a) 9. (c) 10. (b)
11.(d) 12. (c) 13. (a) 14. (d) 15. (c) 16. (a) 17. (d) 18. (b) 19. (d) 20. (b)
21.(c) 22. (c) 23. (a) 24. (a) 25. (b) 26. (a) 27. (b) 28. (c) 29. (c) 30. (b)
ISI: Year 2012
1. (c) 2. (c) 3. (c) 4. (b) 5. (c) 6. (a) 7. (d) 8. (a) 9. (a) 10. (c)
11.(a) 12. (b) 13. (a) 14. (c) 15. (a) 16. (a) 17. (d) 18. (b) 19. (c) 20. (d)
21.(c) 22. (d) 23. (a) 24. (d) 25. (c) 26. (a) 27. (d) 28. (b) 29. (b) 30. (a)
ISI: Year 2013
1. (d) 2. (c) 3. (a) 4. (b) 5. (a) 6. (c) 7. (a) 8. (c) 9. (a) 10. (a)
11.(c) 12. (b) 13. (d) 14. (d) 15. (d) 16. (c) 17. (c) 18. (a) 19. (c) 20. (d)
21.(c) 22. (d) 23. (a) 24. (c) 25. (c) 26. (a) 27. (b) 28. (d) 29. (b) 30. (d)
2
ISI MSQE ME(I) PREVIOUS YEAR’SANSWER KEY
ISI: Year 2014
1. (d) 2. (b) 3. (b) 4. (c) 5. (b) 6. (a) 7. (a) 8. (a) 9. (c) 10. (b)
11.(c) 12. (a) 13. (d) 14. (a) 15. (b) 16. (a) 17. (b) 18. (c) 19. (a) 20. (a)
21.(b) 22. (b) 23. (a) 24. (b) 25. (c) 26. (a) 27. (d) 28. (d) 29. (d) 30. (b)
ISI : YEAR 2015
1. (b) 2. (d) 3. (c) 4. (b) 5. (d) 6. (c) 7. (a) 8. (c) 9. (a) 10. (c)
11. (d) 12. (c) 13. (b) 14. (d) 15. (c) 16. (c) 17. (b) 18. (a) 19. (a) 20. (d)
21. (d) 22. (b) 23. (b) 24. (b) 25. (a) 26. (d) 27. (a) 28. (b) 29. (b) 30. (d)
ISI : YEAR 2016
1. (c) 2. (b) 3. (d) 4. (a) 5. (c) 6. (d) 7. (d) 8. (b) 9. (a) 10. (a)
11. (d) 12. (b) 13. (a) 14. (a) 15. (c) 16. (d) 17. (c) 18. (b) 19. (c) 20. (c)
21. (d) 22. (a) 23. (b) 24. (a) 25. (c) 26. (d) 27. (a) 28. (a) 29. (a) 30. (c)
3
ISI MSQE ME(I) PREVIOUS YEAR’SANSWER KEY
Detailed solutions are given below.
4
ISI MSQE ME(I) PREVIOUS YEAR’SANSWER KEY
ISI MSQE MEI 2004 SOLVED PAPER
1. (a)X ~ B(n, p), then E(X)= np and Var(X)= np(1- p), where o < p <1.
The product of two positive quantities is maximum when they are equal.
And maximizing np(1- p) is similar to maximizing p(1- p).
1
n
4
4
max{ p(1- p) } = . So, Var(X) ≤ .
2. (b)P(x) = ax2 + bx + c ,
P(1) = - P(2) gives a + b + c = - 4a – 2b - c
» 5a + 3b + 2c = 0 .
Also given that P(-1) = a – b + c = 0 .
Choosing c = 1, then solving above two equations we will get
−5
3
a=
,b = .
8
8
8
So, another root is .
5
3. (c) Let f(x) = (x – a)3+(x – b)3+(x – c)3 ,
Here f′(x)>0, so f′(x)=0 has no real roots.
Hence f(x) = 0 has two imaginary and one real root.
4. (a) Probability that the problem will not be solved = P(Ac∩Bc∩Cc)=
2
P(Ac)×P(Bc)×P(Cc) = , due to independence .
5
2
3
5
5
So, the probability that the problem will be solved is 1 - = .
5. (c) Calculate correlation coefficient from each equation & check yourself.
6. (d)Ans is Cov(x,y)/V(y).
5
ISI MSQE ME(I) PREVIOUS YEAR’SANSWER KEY
7. (b) Consider AM > GM inequality for values ex1 and ex2 .
f(x1 )+f(x2 )
Now
2
=
ex1 +ex2
2
>
√ex1 √ex2
x1 +x2
2
=e
= f(
x1 +x2
2
).
n
8. (c) (i) ∑ n . Apply D’Alembert’s Ratio Test :
2
If
a
∑ an is a positive term series, such that lim n+1 = l, then the series converges if l <
n→∞ an
1.
Here lim
an+1
n→∞ an
(ii) Sn =
1
2
1−
1
2
1
= . So, the series is convergent .
2
= 1 . The series is convergent with finite sum .
9. (c)f ′(x)= 2x, when x ≥ 0
= -2x, when x ≤0
So, we can claim that f ′(0) exists. So, f(x) is differentiable at x = 0.
10. (a) The sequence has no limit.
1
11. (d)∑∞
n=1 .
1
n n+1
1
1
n
n+1
= ∑∞
n=1( −
) = 1.
12. (c)
13. (c) Let u = sin−1 (
x2 +y2
x & y respectively,
cosu
So, xcosu
∂u
∂x
∂u
∂x
=
x+y
x2 −y2 +2xy
(x+y)2
+ ycosu
∂u
∂y
&
=
) » sinu = (
𝑐𝑜𝑠𝑢
x2 +y2
x+y
∂u
∂y
x2 +y2
=
x+y
) , differentiating both sides w.r.t.
y2 −x2 +2xy
(x+y)2
(Check Tourself)
= sinu
Hence x
∂u
∂x
+y
∂u
∂y
= tanu .
6
.
ISI MSQE ME(I) PREVIOUS YEAR’SANSWER KEY
14.(d)y = tan−1 (
15.(c)1 +
4−1
4
+
cosx
1+sinx
8−1
42
+
) = tan−1 (
16−1
43
1
+
32−1
44
1
1
x
1−tan2 ( )
2
x
1+tan2 ( )
2
x
2tan2
1+
x
1+tan2 ( )
2
4
x
∂y
4
2
∂x
1
1
1
1
8
4
42
3
3
= .
2
+⋯
= (1 + 1 + + + + ⋯ ) − ( +
2
π
) = tan−1 (tan( − ) »
8
+
1
43
+ ⋯) = 3 − = .
16.(a) Take four girls as a group, then there are total 6 members among
which 4 girls are chosen such that they need to be seated such that no two
girls sit together. Such arrangements are = 6P 4 . And 5 boys can sit in 5!
Ways. Then total number if such arrangements is = 6P 4 × 5! .
17. (b)
18. (c)We are to find remainder when 1!+2!+3!+....+100! is divided by 36.
We know 6! is divisible by 36. Then 7!,8!,..... will be divisible by 36. So,
sum will also be divisible by 36.
their
So, now we need to calculate the remainder when 1!+2!+3!+4!+5! is divisible
by 36. 1!+2!+3!+4!+5!=153. Remainder when 153 is divided by 36 is 153 –
4×36 = 9.
19. (b)
20. (a) The given inequality holds when x < 0.
21.(b) Draw the graphs of both functions & check yourself.
22. (d)
23. (d)
Take ∣x∣= a , then a2 −3a+2 = 0 has two distinct real roots, i.e., a = 1, 2.
Then x = ±1, ±2. There are four real roots of the equation.
7
ISI MSQE ME(I) PREVIOUS YEAR’SANSWER KEY
24. (a)
25. (c)
1
26.
1
( −
)2
99 n n+1
(c)∑n=1 1 1
.
1
= ∑99
n=1 .
1
n n+1
n n+1
1
1
n
n+1
= ∑99
n=1( −
)=1−
1
100
=
1−λ
−1
27. (a) We know ∣ A − λI ∣ = 0 = |
| = λ2 + λ
2
−2 − λ
Using Cayley-Hamilton Theorem, we have
A2 + A= 0 ⇒ A5 = - A ⇒ A100= A . So, A100+ A5 = 0.
28.(d)lim(
x→0
1
29.(d)|1
2
sinx 12
x
)x = lim(
x→0
x−
x3 1
3! x2
)
x
x2 1
1
= lim(1 − )x2 = e−3! .
3!
x→0
1 1
a −1| ≠ 0 » (a − 1)(b − 2) ≠ 0 » ab ≠ 2a + b − 2 .
2 b
3
3
3
3
2
4
2
4
30.(b)y = x2 – 3x + 3 = (x – )2 + ⇔ (x – )2 = (y – ) .
8
99
100
.
ISI MSQE ME(I) PREVIOUS YEAR’SANSWER KEY
ISI MSQE MEI 2005 SOLVED PAPER
1. (A)𝑋 ~ 𝐵𝑖𝑛𝑜𝑚𝑖𝑎𝑙 (𝑛, 𝑝)
𝐸(𝑋) = 𝑛𝑝
2
𝑉(𝑋) = 𝑛𝑝(1 − 𝑝) ≤ 𝑛 (
≤
𝑝 + (1 − 𝑝)
) ;
2
𝑏𝑦 𝐴𝑀 ≥ 𝐺𝑀
𝑛
4
2. (C) lim+ 𝑓(𝑥) = lim− 𝑓(𝑥) = 0
𝑥→0
𝑆𝑜,
𝑥→0
𝑥2, 𝑥 ≥ 0
𝑓(𝑥) = { 2
𝑖𝑠 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑡𝑖𝑎𝑏𝑙𝑒 𝑎𝑡 𝑥 = 0
−𝑥 , 𝑥 < 0
3. (A)
1 −1
𝐴=(
)
2 −2
−1 1
1 −1
𝐴2 = (
) ; 𝐴3 = (
)=𝐴
−2 2
2 −2
5
𝐴3 . 𝐴2 = +𝐴
{𝐴 =100
𝐴
= −𝐴
⇒ 𝐴5 + 𝐴100 = 0
4. (C)𝑓(𝑥)𝑚𝑎𝑥 = 𝑓(4) = |16 + 8 − 3| + 1.5𝑙𝑜𝑔𝑒 22 = 21 + 3𝑙𝑜𝑔𝑒 2
𝑓(𝑥)𝑚𝑖𝑛 = 0
5. (A)𝑥𝑛+1 = 𝛼 𝑛+1 + 𝛽 𝑛+1 = (𝛼 + 𝛽)[𝛼 𝑛 + 𝛽 𝑛 ] − 𝛼𝛽[𝛼 𝑛−1 + 𝛽 𝑛−1 ]
= 𝑝𝑥𝑛 − 𝑞𝑛−1
Hence proved.
9
ISI MSQE ME(I) PREVIOUS YEAR’SANSWER KEY
6. (C)𝑓(𝑥) = 2𝑥 2 (let)
𝑓(0) = 0
𝑓 ′ (0) = 0
𝑓 ′′ (0) = 4
𝑙𝑡 2𝑓(𝑥) − 3. 𝑓(2𝑥) + 𝑓(4𝑥)
𝑙𝑡 4𝑥 2 − 24𝑥 2 + 32𝑥 2
=
= 12.
𝑥→0
𝑥2
𝑥→0
𝑥2
𝑇ℎ𝑒𝑛,
7. (A)
𝑙𝑜𝑔𝑒 𝑥2 𝑒 𝑙𝑜𝑔𝑒 𝑥2
𝑥2
[∵ 𝑒 < 𝑥1 < 𝑥2 < ∞]
< 𝑙𝑜𝑔 𝑥 =
𝑒
1
𝑙𝑜𝑔𝑒 𝑥1 𝑒
𝑥1
8. (C)
1
√1 + √2
=
+
1
√2 + √3
+ ⋯+
1 − √2
(1 + √2)(1 − √2)
+
1
√99 + √100
√2 − √3
(√2 + √3)(√2 − √3)
+ ⋯+
√99 − √100
(√99 + √100)(√99 − √100)
= −1 + √2 − √2 + √3 … − √99 + √100
= 10 − 1 = 9 ∈ (0, 10)
9. (D) Total number of possible combinations= 8 + 5 = 13.
10. (A)
𝑅1 =
𝑐𝑜𝑣 (𝑋, 𝑌 + 𝑋)
√𝑉(𝑋)𝑉(𝑌 + 𝑋)
=
𝑐𝑜𝑣 (𝑋, 𝑌) + 𝑉(𝑋)
√𝑉(𝑋)[𝑉(𝑋) + 𝑉(𝑌) + 2 𝑐𝑜𝑣(𝑋, 𝑌)]
Let us assume 𝑐𝑜𝑣 (𝑋, 𝑌) = 0, i.e., X& Y are independent.
𝑆𝑜, 𝑅1 =
𝑉(𝑋)
√𝑉(𝑋)[𝑉(𝑋) + 𝑉(𝑌)]
≥ 0,
𝑠𝑖𝑛𝑐𝑒 𝑉(𝑋), 𝑉(𝑌) 𝑎𝑟𝑒 𝑛𝑜𝑛 − 𝑛𝑒𝑔𝑎𝑡𝑖𝑣𝑒.
10
ISI MSQE ME(I) PREVIOUS YEAR’SANSWER KEY
𝐵𝑢𝑡 𝑅 =
𝑐𝑜𝑣(𝑋, 𝑌)
√𝑉(𝑋)𝑉(𝑌)
= 0; 𝑠𝑜 𝑅1 > 𝑅.
11. (C)
1
1
1
∫(𝑥 + |𝑥|𝑑𝑥) = ∫ 𝑥 𝑑𝑥 + ∫ |𝑥| 𝑑𝑥
−1
−1
−1
𝑥2 1
1
1
=[ ]
+ ×1×1+ ×1×1
2 −1 2
2
1 1
=0+ + =1
2 2
12. (D)
6𝑥1 + 20𝑥2 = 360
6𝑥1 + 4𝑥2 = 120
⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯
𝑥1 = 10,
𝑥2 = 15
𝜋(0,0) = 0
𝜋(0,18) = 990
𝜋(20,0) = 900
11
ISI MSQE ME(I) PREVIOUS YEAR’SANSWER KEY
𝜋(10,15) = 1275
𝑆𝑜, 𝜋𝑚𝑎𝑥 = 1275
ISI MSQE ME 1 2006 SOLVED PAPER
1. (A)
2𝑥
1 + 1+𝑥2
2𝑥
1+𝑥 2
1+𝑥
𝑓(
) = log (
) = log (
) = 2 log (
) = 2 𝑓(𝑥)
2𝑥
2
1+𝑥
1−𝑥
1−𝑥
1−
1+𝑥 2
2. (A)
𝑢 = 𝜙(𝑥 − 𝑦, 𝑦 − 𝑧, 𝑧 − 𝑥) = 𝜙(𝑡, 𝑟, 𝑠)
𝜕𝑢 𝜕𝑢 𝜕𝑟 𝜕𝑢 𝜕𝑠 𝜕𝑢 𝜕𝑡
𝜕𝑢 𝜕𝑢
=
.
+
.
+
.
=−
+
𝜕𝑥 𝜕𝑟 𝜕𝑥 𝜕𝑠 𝜕𝑥 𝜕𝑡 𝜕𝑥
𝜕𝑠 𝜕𝑡
… … . (1)
𝜕𝑢 𝜕𝑢 𝜕𝑢
=
−
… … . (2)
𝜕𝑦 𝜕𝑟 𝜕𝑡
𝜕𝑢
𝜕𝑢 𝜕𝑢
=−
+
… … . (3)
𝜕𝑧
𝜕𝑟 𝜕𝑠
Adding these three equations we will get,
𝜕𝑢 𝜕𝑢 𝜕𝑢
+
+
= 0.
𝜕𝑥 𝜕𝑦 𝜕𝑧
3. (D) C has 2𝑚+𝑛 elements. And 𝑆 ∩ 𝐴 contains 𝑖 elements.
𝑛
Total no. of elements = (𝑚𝑖) × (𝑘−𝑖
)
4. (c) The function is decreasing over(−∞, −2), then (−1, 0) and (1, 2).
𝑓(𝑥) = |
𝑥2
𝑥2
− |𝑥|| 𝑖𝑠 𝑠𝑦𝑚𝑚𝑒𝑡𝑖𝑟𝑐. 𝐴𝑛𝑑 𝑓(𝑥) = | − 𝑥| 𝑓𝑜𝑟 𝑥 > 0.
2
2
12
ISI MSQE ME(I) PREVIOUS YEAR’SANSWER KEY
5. (D) CS Inequality:
(𝑥12 + 𝑥22 + ⋯ + 𝑥𝑛2 )(12 + ⋯ + 12 ) ≥ (𝑥1 + 𝑥2 + ⋯ + 𝑥𝑛 )2
𝑥12 + 𝑥22 + ⋯ + 𝑥𝑛2
𝑥1 + ⋯ + 𝑥𝑛 2
≥(
)
𝑁
𝑛
⇒
⇒ 𝑅𝑀𝑆 ≥ 𝐴𝑀
So, RMS > AM when 𝑥𝑖′ s are not equal.
6. (c) ∆𝑓(𝑥) = 𝑓(𝑥 + 1) − 𝑓(𝑥) = 𝐸𝑓(𝑥) − 𝑓(𝑥) = (𝐸 − 1)𝑓(𝑥)
𝑛
𝑛
𝑛
∆𝐾 𝑓(𝑥) = (𝐸 − 1)𝐾 𝑓(𝑥) = {𝐸 𝐾 − ( ) 𝐸 𝑛−1 + ( ) 𝐸 𝑛−2 + ⋯ + (−1)𝑛 ( )} 𝑓(𝑥)
1
2
𝑛
𝑛
𝑛
= [𝑓(𝑥 + 𝐾) − ( ) 𝑓(𝑥 + 𝐾 − 1) + ( ) 𝑓(𝑥 + 𝐾 − 2) − ⋯ + (−1)𝐾 𝑓(𝑥)]
1
2
𝐾
𝐾
= ∑(−1)𝑗 ( ) 𝑓(𝑥 + 𝐾 − 𝑗)
𝑗
𝑗=0
7. (c)
∞
𝐼𝑛 = ∫ 𝑥 𝑛 𝑒 −𝑥 𝑑𝑥 = Γ(𝑛 + 1) = nΓ(n) = n In−1
0
8. (d)𝑥 3 = 1.
𝑎
∆ = |𝑏
𝑐
𝑏
𝑐
𝑎
𝑐
𝑎| [𝑅1 ↔ 𝑅1 + 𝑥𝑅2 + 𝑥 2 𝑅3 ]
𝑏
13
ISI MSQE ME(I) PREVIOUS YEAR’SANSWER KEY
𝑎 + 𝑏𝑥 + 𝑐𝑥 2
=|
𝑏
𝑐
𝑏 + 𝑐𝑥 + 𝑎𝑥 2
𝑐
𝑎
1
= (𝑎 + 𝑏𝑥 + 𝑐𝑥 2 ) |𝑥 2
𝑥
𝑏
𝑐
𝑎
𝑐 + 𝑎𝑥 + 𝑏𝑥 2
|
𝑎
𝑏
𝑐
𝑎| [𝑈𝑠𝑒 𝑥 3 = 1]
𝑏
9. (B)𝐿𝑒𝑡 𝑚 = 2𝐾 + 1, 𝑛 = 2𝑙 + 1
𝐼 = 4𝐾 2 + 4𝐾 + 1 + 4𝑙 2 + 4𝑙 + 1
[𝐾(𝐾 + 1) 𝑖𝑠 𝑒𝑣𝑒𝑛]
= 4[𝐾(𝐾 + 1) + 𝑙(𝑙 + 1)] + 2
= 8 [… ] + 2 ≡ 2 (𝑚𝑜𝑑 4)
10
2
10. (d)𝐸1 : First ball drawn is red; 𝑃(𝐸1 ) = 15 = 3.
5
1
𝐸2 : First ball drawn is black; 𝑃(𝐸2 ) = 15 = 3
E: Second ball chosen is red; P(E);
𝑃(𝐸|𝐸1 ) =
10 2
10
𝑅
= ; 𝑃(𝐸|𝐸2 ) =
=
5
3
17 𝑅 + 𝐵
2
3
2
3
So, 𝑃(𝐸) = 𝑃(𝐸1 )𝑃(𝐸|𝐸1 ) + 𝑃(𝐸2 )𝑃(𝐸|𝐸2 ) = × +
10
17
1
3
× =
98
153
11. (c) f is continuous at 𝑥 = 0
𝑥
𝑓(0) = lim
𝑥
log (1 + 𝑃) − log (1 − 𝑞)
𝑥
𝑥→0
𝑥
𝑥
log (1 − 𝑞)
log (1 + 𝑃) 1
1
= lim
+
lim
𝑥
𝑥
𝑃 𝑥→0
𝑞 𝑥→0
−
𝑃
=
𝑞
1 1
log(1 + 𝑥)
+ [∵ lim
= 1]
𝑃 𝑞 𝑥→0
𝑥
12. (c)𝑆 = 𝑥1 𝑦2 + 𝑥2 𝑦1 − 2𝑥1 𝑥2
𝑦1 𝑦2 > 𝑥1 𝑥2
𝑦
𝑦
−
= 𝑥1 𝑦2 − 𝑥1 𝑥2 + 𝑥2 𝑦1 − 𝑥1 𝑥2 | 1 2 𝑦1 𝑥2 > 𝑥1 𝑥2 − 𝑦1 𝑥2
𝑦1 (𝑦2 − 𝑥2 ) > 𝑥2 (𝑥1 − 𝑦1 )
= 𝑥1 (𝑦2 − 𝑥2 ) + 𝑥2 (𝑦1 − 𝑥1 ) > 0
So, the expression is always positive.
14
ISI MSQE ME(I) PREVIOUS YEAR’SANSWER KEY
13. (a)𝑅𝑀𝑆 ≥ 𝐴𝑀
𝑥 2 + 3𝑦 2 + 5𝑧 2 𝑥 + 3𝑦 + 5𝑧
√
≥
9
9
′ = ′ holds when 𝑥 = 𝑦 = 𝑧
Given that 𝑥 + 𝑦 + 𝑧 = 12.
So, maximum value of 𝑥 + 3𝑦 + 5𝑧 = 4 + 3 × 4 + 5 × 4 = 36.
14. (c)2𝑥𝑦
⏟
⏟ = 11
⏟ − 4𝑥 2 + 12𝑥 − 5𝑦
𝑒𝑣𝑒𝑛
𝑜𝑑𝑑
⇒ 𝑦 = 𝑜𝑑𝑑
𝑜𝑑𝑑
⇒ (−2𝑥 + 𝑦 + 1)(2𝑥 − 5) = 6 = 1 × 6 = 2 × 3
Since 𝑦 is odd. So, −2𝑥 + 𝑦 + 1 = 𝑒𝑣𝑒𝑛 = 2 𝑜𝑟 6.
2𝑥 − 5 = 3 𝑜𝑟 1.
2𝑥 − 5 = 3
∴ 2𝑥 = 8
∴𝑥=4
𝑦=9
2𝑥 − 5 = 1
∴𝑥=3
∴ 𝑦 = 11
Positive solutions: {(4, 9), (3, 11)}
15. (D) f(0) and f(1) can have values other than 0 and 1. And f(0) + f(1) need not be equal to 1.
Counter example:
15
ISI MSQE ME(I) PREVIOUS YEAR’SANSWER KEY
𝑓(0) 𝑓(1) = 2
ISI MSQE ME 1 2007 SOLVED PAPER
1. (A)
(1 + 𝑥)𝛼 − 1 0
[ 𝑓𝑜𝑟𝑚]
𝑥→0 (1 + 𝑥)𝛽−1 0
lim
Applying L’Hospital Rule,
𝛼(1 + 𝑥)𝛼−1 𝛼
=
𝑥→0 𝛽(1 + 𝑥)𝛽
𝛽
= lim
2. (B)𝑋 = odd; 𝑋 = 2𝑚 + 1
𝑋 2 − 1 = (2𝑚 + 1)2 − 1 = 4𝑚2 + 4𝑚 = 4𝑚(𝑚 + 1) 𝑖𝑠 𝑛𝑜𝑡 𝑝𝑟𝑖𝑚𝑒.
3. (A)
16
ISI MSQE ME(I) PREVIOUS YEAR’SANSWER KEY
1
1
∫ 𝐾𝑒
0
𝐾𝑥
𝑒 𝐾𝑥
𝑑𝑥 = 1 ⇒ 𝐾 [
] =1
𝐾 0
⇒ 𝑒𝐾 − 1 = 1
⇒ 𝐾 = log 2
4. (D)
𝑝2 − 𝑞 2is prime.
So factors are = 1, 𝑝2 − 𝑞 2 .
So, 𝑝2 − 𝑞 2 = (𝑝 + 𝑞)(𝑝 − 𝑞)
So, 𝑝 − 𝑞 = 1, 𝑝 + 𝑞 = 𝑝2 − 𝑞 2
So, the answer is none of the above.
5. (D)𝑓 ′ (𝑥) ≥ 0 ∀𝑥 ∈ [𝑎, 𝑏]
6. (C)
𝑥
|1
1
3 4
2 1| = 0
8 1
𝑥−4 3
⇒| 0
2
0
8
4
1| = 0 [𝑐1′ = 𝑐1 − 𝑐3 ]
1
⇒ 6(𝑥 − 4) = 0
⇒𝑥=4
7. (B)𝑓(𝑥) = √𝑥 + 𝑓(𝑥)
⇒ [𝑓(𝑥)]2 = 𝑥 + 𝑓(𝑥);
After differentiation we will get
2𝑓(𝑥)𝑓 ′ (𝑥) = 1 + 𝑓 ′ (𝑥)
⇒ 𝑓 ′ (𝑥) =
1
2𝑓(𝑥) − 1
17
ISI MSQE ME(I) PREVIOUS YEAR’SANSWER KEY
8. (A)
𝑃 =1+
log 𝑦
log 𝑥
; 𝑄 =1+
log 𝑥
log 𝑦
𝑃+𝑄 =2+
𝑃𝑄 = 1 +
log 𝑥 log 𝑦
+
log 𝑦 log 𝑥
log 𝑥 log 𝑦
+
+1=𝑃+𝑄
log 𝑦 log 𝑥
9. (C)
∫
2𝑥 3 + 1
𝑥 4 + 2𝑥 = 𝑧
𝑑𝑥
𝑥 4 + 2𝑥
2(𝑥 3 + 1)𝑑𝑥 = 𝑑𝑧
1 𝑑𝑧
= ∫
2 𝑧
1
= log|𝑥 4 + 2𝑥| + 𝑐
2
10. (D)𝑥 2 − 3𝑥 + 2 > 0 ⇒ (𝑥 − 2)(𝑥 − 1) > 0
Domain: (−∞, 1) ∪ (2, ∞)
11. (B)
𝑓1 (𝑥) = 𝑥 2 ;
𝑓2 (𝑥) = 4𝑥 3 + 7
𝑓2 (𝑥1 ) = 𝑓2 (𝑥2 )
⇒ 4𝑥13 + 7 = 4𝑥23 + 7
|
⇒ 𝑥1 = 𝑥2
𝐻𝑒𝑛𝑐𝑒 𝑜𝑛𝑒 − 𝑜𝑛𝑒
1/3
𝑓2 (𝑥) − 7
𝑎𝑛𝑑 𝑥 = (
)
4
|
⇒ 𝑥 = √𝑓1 (𝑥)
∴ 𝑥 ℎ𝑎𝑠 𝑣𝑎𝑙𝑢𝑒 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑣𝑎𝑙𝑢𝑒𝑠 𝑜𝑓 𝑓2 (𝑥)
𝑖𝑠 𝑛𝑜𝑡 𝑑𝑒𝑓𝑖𝑛𝑒𝑑 𝑓𝑜𝑟 𝑓1 (𝑥) < 0. 𝑆𝑜, 𝑛𝑜𝑡 𝑜𝑛𝑡𝑜.
𝑠𝑜, 𝑜𝑛𝑡𝑜
𝑓1 (𝑥1 ) = 𝑓1 (𝑥2 )
⇒ 𝑥12 = 𝑥22
⇒ 𝑥1 = ±𝑥2
∴ 𝑁𝑜𝑡 𝑜𝑛𝑒 − 𝑜𝑛𝑒
𝐴𝑛𝑑 𝑓1 (𝑥) = 𝑥 2
12. (B)
log 𝑓(𝑥) = (𝑎 + 𝑏 + 2𝑥)[log(𝑎 + 𝑥) − log(𝑏 + 𝑥)]
𝑓 ′ (𝑥)
𝑎+𝑥
1
1
= 2 [log (
)] + (𝑎 + 𝑏 + 2𝑥) [
−
]
𝑓(𝑥)
𝑏−𝑥
𝑎+𝑥 𝑏+𝑥
18
ISI MSQE ME(I) PREVIOUS YEAR’SANSWER KEY
𝑎 𝑎+𝑏
𝑓(0) = ( )
𝑏
𝑎 𝑎+𝑏
𝑎 (𝑎 + 𝑏)(𝑎 − 𝑏)
𝑓 ′ (0) = ( )
[2 log +
]
𝑏
𝑏
𝑎𝑏
13. (D)
1
3
3
9
2
2
2
2
𝑍(3,9) = × + × =
22
30
3 9
.𝑍
occurs at ( , ).
4 𝑚𝑎𝑥
2 2
3
9
Points are: 𝑥 + 𝑦 = 6 ∩ 𝑥 + 3𝑦 = 16 gives 𝑥 = 2 , 𝑦 = 2
14. (A)
15. (C)
16. (A)
17. (B)
ℎ(𝑥) =
ℎ(ℎ(𝑥)) =
1
1−
ℎ (ℎ(ℎ(𝑥))) =
1
1−𝑥
1
1−𝑥
=
1
1−
𝑥−1
𝑥
18. (D)
19
1−𝑥 𝑥−1
=
.
−𝑥
𝑥
=
𝑥
= 𝑥.
𝑥−𝑥+1
ISI MSQE ME(I) PREVIOUS YEAR’SANSWER KEY
𝑓(𝑥) = 𝑥|𝑥| + (
2
= { 𝑥 2+ 1
−𝑥 + 1
|𝑥|
)
𝑥
𝛽
𝑤ℎ𝑒𝑛 𝑥 > 0
𝑤ℎ𝑒𝑛 𝑥 < 0
The function is continuous at 𝑥 = 0
And 𝐿𝐻𝐷 = 𝑅𝐻𝐷 = 0
So, Differentiable at 𝑥 = 0.
19. (A)
∫
2𝑑𝑥
2+𝑥−𝑥
=∫
𝑑𝑥
(𝑥 − 2)(𝑥 − 1)𝑥
(𝑥 − 2)(𝑥 − 1)𝑥
= −∫
=∫
𝑑𝑥
𝑑𝑥
+∫
(𝑥 − 2)(𝑥 − 1)
𝑥(𝑥 − 1)
𝑑𝑥
𝑑𝑥
𝑑𝑥
𝑑𝑥
−∫
+∫
−∫
𝑥
𝑥−1
𝑥−2
𝑥−1
𝑥(𝑥 − 2)
= log |
|+𝑐
(𝑥 − 1)2
20. (B)
P(a Person comes) = 1/5;
P(a Person doesn’t come) = 4/5
P(all person will be accommodated) = 1 − [𝑃(51 𝑝𝑒𝑟𝑠𝑜𝑛 𝑠ℎ𝑜𝑤𝑠 𝑢𝑝) + 𝑃(52 𝑝𝑒𝑟𝑠𝑜𝑛 𝑐𝑜𝑚𝑒𝑠)]
20
ISI MSQE ME(I) PREVIOUS YEAR’SANSWER KEY
52 1 4 51
4 52
= 1 − [( ) ( ) ( ) + ( ) ]
1 5 5
5
4 52
= 1 − 14 ( )
5
21. (B)
3/2
3/2
∫ [𝑥]𝑑𝑥 + ∫ [𝑥 2 ]𝑑𝑥
0
0
3/2
1
3/2
√2
1
= ∫[𝑥]𝑑𝑥 + ∫ 1. 𝑑𝑥 + ∫[𝑥 2 ]𝑑𝑥 + ∫ [𝑥 2 ]𝑑𝑥 + ∫ [𝑥 2 ]𝑑𝑥
0
1
1
0
3/2
1
1
√2
√2
3/2
= ∫ 0𝑑𝑥 + ∫ 1𝑑𝑥 + ∫ 0 𝑑𝑥 + ∫ 𝑑𝑥 + ∫ 2 𝑑𝑥
0
1
0
1
√2
5 − 2√2
2
=
22. (D) (𝑛3 − 𝑛)(𝑛2 − 4) = (𝑛 − 2)(𝑛 − 1)𝑛(𝑛 + 1)(𝑛 + 2)
Product of 5 consecutive integers are divisible by 120 but not 720.
23. (C)
Buyer 1: Average expenditure per mango =
Buyer 2: Average expenditure per mango =
5𝑃1 + 10𝑃2
15
𝑥
250
100 150
+
𝑃1 𝑃2
𝑥
∴
𝑃1 + 2𝑃2
=
3
5
2
𝑃1
⇒ 15 = 2 + 6 + 4.
⇒
3
+𝑃
2
𝑃2
𝑃1
+ 3.
𝑃1
𝑃2
𝑃1 4
= 𝑜𝑟 1
𝑃2 3
3
⇒ 𝑃1 = 𝑃2 𝑜𝑟, 𝑃2 = 𝑃1
4
21
ISI MSQE ME(I) PREVIOUS YEAR’SANSWER KEY
24. (D)
25. (B)
26. (D)𝑓(𝑥) = 𝑥𝑓(𝑥)
𝑓(𝑥)(𝑥 − 1) = 0
When 𝑥 ≠ 1, 𝑓(𝑥) = 0 ⇒ 𝑓(2) = 0
When 𝑥 = 1, f(x) is continuous 𝑓(1) = 1. And 𝑓(2) = 𝑓(1)
27. (C)
28. (D)
29. (D)
𝑓(𝑥)𝑓 ′ (𝑥) < 0
𝑓(𝑥) > 0 𝑓(𝑥) < 0
|
|
𝑓 ′ (𝑥) < 0 𝑓 ′ (𝑥) > 0
22
ISI MSQE ME(I) PREVIOUS YEAR’SANSWER KEY
𝑠𝑙𝑜𝑝𝑒 𝑜𝑓 𝑓(𝑥) =
|𝑓(𝑥)| ′
. 𝑓 (𝑥) < 0
𝑓(𝑥)
So, |f(x)| is a decreasing function.
Alternative Method:
𝑑
2
(𝑓(𝑥)) < 0
𝑑𝑥
2
⇒ (𝑓(𝑥)) 𝑖𝑠 𝑑𝑒𝑐𝑟𝑒𝑎𝑠𝑖𝑛𝑔
⇒ |𝑓(𝑥)| 𝑖𝑠 𝑑𝑒𝑐𝑟𝑒𝑎𝑠𝑖𝑛𝑔
30. (C)
𝑝2 + 𝑝 + 1
≥𝑝;
3
𝑆𝑜,
𝑞2 + 𝑞 + 1
𝑟2 + 𝑟 + 1
𝑠2 + 𝑠 + 1
≥𝑞;
≥𝑟 &
≥𝑠
3
3
3
(𝑝2 + 𝑝 + 1)(𝑞 2 + 𝑞 + 1)(𝑟 2 + 𝑟 + 1)(𝑠 2 + 𝑠 + 1)
≥ 81
𝑝𝑞𝑟𝑠
ISI MSQE ME 1 2008 SOLVED PAPER
1. (d)
1 + log 𝑥 = 𝑧
𝑑𝑥
𝑑𝑥
∫
=∫
| 𝑑𝑥
𝑥 + 𝑥 log 𝑥
𝑥(1 + log 𝑥)
= 𝑑𝑧
𝑥
=∫
𝑑𝑧
𝑧
= log|1 + log 𝑥| + 𝑐
2. (b)
23
ISI MSQE ME(I) PREVIOUS YEAR’SANSWER KEY
𝑦 = √−1 + 𝑥
⇒ 𝑦 = −1 + 𝑥
⇒ 𝑥 = 1 + 𝑦2
2
|
𝑥 = 𝑦 2 + 1; 𝑥 ∈ (1, ∞)
∵ 𝑓 𝑖𝑠 𝑜𝑛𝑡𝑜, 𝑜𝑛𝑒 − 𝑡𝑜 − 𝑜𝑛𝑒
𝑓 𝑖𝑠 𝑖𝑛𝑣𝑒𝑟𝑡𝑖𝑏𝑙𝑒
Inverse of the function is 𝑥 2 + 1 = 𝑓 −1 (𝑥)
3. (C)
𝑥 − 1 < 0gives x < 1. And √𝑥 is there, so, 𝑥 ≥ 0.
At 𝑥 = 1, 𝑓(𝑥) is discontinuous. So, domain of continuity of f(x) is [0, 1] ∪ (1, ∞)
4. (b)
𝑧 = 𝑥 − 2𝑦
𝑧(0,3) = −6
𝑧(3,0) = 3
So, 𝑧𝑚𝑖𝑛 = −6 at (0, 3)
5. (c)
24
ISI MSQE ME(I) PREVIOUS YEAR’SANSWER KEY
1
𝑡
6. (c)𝑥 = 𝑡 𝑡−1 , 𝑦 = 𝑡 𝑡−1
log 𝑥 =
log 𝑡
𝑡 log 𝑦
; log 𝑦 =
𝑡−1
𝑡−1
𝑥 log 𝑦 =
𝑡
𝑡 log 𝑡 1
log 𝑡
. 𝑡 𝑡−1 = 𝑡 𝑡−1 .
… … … (1)
𝑡−1
𝑡−1
𝑡
𝑦 log 𝑥 = 𝑡 𝑡−1 .
log 𝑡
… … . (2)
𝑡−1
(1) = (2)
𝑥𝑦 = 𝑦𝑥
𝑜𝑟,
7. (b)20𝑥𝐵 + 15𝑥𝑆 = 900
𝑥 = 45 𝑤ℎ𝑒𝑛 𝑥𝑆 = 0
⇒{ 𝐵
𝑥𝑆 = 60 𝑤ℎ𝑒𝑛 𝑥𝐵 = 0
So, (45, 0) and (0, 60) are two optimum points.
So, 𝑇𝑚𝑎𝑥 = 2 × 0 + 3 × 60 = 180.
8. (d)
2
1
𝑛 ′ (𝑥)𝑑𝑥
∫[𝑥] 𝑓
0
2
= ∫ 0𝑓
′ (𝑥)𝑑𝑥
0
+ ∫ 𝑓 ′ (𝑥)𝑑𝑥
1
= 𝑓(2) − 𝑓(1)
9. (d) Total : 10000 ; College : 2100 ; University : 4200 ; School : 3700
Went to school & read news =
70×37
%
100
= 25.9%.
25
ISI MSQE ME(I) PREVIOUS YEAR’SANSWER KEY
𝑥 2 𝑒 −𝑥 ; 𝑥 ≥ 0
10. (c)𝑓(𝑥) = { 2 −𝑥
−𝑥 𝑒
; 𝑥<0
lim 𝑓(𝑥) = lim− 𝑓(𝑥) = 𝑓(0) = 0
𝑥→0+
𝑥→0
2𝑥𝑒 −2 − 𝑥 2 𝑒 −𝑥 ;
𝑓 ′ (𝑥) = {
−2𝑥𝑒 −𝑥 + 𝑥 2 𝑒 −𝑥 ;
𝑥≥0
𝑥<0
lim 𝑓 ′ (𝑥) = lim− 𝑓 ′ (𝑥) = 0.
𝑥→0+
𝑥→0
So, differentiable everywhere.
11. (c)
12. (b)
𝑋𝑚 =
𝑥1 + 𝑥2 + ⋯ + 𝑥5𝑛
5𝑛
(𝑥1 +𝑥2 +⋯+𝑥𝑛 )
=
𝑛
+
(𝑥𝑛+1 +⋯+𝑥2𝑛 )
𝑛
+ ⋯+
(𝑥4𝑛+1 +⋯+𝑥5𝑛 )
𝑛
5
⇒ 𝑋 ′ 𝑠method is correct.
But in Y’s and Z’s, the partitions do not contain equal number of elements.
Hence X’s method is correct.
13. (b)
𝑇𝑜𝑡𝑎𝑙 𝑛𝑜. 𝑜𝑓 𝑤𝑎𝑦𝑠 =
=
14. (c)
6!
6
= ( ) 𝑤𝑎𝑦𝑠
4! 2!
2
6×5
= 15
2×1
𝑥 > −5
} 𝑤𝑖𝑙𝑙 𝑠𝑎𝑡𝑖𝑠𝑓𝑦 𝑡ℎ𝑒 𝑖𝑛𝑒𝑞𝑢𝑎𝑙𝑖𝑡𝑦 𝑔𝑖𝑣𝑒𝑛 𝑎𝑠 |𝑥 − 3| + |𝑥 + 2| < 11
𝑥<6
15. (b)𝑓 ′ (𝑥) = 4𝑥 3 − 12𝑥 2 + 16 = 0
⇒ 𝑥 3 − 3𝑥 2 + 4 = 0
26
ISI MSQE ME(I) PREVIOUS YEAR’SANSWER KEY
⇒ (𝑥 + 1)(𝑥 2 − 4𝑥 + 4) = 0
⇒ 𝑥 = −1; (𝑥 − 2)2 = 0
⇒ 𝑥 = −1,
𝑥=2
𝑓 ′′ (𝑥) = 12𝑥 2 − 24𝑥
𝑓 ′′ (𝑥)|𝑥=−1 = 36 > 0; 𝑓 ′′ (𝑥)|𝑥=2 = 0
So, at 𝑥 = −1, f(x) is minimum, but f(x) has no maximum.
16. (c)𝐾(𝑛) = (𝑛 + 3)(𝑛2 + 6𝑛 + 8)
= (𝑛 + 2)(𝑛 + 3)(𝑛 + 4)
For any integer n, K(n) will be divisible by 6.
[∵ 𝐾(0) = 4! = 24 𝑑𝑖𝑣𝑖𝑠𝑖𝑏𝑙𝑒 𝑏𝑦 6]
17. (a) 25 books can be arranged in 25! ways.
Consider the two books taken as a pair then number of favorable ways of getting those two books
together is 24! 2!.
Required probability
=
24! × 2!
2
=
25!
25
18. (b)𝑃(𝑥) = 𝑎𝑥 2 + 𝑏𝑥 + 𝑐
𝑃(−1) = 𝑎 − 𝑏 + 𝑐 = 0
⇒ 2𝑎 − 2𝑏 + 2𝑐 = 0 … … . . (1)
𝑃(1) = −𝑃(2) ⇒ 𝑎 + 𝑏 + 𝑐 = −4𝑎 − 2𝑏 − 𝑐
⇒ 5𝑎 + 3𝑏 + 2𝑐 = 0 … … … (2)
(1) − (2)
⇒ 3𝑎 + 5𝑏 = 0 ⇒ −
𝑏 3
=
𝑎 5
∴Sum of the roots = 3/5, one root is = −1.
27
ISI MSQE ME(I) PREVIOUS YEAR’SANSWER KEY
3
8
So, another root is = 5 + 1 = 5 .
19. (a)
20. (d)𝐴𝑀 ≥ 𝐺𝑀
𝑎
𝑏
𝑏
𝑐
𝑑
+𝑐 +𝑑+𝑎
4
⇒
4
≥ √1
𝑎 𝑏 𝑐 𝑑
+ + + ≥4
𝑏 𝑐 𝑑 𝑎
21. (d)𝑙𝑜𝑔(2−𝑥) (𝑥 − 3) ≥ −1
⇒
log(𝑥 − 3)
≥ −1
log(2 − 𝑥)
1
⇒ log(𝑥 − 3) ≥ log (
)
2−𝑥
⇒ (𝑥 − 3)(2 − 𝑥) ≥ 1
⇒ −𝑥 2 − 𝑥 − 6 ≥ 1
No such x exists.
22. (c)𝑓(𝑥) = 5𝑥 3 − 5𝑥 2 + 2𝑥 − 1
𝑓(−2) = −65 < 0
𝑓(−1) = −13 < 0
𝑓(0) = −1 < 0
} 𝑠𝑖𝑔𝑛. 𝑐ℎ𝑎𝑛𝑔𝑒
𝑓(1) = 1 > 0
So, it has a root between 0 and 1.
𝑓(2) = 23 > 0
23. (a)
28
ISI MSQE ME(I) PREVIOUS YEAR’SANSWER KEY
∞
𝑎 ∫ 𝑥 2 𝑒 −𝐾𝑥 𝑑𝑥 = 1
0
∞
𝑎
⇒ 3 ∫ 𝑧 2 𝑒 −𝑧 𝑑𝑧 = 1
𝐾
0
2𝑎
⇒ 3=1
𝐾
⇒𝑎=
∞
[∵ ∫ 𝑧 3−1 𝑒 −𝑧 𝑑𝑧 = ⌈3]
0
𝐾3
2
24. (a)
25. (b)
26. (a)
27. (c) Let 𝑦𝑛×1
𝑦1
𝑥1
𝑦2
𝑥2
= ( ⋮ ) and 𝑥𝑛×1 = ( ⋮ )
𝑦𝑛
𝑥𝑛
𝑦𝑛× = 𝑃𝑛×𝑛 𝑥𝑛×1 ;
𝑦𝐾 = 𝑎𝐾1 𝑥1 + 𝑎𝐾2 + ⋯ + 𝑎𝐾𝑛 𝑥𝑛
𝑎11
𝑎21
𝑃=[ ⋮
𝑎𝑛1
𝑎12
𝑎22
⋮
𝑎𝑛2
∑ 𝑦𝐾 = (∑ 𝑎𝐾1 ) 𝑥1 + (∑ 𝑎𝐾2 ) 𝑥2 + ⋯ + (∑ 𝑎𝐾𝑛 ) 𝑥𝑛
∑ 𝑦𝐾 = ∑ 𝑥𝐾 ⇒ ∑ 𝑎𝐾𝑖 𝑥 = 1 𝑓𝑜𝑟 1 ≤ 𝑖 ≤ 𝑛
This is possible if P is bi−stochastic with elements 0 and 1.
28. (d)
𝑓1 (𝑥) =
𝑥
𝑥+1
29
⋯ 𝑎1𝑛
⋯ 𝑎2𝑛
⋮ ]
⋱
⋯ 𝑎𝑛𝑛
ISI MSQE ME(I) PREVIOUS YEAR’SANSWER KEY
𝑓𝑛 (𝑥) = 𝑓1 (𝑓𝑛−1 (𝑥)); 𝑛 ≥ 2
𝑓𝑛 (𝑥) =
𝑓𝑛−1 (𝑥)
1 + 𝑓𝑛−1 (𝑥)
𝑓2 (𝑥) =
𝑥
𝑥+1
𝑥
1 + 𝑥+1
∴ 𝑓𝑛 (𝑥) =
𝑥
𝑥
𝑥
=
; 𝑓3 (𝑥) = 2𝑥+1𝑥 =
2𝑥 + 1
1 + 2𝑥+1 3𝑥 + 1
𝑥
1
′ (𝑥)
; 𝑛 ≥ 1. 𝑓𝑛 (𝑥) =
=−
1 ; 𝑓𝑛
𝑛𝑥 + 1
𝑛+
𝑥
1
1 2
<0
(𝑛 + 𝑥)
So, 𝑓𝑛 (𝑥) is decreasing in n.
29. (d)
If 𝑥 = 1, 𝐿 = 0
If 𝑥 > 1, 𝐿 = 1
𝐼𝑓 𝑥 ∈ (0, 1),
∴ lim
𝑛→∞
1−ℎ
= −1
𝑛→∞ 1 + ℎ
lim
−1; 0 < 𝑥 < 1
𝑥 2𝑛 − 1
0;
𝑥 = 1 𝑆𝑜, 𝑙𝑖𝑚𝑖𝑡 𝑑𝑜𝑒𝑠𝑛′ 𝑡 𝑒𝑥𝑖𝑠𝑡
=
{
𝑥 2𝑛 + 1
1;
𝑥>1
30. (c)𝑥1 + 𝑥2 = 21
𝑠𝑜, min 𝑓(𝑥1 , 𝑥2 )
𝑥2 = 21 − 𝑥1
⇒ max{6 − 𝑥1 ,
7 − 𝑥2 = 7 − 21 + 𝑥1
⇒ 7 − 𝑥2 = 𝑥1 − 14
Minimum occurs at 𝑥1 = 10, 𝑥2 = 21 − 𝑥1 = 11
(∴ 𝑥1∗ = 10,
30
𝑥2∗ = 11)
𝑥1 − 14}
ISI MSQE ME(I) PREVIOUS YEAR’SANSWER KEY
MSQE 2009 ME I (Mathematics) SOLVED PAPER
1. (B) Here it is given that : a + ar + ar2 + ar3 + ......... =
a
1−r
= 4 , for an infinite
geometric series, where, first term = a = 1 , and r be the common ratio.
So,
a
1−r
=4⇒
1
1−r
3
=4 ⇒ r= .
4
31
ISI MSQE ME(I) PREVIOUS YEAR’SANSWER KEY
2. (D) X be a continuous r.v. with pdf f(x) = 3x2 , 0≤ x ≤1 .
Now it is also given that P(X ≤ a)=P(X > a) ⇒ 2P(X ≤ a)=1
a
1
⇒ 2∫0 3x2 = 1 ⇒ a3 = ⇒ a = (½)1/3 .
2
3. (B)Let y = f(x) . Now squaring both sides we have y2 = ex + y . Now
differentiating both sides w.r.t. x, we have 2f(x). f′(x) = ex + f′(x)
⇒ f′(x) =
ex
2f(x)−1
f2 (x)−f(x)
⇒ f′(x) =
2f(x)−1
4. (A) To find limx→4
√x+5−3
limx→4
=
x−4
.
√x+5−3
x−4
limx→4
, use L’Hospital’s Rule :
−1
1
(x+5) 2
2
1
1
= .
6
5. (D)Here Y= 264 + 263 + . . . . . . + 21 + 20 =
1 ex
6. (C)∫0
ex +1
1 ex +1−1
dx = ∫0
ex +1
1
dx = ∫0 (1 −
20 (265 −1)
1
ex +1
2−1
= 265 – 1 = X – 1.
1+e
)dx = log
2
.
7. (C) Probability of choosing two balls with odd numbers =
5×4
2
= .
9×10
9
Since there are total 10 balls among with 2 balls can be chosen randomly in 10×9
ways, and also there are 5 odd numbered balls. Same logic applied for finding
favorable number of cases.
8. (C) Total No of balls = 100. If X increases the Y decreases. Now, if we plot the
scatter diagram, we will get X and Y are negatively correlated, i.e. , X and Y are
exactly linearly related with negative slope. So, rXY = - 1.
32
ISI MSQE ME(I) PREVIOUS YEAR’SANSWER KEY
x
9. (C) Here, h(x) = g(x) = ,
if 0 ≤ x ≤
2
1
1
2
= f(x) = x(1 - x), if ≤ x ≤1 .
2
It is clear that h(x) is continuous but not differentiable.
10.(C)Here there are 3 persons, each throwing a single die once, the score is 8.
Now 8 = 6+1+1 , 8 = 4+2+2 , 8 = 4+3+1 , 8 = 5+2+1 , 8 = 3+3+2 . These cases can
be happened in a total of (3 + 3 + 3! + 3! + 3) number of ways. So, total number of
ways is equal to 21.
11.(B) It is given that
2f(x) + 3f(-x) = 55 – 7x --------------(1)
Putting x = -x , we get 2f(-x) + 3f(x) = 55 + 7x --------------(2)
Solving (1) and (2) for f(x) , we get , f(x) = 11 + 7x . So, f(3) = 32 .
12.(B)We are to find P [∣x - y∣≤15], where 0 ≤ x,y ≤ 60 .
60×60−45×45
Applying Geometric probability, the required probability is =
=
60×60
13. (C) We know A.M. ≥ G.M. inequality:
14.(B)log n→∞
12 +22 +32 +⋯+n2
n3
x₁
x₂
+
x₂
x₃
x₃
x₁ x₂ x₃
x₁
x₂ x₃ x₁
n
.
+ ≥ 3.∛( . . ) =3.
1 12 +22 +32 +⋯+n2
= log n→∞ .
7
16
n2
1
1
= ∫0 x 2 = .
3
15.(D) The roots of the equation x2 – 7x + 12 = 0 are 3, 4. Let 3 be the common
root, the (x - 3)(x - k) = x2 – 8x +b ⇒ k = 5 & b = 15, which is an odd integer.
So, the answer is 5.
1
1
1
16.(B) Given n ≥ 9, μ = n2 + n3 + n4 .
1
1
1
Clearly, at n = 9, μ < n, because n4 <n3 <n2 = 3 .
33
ISI MSQE ME(I) PREVIOUS YEAR’SANSWER KEY
dμ
∴
dn
−1
1
1
−2
1
−3
1
−1
1
−1
1
−1
1
= n 2 + n 3 + n 4 < n 2 + n 2 + n 2 ≤ for all n ≥ 9
Since
2
dμ
3
4
2
2
2
2
< 1 at all n ≥ 9 and at n = 9, μ < 9 therefore, μ < n.
dn
17.(B) The given functional equation is linear in x & y.
So, f (z) = az for some real a must satisfy the functional equation.
x a 2
x−4 a−2 2
18. (A)|2 x 0| = | 2
x
0| = x(x - 4) - 2(a - 2) = 0
0
0
1
2 1 1
⇒x2 – 4x + 4 – 2a = 0,
⇒ (x – 2)2 = 0,
if 2a=0 .⇒ x=2 is the unique solution iff a=0 .
19. (C) While we will find the value of the determinant , the value will be looked
like this :
y = αf(x) – βg(x) + γh(x) ,
So, differentiating,
dy
dx
= αf′(x) – βg′(x) + γh′(x).
20.(A) It is very easy to show that f(x) is continuous at x=1,2,3 but not
differentiable at any point on the real line.
21.(C) While calculating
as zero. So,
dy
dx
dy
dx
from the given equation, we get the right hand side
will be independent of c.
22.(A) A student finds n books of his interest in a second hand shop. The shop has
m copies of each of these n books.
From m like objects we can choose 0,1,2,...,m objects, i.e., there are m+1 choices.
Hence by multiplication principle, there are in all (m+1)(m+1)......(m+1)
34
ISI MSQE ME(I) PREVIOUS YEAR’SANSWER KEY
collections. Since he selects at least one book, so, the number of non-empty
collections is (m+1)(m+1).......(m+1) – 1 = (m+1)n –1.
23.(B) Let us examine the continuity of f(x) at x=-1 & x=1 .
At x= -1, LHL=RHL gives 2 - 3A + B= -A –B ⇒ A – B = 1.
At x = 1, LHL=RHL gives 2 + 3A+ B= 4
⇒ 3A + B = 2.
3
1
Solving we get A= , B= - .
4
4
1
1
24.(D) lim ( 3x + 32x )x = 3 lim ( 1 + 3x )x = 3.3 =9.
n→∞
n→∞
25.(B)rxy =
√
1
̅Y
̅
∑ XY−X
n
̅ 2 ∑ Y2 −n Y
̅2
∑ X2 −n X
√
n
.
n
Let us calculate the following values for the corrected data set.
Here n=25. ∑ X = 125 ,∑ X 2 = 650 , ̅
X=5
2
2
2
∑ Y = 100 ,∑ Y = 460 + 12 - 14 + 82 – 62 = 436 , ̅
Y= 4
∑ XY = 508 – (14× 6) + (12× 8) = 520 .
3
Putting these values we shall get rxy = 0.73 < .
4
26.(A) If you draw the curve of y = x2 – 1, you will see that the given point is
nearest to (1,0).
2
27.(B) E(X)= ∫0 xkx( 2 − x)dx =
4k
3
,
2
3
Now, we know ∫0 kx( 2 − x)dx = 1 ⇒ k = . Then, E(X) = 1.
4
28.(B)f(x+y) = f(xy) . Put x=y=4, then f(8)= f(16)= 9, so it is clear that f(x) is a
constant function which takes value 9 at each points ≥ 4.
So, f(9) = 9 .
35
ISI MSQE ME(I) PREVIOUS YEAR’SANSWER KEY
29.(A) The function f(x) has roots at x = 1, 2 ,3. If you draw the graph, you will get
it increases from [1, 1.5], then decreases from [1.5, 2.5], then increases from
−1
[2.5, 3] . Here 3 2 is equal to 0.5 .
5
30.(D) V(x) = x(5 – 2x)(8 – 2x) ; 0< x < .
2
3
2
= 4x – 26x + 40x
20
V′(X) =12x3 – 52x + 40 = 0 ⇒ x = 1, x= (rejected)
6
∴ The least possible volume is V(1) = 18 m3 .
MSQE 2010 PEA (Mathematics) SOLVED PAPER
1. (a) 100 [
1
1.2
+
1
2.3
1
+
1
+ ⋯+
3.4
1
1
1
= 100{(1 − ) + ( − ) + ⋯ + (
2
2
3
]
99.100
1
99
−
1
100
)} = 100(1 −
2. (a) Check Continuity & Differentiability at x = 0.
36
1
100
) = 99.
ISI MSQE ME(I) PREVIOUS YEAR’SANSWER KEY
3. (a) For GP Series : First Term = a = 1 , Common Ratio = r (>1). t3 = r2 .
For AP Series : First Term = a = 1 . (r+2)thterm is tr+2=1+(r+2-1)d .
So, 1+(r+2-1)d = r2 » d = r – 1.
4. (d)(1 + x)n = 1 +
nx
So, nx = -9, {n2 - n}x2 =
» 81 + 9x =
»9+x=
33
4
297
4
+
1!
297
4
n(n−1)x2
2!
+⋯
» n2 x2 – nx.x =
297
4
3
» x = - . So, n = 12 .
4
p
1
q
α
5. (a)log x = log x p − log x q =
1
β−α
β
αβ
− =
, So ,log p x =
q
αβ
.
β−α
6. (c) Since X∩Q = {2} , So, we can’t choose 1 from P. So, 2 will be common
while making subsets of P among 2,3,4,5.
So, total number of such subsets is = 23 = 8 .
7. (b) Here R.V. X is following Geometric Distribution with parameter = p = ½ So,
1
E(X) = = 2 .
p
∞
8. (b) E(X) = ∫C x.
c
x2
dx = ∞ .
9. (c) Let∣x∣ = a , Now, a2 – 5a +4 = 0 » a = 1 , 4 . So, x = ±1, ±4 . So, there are 4
real solutions.
10.(a) Here∣x2∣≤ 1 » 1+ x2≤ 2 »
that
x2
1+x2
1
1+x2
1
≥ »12
1
1
x2
2
1+x2
≤ »
2
1+x
≥ 0 . So, f(x)∈ [0,1) .
x + 2 x + 3 x + 2a
11.(c) It is given that b−c=a-b , so, |x + 3 x + 4 x + 2b|
x + 4 x + 5 x + 2c
37
1
≤ , and also it is clear
2
ISI MSQE ME(I) PREVIOUS YEAR’SANSWER KEY
Applying R3'= R3 – R2 , R2'= R2 - R1 ,
x + 2 x + 3 x + 2a
1
2(b − a)|
=| 1
1
1
2(c − b)
Applying C2'= C2 - C1 ,
x + 2 1 x + 2a
0 2(b − a)|
=| 1
1
0 2(c − b)
= 2(c − b) − 2(b − a)
12. (c)Let f(x) = (x – a)(x – b) + 2(x – c)(x – d)
Here f(a) > 0 , f(b) < 0 , f(c) < 0 , f(d) > 0 .
So, there exists two real and distinct roots one in the interval (a,b) and
other in (c,d) .
13. (b) Let f(x) = 4x + 2 & g(x) = 2x .Then f′(θ) = 2g′(θ).
1
1
14. (b) Let log x a = m , log a x= , minimum value of m + is 2 ( By the logic of
m
m
maxima & minima ).
15. (c) Take (1 + √x) = z ,
9
∫4
4
dx
2x(1+√x)
16. (c) y =
= ∫3
1
1+x
4 dz
dz
z(z−1)
»x=
= ∫3
1−y
y
z−1
4 dz
− ∫3
z
= 2log e 3 − 3 log e 2 .
. So, the inverse function of f(x) is
1−x
x
.
17. (b)
18. (c) X∼N(0,1) . If Φ(X) is cumulative distribution function of the variable X, then
dΦ(X)
φ(x) =
is the peobability density function.
dx
1
Now, E[Φ(X)] = ∫0 zdz =
1
2
. [ Let Φ(X)=z ]
38
ISI MSQE ME(I) PREVIOUS YEAR’SANSWER KEY
19. (b)L = lim
x→0
1−xr+1
)
1−x
x
e −1
loge (
= lim
x→0
− loge (1−x)
ex −1
= lim
x→0
loge(1−x)
x
ex −1
x
−
= 1.
20. (d)
21. (d)
22. (b) The correlation coefficient between α̂ and β̂ is
Cor(α̂ , β̂ ) = Cov(α̂ , β̂ )/√Var(α̂ )Var(β̂ ) .
Now, the OLS estimates of α, β are α̂ =
∑n
i=1 yi
n
, β̂ =
∑n
̅)
i=1 yi (xi −x
∑n
̅)2
i=1(xi −x
.
Using yi = α + β(xi − x̅) + ℰi , we can write α, β as
∑ni=1 ℰi
∑ni=1 ℰi (xi − x̅)
α̂ = α +
&β̂ = β + n
∑i=1(xi − x̅)2
n
Therefore, E(α̂ ) = α and E(β̂ ) = β, because E(ℰi ) = 0.
Cov(α̂ , β̂ ) = E((α̂ – E(α̂ ))(β̂ – E(β̂ ))
∑ni=1 yi ∑ni=1 yi (xi − x̅)
= E(
)
n ∑ni=1(xi − x̅)2
n n
1
=
E(∑ ∑ yi yj (xj − x̅)
n ∑ni=1(xi − x̅)2
n
=
i=1 j=1
n
1
∑ ∑ E(yi yj (xj − x̅))
n ∑ni=1(xi − x̅)2
i=1 j=1
n n
=
1
∑ ∑(xj − x̅) E(yi yj )
n ∑ni=1(xi − x̅)2
i=1 j=1
n n
=
1
∑ ∑(xj − x̅) E(yi 2 )
n ∑ni=1(xi − x̅)2
i=1 j=1
39
ISI MSQE ME(I) PREVIOUS YEAR’SANSWER KEY
[E(yi yj ) = 0 for i ≠ j because yi s are iid with E(yi ) = 0]
=
1
̅)2
n ∑n
i=1(xi −x
∑ni=1 ∑nj=1(xj − x̅)[becauseE(yi 2 ) = 1]
=0.
23. (d)
w2
√(λx1 )2 +(λx2 )2 −((λx )2+(λx )2 )
1
2
e
dw
∫0
24. (c)f(λx1 , λx2 ) =
Substituting
w
λ
f(λx1 , λx2 ) =
= v,
v2
)
√x 2 +x 2 −(
λ ∫0 1 2 e (x1)2+(x2)2 dv
= λf(x1 , x2 ).
25. (a)
26. (b)
27. (c)
28. (d)
29. (b) Put x = - x in the given equation, we have a1f(-x) + a2f(x) = b1+b2x .
Now comparing the above equation with the given equation, we have
2a b x
(x) = −( 22 2 2 ) .
a1 −a2
30. (c)
ISI MSQE MEI 2011 SOLVED PAPER
40
ISI MSQE ME(I) PREVIOUS YEAR’SANSWER KEY
1. (d)
𝐾 = √13 + 3√23/3 + √13 − 3√23/3is an irrational number exceeding 6.
To show, we apply 𝐴𝑀 ≥ 𝐺𝑀 inequality
√13 + 3√23/3 + √13 − 3√23/3
2
≥ √10
𝐾 ≥ 2√10; 𝑠𝑜 𝐾 𝑖𝑠 𝑖𝑟𝑟𝑎𝑡𝑖𝑜𝑛𝑎𝑙.
≥ 2√9 = 6.
2. (d)𝑥 > −3 is the domain of definition of the function 𝑥 ∈ (−3, ∞) − {−1}
[∵ (𝑥 + 1)(𝑥 + 4) = 0]
⇒ 𝑥 ≠ −4, −1
3. (b)
2
𝑙𝑜𝑔42 − 𝑙𝑜𝑔82 + 𝑙𝑜𝑔16
−⋯
2
2
2
= 𝑙𝑜𝑔(2
2 ) − 𝑙𝑜𝑔(23 ) + 𝑙𝑜𝑔(24 ) …
1
1
1
= 𝑙𝑜𝑔22 − 𝑙𝑜𝑔22 + 𝑙𝑜𝑔22 …
2
3
4
1 1 1
= ( − + …)
2 3 4
= 1 − 𝑙𝑜𝑔𝑒 2
4. (d)
1
𝑓(𝑥) = { 2
𝑥
𝑤ℎ𝑒𝑛 − 1 < 𝑥 < 1
𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒
41
ISI MSQE ME(I) PREVIOUS YEAR’SANSWER KEY
5. (c) x, y, z, are in H.P.; So
1 1 2
+ =
𝑥 𝑧 𝑦
⇒
1 1 1 1
− = −
𝑥 𝑦 𝑦 𝑧
⇒
𝑥−𝑦 𝑥
=
𝑦−𝑧 𝑧
6. (c)
𝑥
; 𝑥<0
𝑥
𝑓(𝑥) = {1 −
𝑥
; 𝑥>0
1+𝑥
𝑙𝑡
𝑙𝑡
𝑓(𝑥)
=
𝑓(𝑥) = 𝑓(0) = 0
𝑥 → 0+
𝑥 → 0−
𝑙𝑡 𝑓(ℎ) − 𝑓(0)
𝑙𝑡 𝑓(ℎ) − 𝑓(0)
=
=1
+
ℎ→0
ℎ
ℎ → 0−
ℎ
So, f(x) is continuously differentiable everywhere.
7. (d)
𝑓(𝑥) = 𝑎𝑥 + 𝑏
𝑓(1) = 𝑎 + 𝑏 = 3
𝑓 ′ (𝑥) = 𝑎
𝑓 ′ (1) = 9 = 𝑎
𝑆𝑜, 𝑏 = −6
𝑆𝑜, 𝑓(𝑥) = 9𝑥 − 6and𝑓(1 + 𝑥) = 9𝑥 + 3
42
ISI MSQE ME(I) PREVIOUS YEAR’SANSWER KEY
1/𝑥
𝑓(1 + 𝑥)
lim (
)
𝑥→0
𝑓(1)
= lim (1 + 3𝑥)1/𝑥 = 𝑒 3
𝑥→0
8. (a)
𝑓(𝑥) ↓ 𝑓, 𝑔: [0, ∞) → [0, ∞)
𝑔(𝑥) ↑
ℎ(0) = 0; ℎ(𝑥) ↓
ℎ(1) > ℎ(2) > ℎ(3) > ⋯
ℎ(𝑥) − ℎ(−1) ≤ 0 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑥 ≥ 1
for𝑥 = 1, ℎ(𝑥) − ℎ(1) = 0
ℎ(𝑥) − ℎ(1) > 0 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑥 < 1
9. (c) Total number of ways= 6𝐶3 × 4𝐶2 − 5𝐶2 × 3𝐶1 = 120 − 30 = 90
10. (b) Assuming that 𝑓 ∶ ℝ → ℝ
𝑓(1) + 𝑓(1) = (1 + 1)𝑓(1)𝑓(1)
⇒ 𝑓(1) = 0 𝑜𝑟 1
If f(1) = 1, then for any 𝑥 ≠ 0;
𝑥 + 𝑓(𝑥) = (𝑥 + 1)𝑓(𝑥) ⇒ 𝑓(𝑥) = 1
By continuity 𝑓 ≡ 1.
Similarly if f(1)= 0, 𝑡ℎ𝑒𝑛 𝑓 ≡ 0.
Hence answer is 2.
11. (d)
𝐸𝑥𝑝𝑟𝑒𝑠𝑠𝑖𝑜𝑛 =
=
√𝑥𝑛 − 1 − √𝑥𝑛
√𝑥1 − √𝑥2
+ ⋯+
𝑥1 − 𝑥2
𝑥𝑛−1 − 𝑥𝑛
1
[√𝑥1 − √𝑥𝑛 ][𝑤ℎ𝑒𝑟𝑒, 𝑑 𝑖𝑠 𝑡ℎ𝑒 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 𝑖𝑛 𝑡ℎ𝑒 𝑔𝑖𝑣𝑒𝑛 𝐴. 𝑃. ]
𝑑
43
ISI MSQE ME(I) PREVIOUS YEAR’SANSWER KEY
=
=
=
𝑥1 − 𝑥𝑛
𝑑(√𝑥1 + √𝑥𝑛 )
𝑥1 − (𝑥1 + (𝑛 − 1)𝑑)
𝑑(√𝑥1 + √𝑥𝑛 )
𝑛−1
√𝑥1 + √𝑥𝑛
12. (c) If 𝑥 ≥ 𝑦, max(𝑥, 𝑦) = 𝑥 and
𝑥 + 𝑦 + |𝑥 − 𝑦| 𝑥 + 𝑦 + 𝑥 − 𝑦 2𝑥
=
=
=𝑥
2
2
2
Else, then max(𝑥, 𝑦) = 𝑦and
𝑥 + 𝑦 + |𝑥 − 𝑦| 𝑥 + 𝑦 + 𝑦 − 𝑥 2𝑦
=
=
=𝑦
2
2
2
𝑇ℎ𝑢𝑠, max(𝑥, 𝑦) =
𝑥 + 𝑦 + |𝑥 − 𝑦|
2
13. (a) Since 𝑥1 , 𝑥2 , 𝑥3 , 𝑥4 > 0
So, P > 0
Then apply 𝐴𝑀 ≥ 𝐺𝑀 inequality
(𝑥1 + 𝑥2 ) + (𝑥3 + 𝑥4 )
≥ √(𝑥1 + 𝑥2 )(𝑥3 + 𝑥4 )
2
𝑃 = (𝑥1 + 𝑥2 )(𝑥3 + 𝑥4 ) ≤ 1
So, P is bounded between 0 and 1.
14. (d) Total number of handshakes = 𝑛𝐶2 =
𝑛(𝑛−1)
2
= 91
⇒ 𝑛 = 14
44
ISI MSQE ME(I) PREVIOUS YEAR’SANSWER KEY
15. (c) Such combinations are:
(1, 5) → 6 ways
(5, 1) → 6 ways
(3, 3) → 20 ways
(2, 4) → 15 ways
(4, 2) → 15 ways
Total number of ways = 62.
16. (a) Do yourself.
17. (d)𝑓(𝑥) = 𝑝𝑥 + 𝑞𝑥 2
𝑓(𝑦) = 𝑝𝑦 + 𝑞𝑦 2
1
𝑓(𝑥) = 𝑥 + ∫(𝑥𝑦 2 + 𝑥 2 𝑦)(𝑝𝑦 + 𝑞𝑦 2 )𝑑𝑦 = 𝑝𝑥 + 𝑞𝑥 2
0
Solve this and find p and q.
18. (b)(|𝑥| + |𝑦|)2 = 1 + 2|𝑥||𝑦|
|x||y| is maximum when |𝑥| = |𝑦| = 1/√2
So, maximum value of |𝑥| + |𝑦| = √2
19. (d) The number of onto function from A to B is
𝑛
𝑛
𝑛
= ( ) 𝑛𝑚 − ( ) (𝑛 − 1)𝑚 + ( ) (𝑛 − 2)𝑚 .
0
1
2
2
2
= ( ) 24 − ( ) (2 − 1)4 [∵ 𝑛 = 2, 𝑚 = 4]
0
1
= 16 − 2 = 14
𝑛
𝑛
𝐺𝑒𝑛𝑒𝑟𝑎𝑙 𝐹𝑜𝑟𝑚𝑢𝑙𝑎 ∶ ∑(−1)𝐾 ( ) (𝑛 − 𝐾)𝑚
𝐾
𝐾=0
45
ISI MSQE ME(I) PREVIOUS YEAR’SANSWER KEY
20. (b)2𝑟 + 5 + 𝑟 − 6 = 39 + 2
⇒ 𝑟 = 14
𝑛𝐶12 = 14𝐶12 =
14 × 13
= 91
2
21. (c)det(𝑋) = 𝑐 2 − 2
(𝑐 2 − 2)7 = 128 = 27
∴ 𝑐2 − 2 = 2
⇒ 𝑐2 = 4
⇒ 𝑐 = ±2
22. (c) Do yourself.
23. (a)
1st person can get out at any of the 6 floors.
2nd person can get out at any of the 5 floors.
3rd person can get out at any of the 4 floors.
4th person can get out at any of the 3 floors.
Total number of ways = 6 × 5 × 4 × 3 = 360
24. (a) Do yourself.
25. (b) Vaccination event : X,
Attacked event : Y
𝑛(𝑋 ∪ 𝑌) = 𝑛(𝑋) + 𝑛(𝑌) − 𝑛(𝑋 ∩ 𝑌)
100 = 70 + 85 − 𝑛(𝑋 ∩ 𝑌)
𝑛(𝑋 ∩ 𝑌) = 55
So, out of 85 vaccinated, minimum no. of attacked = 55
So, out of 100 vaccinated, minimum no. of attacked = 65
46
ISI MSQE ME(I) PREVIOUS YEAR’SANSWER KEY
26. (a) Do yourself.
27. (a) Do yourself.
28. (c) Calculate F(x) yourself.
∞
𝐸(𝑋) = ∫
𝑥0
𝛼. 𝑥0𝛼
𝑑𝑥
𝑥𝛼
= 𝛼. 𝑥0𝛼 [
=
𝑥 −𝛼+1 ∞
]
−𝛼 + 1 𝑥0
𝛼
. 𝑥0
𝛼−1
29. (c)
𝑛
𝑛
𝑛 1
𝜇 = ∑ 𝐾. ( ) ( )
𝐾 2
𝐾=0
∞
(𝑛 − 1)!
𝑛
1 𝑛−1
1
𝑛
= ∑
( )
= 𝐾. 2𝑛−1 . 𝑛 = .
(𝐾 − 1)! (𝑛 − 𝐾)! 2
2
2
2
𝐾=1
𝑛
𝐸(𝑋
2)
𝑛
1
𝑛2 + 𝑛
𝑛 1
𝑛−2
= ∑ 𝐾 ( ) ( ) = 𝑛(𝑛 + 1). 2 . 𝑛 =
𝐾 2
2
4
2
𝐾=0
𝑆𝑜 𝑉(𝑋) =
𝑛2 + 𝑛 𝑛2 𝑛
−
= .
4
4
4
30. (b)𝑋𝑖 ~ Bernoulli (p)
𝑛
𝐸(𝑦
2)
2
= 1 . 𝑃 (∑ 𝑋𝑖 = 100)
𝑖=1
𝑛
=(
) 𝑃100 (1 − 𝑝)𝑛−100
100
47
ISI MSQE ME(I) PREVIOUS YEAR’SANSWER KEY
ISI MSQE MEI 2012 SOLVED PAPER
1. (c) The frog will complete 10 feet in 5 days. Then in the 6th day morning the
frog will climb up 5 feet in the light and easily she can see the world then. So, it
will take total (5+1) = 6 days.
2. (c) f′(x) = 2x , f′(0)=0 .
3. (c) A1 = { 2,4,6,....} , A2= {3,6,9,....},
So A1∩A2 ={6,12,18,...}={6k , k∈N} = A5.
4. (b) Let us take an example : choosing a< b< c, then max{a,b}=b, max{a,c}= c,
max{b,c}=c .Then min{max{a,b}, max{a,c}, max{b,c}}= min{b, c} = b , which is
the second highest number in S.
5. (c)
xdx
1
6. (a)∫ 2 = ln(7x 2 + 2) + c , let 7x 2 + 2 = z .
7x +2
14
7. (d) f(x) = (x – 3)2 – (x – 1)2 + (x + 1)2 – (x – 1)2 – 8 = 0
» Put x = - 1, 1, 3, 0, then the equation will be satisfied.
8. (a) [y+1000z ,x+1000z]=[0,0] » x = y = -1000z .
So, three vectors are linearly dependent .
9. (a)
10. (c) Apply R1'= R1+ R3- 2R2 . Then 1st row of the determinant is zero, so the
answer is zero.
11. (a) If a +b = k, given and positive quantity then ab is maximum when a = b.
48
ISI MSQE ME(I) PREVIOUS YEAR’SANSWER KEY
12. (b)
13. (a)
14. (c) x ≥ 0 , 3 – x ≥ 0 , x2 – 4x ≥ 0 , these 3 conditions satisfies only when x = 0 .
15. (a) Let us choose P(x) = ax2 + c since P(1) = P(-1). So, two roots sum is zero.
16. (a)
17. (d) The maximum value of a(1 – a) is ¼ , since a+1 – a = 1, so, product is
maximum when values are equal that is a = ½ .
So, here the max{a(1 – a) b(1 – b) c(1 – c)} is ¼ × ¼× ¼ = 1/64 .
18. (b)
19. (c) If we choose a = b = 1 and c = 2 then we shall get the least positive value of
the given expression. So, the answer is 4.
20. (d) (a+b+c)2 = 1+ 2(ab+bc+ca) ≥ 0 » (ab+bc+ca) ≥ - ½.
21. (c)
−1 dx
22. (d)∫−4
x
1
= log ( ) = − ln 4 .
4
23. (a) Here x + y + z = 9 , then max(x,y,z) = 3. So, x + 3y + 5z = 27.
24. (d)
49
ISI MSQE ME(I) PREVIOUS YEAR’SANSWER KEY
25. (c)
26. (a)
27. (d) Just give counter example for each options.
28. (b) Take ∣x∣= a , then a2 −3a−10 = 0 has two distinct real roots. a = 5, -2 . Then
x = ±5. There are one non-negative real roots of the equation.
29. (b) If one of the two sequences is divergent then the sum of two sequences
will also diverge.
30. (a) f(2) = 2/3 , f(3) = 3/4 ; here f(3) > f(2)
So, the function is increasing when x is positive.
50
ISI MSQE ME(I) PREVIOUS YEAR’SANSWER KEY
ISI MSQE MEI 2013 SOLVED PAPER
1
x−1
1
1. (d)f ( ) =
, f (f ( )) =
x
x+1
x
2. (𝐜) lim
x−1
x+1
x−1
1+
x+1
1−
1.2+2.3+3.4+⋯+n(n+1)
n3
n→∞
1
= .
= lim
x
n(n+1)(n+2)
3n3
n→∞
1
= .
3
1+a
3. (a) By AM ≥ GM Inequality, ( i) ≥ √ai , where i = 1,2,3,...,n.
2
Multiplying all these we have (1+a1)(1+a2)....(1+an) ≥ 2n , since a1a2...an=1 .
4. (b) P[X=0]=P[X=1] , where X∼Bin(n,p) , 0<p<1.
1−p
1
So, (1- p)n= np(1- p)n-1 » n =
»p=
.
p
n+1
5. (a) E(X100)=E(X2.X2.X2....X2)= E(X2) E(X2).... E(X2)=1.
6. (c) So, α+β=a , αβ=b, so a+b,b-a are the roots of the equation.
The equation becomes x2- (a+b+b-a)x + (a+b)(b-a)= 0 » x2 - 2bx + b2- a2 = 0.
1 2
1
1
7. (a) f(x) = 2[(x + ) − 2] − 3 (x + ) − 1 = 0 , Let (x + ) = a
x
x
x
5
So, 2a2 – 3a – 5 = 0 » a = - 1, .
2
2
1
So, 2x – 5x + 2 = 0 » x = 2 , . So, their product is 1.
2
8. (c) Total number of cases of [Number of Head > Number of Tail + Number of
Tail > Number of Head] = 243 , where Number of Head>Number of Tail =
Number of Tail > Number of Head. So, answer is
243
2
= 242 .
9. (a)
10.(a)f (x,y) = m when y = 0. f(x+k,y) = f(x,y)=m when y=0
f (x,y+k)= f(x,y)+kx= f(x,0) + kx= m+kx , when y=0
So, f (x,k) = m+kx , f (x,y)=m+xy .
51
ISI MSQE ME(I) PREVIOUS YEAR’SANSWER KEY
11.(c)
n
n
12. (b)(r−1
) = 165 , (nr) = 330 , (r+1
) = 462 . Solve these 3 equations to find n .
1
13. (d) (a+b+c)2 ≥ 0 » a2+b2+c2 + 2(ab+bc+ca) ≥ 0 » (ab+bc+ca) ≥ - .
2
14. (d)f(x) is not differentiable at x = 4 and 5 .
15. (d)
16.(c) A1 = { 2,4,6,....} , A2= {3,6,9,....}, so A1∩A2 ={6,12,18,...}={6k, k∈N} = A5.
1
1
17.(c)lim{ (√1 + x + x 2 − 1)} = lim{ (
1+x+x2 −1
x→0 x √1+x+x2 +1
1+x
x→0 x
= lim(
x→0
)}
1
√1+x+x2 +1
)= .
2
1
n
18.(a) For Binomial distribution E(K+1)= ∑nk=0(k + 1)(nk) n = +1 ,
2
2
where, p = ½ . So, the value of the given sum is = n2n-1 + 2n.
19.(c)Perform these elementary row operations to the given matrix to reduce it into
row-reduced matrix form : (i) R3' = R3 – R2 + R4 ,
(ii) R2 is interchanged by R4,
(iii) R4' = R4 - R3 ,
Then R4 of the given matrix will vanish. So, rank is 3.
20.(d) The two integers with product is maximum is of the form
2n+1−1 2n+1+1
(
2
,
2
) = (n, n + 1).
21. (c) S = {a2+a4+a6+.......}+{ab+a2b2+a3b3+.....} =
52
a2
1−a2
+
ab
1−ab
.
ISI MSQE ME(I) PREVIOUS YEAR’SANSWER KEY
22.(d) The necessary condition to exist maxima and minima at any point x=a is
that f '(a)=0 . But here is no real ‘a’ for which f '(x)=0.
[Since x2 – 4x + 8 =0 doesn’t have any real solution]
23.(a)
24.(c)
25.(c) On integration, we have log(f′(x))= x + c » f′(x) =kex » f(x)= kex + m.
2
Given f(0)=e2 , f(1)=e3 gives k = e2 . ∫−2 ex+2 dx = e4 − 1 .
26.(a)
27. (b) Compute A2, then put the value of A2, A, I in the given form to get the
answer.
n
4!
2
2
28.(d) The number of permutations is = − 1 =
− 1 = 11.
29.(b) Mean deviation about mean can’t exceed the standard deviation. To prove
this statement use Cauchy-Schwarz inequality & choose ai =∣ xi - x̅ ∣ and bi = 1.
30.(d) A same kind of problem has been done earlier.
53
ISI MSQE ME(I) PREVIOUS YEAR’SANSWER KEY
ISI MSQE MEI 2014 SOLVED PAPER
1
𝑥−1
1
1. (d)𝑓 ( ) =
, 𝑓 (𝑓 ( )) =
𝑥
𝑥+1
𝑥
𝑥−1
𝑥+1
𝑥−1
1+
𝑥+1
1−
1
= .
𝑥
𝐼
2. (b) = ∫ 𝑒 𝑥 cos(𝑥) 𝑑𝑥 = 𝑒 𝑥 𝑐𝑜𝑠𝑥 + ∫ 𝑒 𝑥 𝑠𝑖𝑛𝑥𝑑𝑥 = 𝑒 𝑥 𝑐𝑜𝑠𝑥 + 𝑒 𝑥 𝑠𝑖𝑛𝑥 −
2
𝑥
п
2
𝑥
𝐼
2
So, I = (𝑒 𝑐𝑜𝑠𝑥 + 𝑒 𝑠𝑖𝑛𝑥) Now put the limits, answer will be 𝑒 − 1.
3. (b)f(2) = 2, but f(1) = 0. So, f(2) ≠ 2f(1)
4. (c)
Let there are 100 people in the city among which 40 are males & 60 are females.
And there are 20 males who smoke & 18 females who smoke. So, the probability
20
that a smoker is male is = 0.526.
38
5. (b)Pr(E) = Pr(the sum of the numbers on the two dice is 7) = 1/6
Pr(F) = Pr(the number on the blue die equals 4) = 1/6
Pr(E. F) = Pr(F. G) = 1/36
Since, we also have Pr(E)Pr(F) = 1/36
Therefore, E and F are independent.
6. (a) The matrix is called p-special if det(A) is not divisible by p. det(A) = - ab
Now any number can be broken down in two products of prime numbers.
This number will be divisible by p only if p is one of the factors of – ab = 0,
or, ab = 0.
So, in order to form a p-special matrix we will have to select a and b in such a way
that none of them has p as a factor. So, we can choose (p – 1) elements for a &
similarly(p – 1) elements for b. So, a×b can be taken in (p – 1)(p – 1) ways.
7. (a) We have to count the number of ways in which we can fill these seven
spots using letters from {A, B, C} so that A cannot be followed by B, B cannot
be followed by C, and C cannot be followed by A.
First spot can be filled by any of the three letters, i.e. in three ways. Once the
first spot is filled, we have only two ways to fill the second spot. For example, if
54
ISI MSQE ME(I) PREVIOUS YEAR’SANSWER KEY
the first spot is filled by A, then second spot will be taken by either A or C.
Similarly for the third spot and so on. So there are 3×26 = 192 ways.
8. (a) Both elements of the given equation is of odd degree. So, their sum can be
0 only if they are equal & opposite in sign. This can occur only if x lies in [a,b]
𝑎+𝑏
such that it is equidistant from both a and b. Thus x =
is the only solution
2
of the given equation.
9. (c) Simple Linear Programming Problem.
10.(b) The easiest way to do it is to recognize that F(x) ≤ G(x) for all x means that
for any x, X takes smaller values less than x is more likely to occur in G than in
F. Or in other words, X takes smaller values than x in G more often than it
takes in F. So, expectation of X in F must be greater than or equal to
expectation of X in G. More formal proof is attached.
11.(c)f(x) is a pdf. Therefore, total area under f(x) is 1.
𝛼
21
∫0 𝑥𝑑𝑥 + ∫𝛼 2 𝑑𝑥 = 1implying α = 0, 1. Now α can’t be 0, so α equals to 1.
21
1
2
2
So, P[X ≥ 1] = ∫1 𝑑𝑥 = .
1 𝑥𝑘
12.(a)∑100
𝑘=1 ∫0
𝑘
dx = ∑100
𝑘=1
1
𝑘(𝑘+1)
1
1
𝑘
𝑘+1
= ∑100
𝑘=1( −
)=1−
1
101
=
100
101
.
13.(d) Add 2nd& 4th equation, you will get x4 – x3≤ a – 2,
From 5th equation x4 – x3≤ - 4.
Since, this system of equations has solution, so these two inequalities needs to be
consistent. So, a – 2 = - 4 => a = - 2.
14.(a) One of the ways to do this problem is:
55
ISI MSQE ME(I) PREVIOUS YEAR’SANSWER KEY
Pr(Head turns up for the first time after even number of tosses) = Pr(TH) +
Pr(TTTH)+ Pr(TTTTTH) + ......
where TH is the event that first toss is tail and second is head, TTTH is the event
that first three tosses are tails and forth toss is heads, etc.
Thus, Pr(Head turns up for the first time after even number of tosses)
= [1/(22)] + [1/(24)] + [1/(26)] + ...
This is a geometric series that sums to 1/3.
15.(b) Use Binomial Probability.
16. (a) Sum of the infinite series =
17.(b) Leibnitz’s Rule:
𝑑
𝑎
1−𝑟
=
(𝑘−1)
𝑘!
1
1−
𝑘
𝑣(𝑥)
𝑓(𝑡)𝑑𝑡} = 𝑓(𝑣(𝑥))
{∫
𝑑𝑥 𝑢(𝑥)
𝑑𝑣(𝑥)
𝑑𝑥
− 𝑓(𝑢(𝑥))
𝑑𝑢(𝑥)
𝑑𝑥
.
1
So, G’(x) = xe2x (1+x). So, lim 𝐺 ′ (𝑥) = lim {𝑒 2𝑥 (1 + x)} = 1.
𝑥→0 𝑥
𝑥→0
1
18.(c) f’(x) = 0 gives x = 1/2 . f(1/2) = 𝛼−1. Since α ∈ (0,1), so maximum value of f
2
will be in (1,2).
19.(a)∑𝑛𝑘=1 𝑘.
𝑛!
𝑘!(𝑛−𝑘)!
(𝑛−1)!
𝑛−1
= 𝑛 ∑𝑛−1
= 𝑛2𝑛−1 .
𝑘=1 (𝑘−1)!(𝑛−𝑘)! = 𝑛(1 + 1)
𝑎
20.(a) k-th term of the A.P. = tk = a + (k – 1)×d =
,
1−𝑑
where a > 2 is a prime & d ∈ (0,1).
Put d = ½ , you will get finite satisfactory solution of a & k which satisfies the given
conditions.
21.(b) 2k + 3 = 31 gives k = 14, so, maximum value of T is (k+1) = 15.
Calculate minimum value yourself.
56
ISI MSQE ME(I) PREVIOUS YEAR’SANSWER KEY
22.(b)Det(A) = 0 gives p =
𝑎2
2
, So, p =2 gives a = ±2, which is a possible value.
23.(a) (𝑥 − 1)(𝑥 2 − 7𝑥 + 𝛼) = 0.
So, x =1 is one root.
And if (𝑥 2 − 7𝑥 + 𝛼) = 0 then putting x =1 there we get α = 6.
If α = 6 then (𝑥 − 1)(𝑥 2 − 7𝑥 + 𝛼) = (𝑥 − 1)(𝑥 − 1)(𝑥 − 6) = 0which implies x
= 1, 6.
24.(b) Sum of the infinite series = log 𝑒 3 −
1
1
log𝑒 3
2
+
log𝑒 3
3
− ⋯∞
= log 𝑒 3(1 − + − ⋯ ∞)
2
3
= log 𝑒 3 log 𝑒 2.
25.(c) 20 persons can shook hands with 19 number of persons.
26.(a) P[max(X,Y) ≤ z] = P[min(X,Y) ≤ (1-z)]
 P[X≤ 𝑧] P[Y≤ 𝑧] = P[X≤ (1 − 𝑧)] P[Y≤ (1 − 𝑧)]
 z2 = (1 – z)2 gives z = ½ .
27.(d) Put x = y = 0, then f(0) = 0.
Put x = y = 1, then f(1) = 0 or 1.
Similarly, calculate the value of f(2), f(3) & f(5) & check yourself.
28.(d) Draw the graph of g(x).
29.(d) Here (x+1)→0, so following limit doesn’t exist.
30.(b) Here a2 – 3a =2 doesn’t have any real solution, so f is not continuous.
57
ISI MSQE ME(I) PREVIOUS YEAR’SANSWER KEY
ISI MSQE PEA (Mathematics) 2015 SOLVED PAPER
sin{√𝑥}
{√𝑥}
𝑥→0
1. (b) lim+
= lim+
sin √𝑥
𝑥→0
√𝑥
₌ 1.
2. (d) By Rolle’s theorem, there exists f(x) = x for at least one x ∈ [0, 1] where f is
continuous.
6 − 2𝑥 ,
,
3. (c) f(x) = {2
2𝑥 − 6 ,
𝑙𝑡
f(x)
𝑥 → 2−
=
𝑙𝑡
𝑥 → 2−
𝑙𝑡
f(x)
𝑥 → 2+
= 2.
𝑥 ≤2
2 <𝑥 <4
𝑥 ≥ 4
(6 – 2x) = 2
4. (b)
Number of students passed inexactly one
subject = 10 + 10 + 5 = 25
𝑎
5. (d) A = [
𝑐
𝑏
]
0
2
𝐴2 = [𝑎 + 𝑏𝑐
𝑎𝑐
So, a = 0 and either b or c equal to 0.
58
𝑎𝑏] = [0 0]
0 0
𝑏𝑐
ISI MSQE ME(I) PREVIOUS YEAR’SANSWER KEY
2
1
𝑦
𝑥
6. (c) =
∴ 2y =
1
+
𝑧
4 𝑥𝑧
𝑥+𝑧
log (x + z) + log (x + z – 2y) = log (x + z) + log {
(𝑥+𝑧)2 −4𝑥𝑧
(𝑥+𝑧)
=log ((𝑥 + 𝑧)2 – 4xz) = log (𝑥 – 𝑧)2 = 2 log (x – z).
7. (a) x + 2y = x’ , x – 2y =y’
∴x =
𝑥 ′ + 𝑦′
2
,y=
∴f (x, y) = (
𝑥+𝑦
2
𝑥 ′ − 𝑦′
4
𝑥 −𝑦
)(
4
)=
𝑥2 − 𝑦2
8
.
8. (c) Do yourself.
9. (a) log (1 – 2x + 𝑥 2 ) = 0 = log 1
∴ 𝑥 2 −2x = 0
∴x(x – 2) = 0
∴x= 0, x = 2.
10.(c ) x + y = 100
xy≤ 2500, since 𝑥𝑚𝑎𝑥 = 𝑦𝑚𝑎𝑥 = 50 for maximizing xy.
1
1
∴( + )
𝑥
𝑦
𝑚𝑖𝑛
=
(𝑥+𝑦)
𝑥𝑦𝑚𝑎𝑥
=
100
2500
=
1
25
.
59
}, putting 2y =
4 𝑥𝑧
𝑥+𝑧
ISI MSQE ME(I) PREVIOUS YEAR’SANSWER KEY
11.(d) AM ≥ GM
⇒
𝑥+
1
𝑥
2
∴𝑥+
≥ √𝑥 .
1
𝑥
1
≥2
𝑥
So, f(x) = 4𝑥 + 2𝑥 + 4−𝑥 + 2−𝑥 + 3
= (4𝑥 +
1
4𝑥
) + (2𝑥 +
1
2𝑥
)+ 3≥7
∴f(x) ∈ [7, ∞).
12.(c) putting x = 1, y = 0, then f(1) = f(1) + f(0)
⇒f(0) = 0, f(1) = 7.
Again putting x = 1, y = 1, then f(2) = 2f (1) = 14.
Similarly, f(3) = 21 and so on.
∑𝑛𝑟 =1 𝑓(𝑟) = 7{1 + 2 + 3 + …. + n} =
7𝑛(𝑛+1)
2
13.(b) f(x) = 2g(x) + c
f(0)= 2g(0) + c
⇒c= 0
∴f(1) = 2g(1)
⇒ 6 =2g(1)
∴ g(1) = 3.
14.(d) k = 2ax + b
60
.
ISI MSQE ME(I) PREVIOUS YEAR’SANSWER KEY
15.(c)1⁄3 ,
1⁄
3
1+ 1⁄3
1
⁄
= 1⁄4 , 14 =
1+ ⁄4
1
5
, ….
This is a Harmonic progression.
So, 𝑡500 =
1⁄
501
1+ 1⁄501
=
1
502
.
16. (d)𝑥1 + 𝑥2 + 𝑥3 = 10
1 ≤ 𝑥1 , 𝑥2 , 𝑥3 ≤ 6
No. of way = 𝑛𝐶𝑛−𝑟+1 = 10𝐶6−1 = 9𝐶5 =
(𝑥 –𝑎)(3𝑥 –𝑎)
17.(b)
2
10×9×8×7×6
2×3×4×5
=0
𝑎
∴x= a, x = .
3
18.(a) Do yourself.
19.(a) Do yourself.
2𝑥 − (𝑎1 + 𝑎2 ); 𝑥 ≤ 𝑎1
𝑎1 < 𝑥 < 𝑎2
20.(d) f(x) = { 𝑎1 − 𝑎2 ;
(𝑎1 + 𝑎2 ) − 2𝑥; 𝑥 ≥ 𝑎2
𝑥 ∈ [𝑎1 , 𝑎2 ]
21.(d) P (A∕B) =
𝑃(𝐴 ⋂ 𝐵)
𝑃(𝐵)
< 𝑃 (𝐵)
22. (d) P(𝐴 ∩ 𝐵) = 0.
61
= 36.
ISI MSQE ME(I) PREVIOUS YEAR’SANSWER KEY
We know P(A), P(B) are positive.
Since P (𝐴 ⋂ 𝐵) ≠ 𝑃(𝐴)𝑃(𝐵), so not independent
And A⋂ 𝐵 = 𝜑, not dependent
P(A) not necessarily exact to P(B).
So, none of the above.
23.(b) Both mean & median will be increased by 200,000.
24.(b) Put n = 0, 𝜇 = E(X).
So, 𝜇2𝑛 = 0.
25.(a)
26.(d) Gambler’s Ruin Problem.
So, P(Puja’s chance of winning Priya’s money) =
27.(c) P(𝑊1 ) =
P(𝑊2 ) =
𝑤−1
𝑤+𝑏−1
𝑤
𝑤+𝑏
.
, since WOR case.
62
𝑎
𝑎+𝑏
.
ISI MSQE ME(I) PREVIOUS YEAR’SANSWER KEY
28.(b)P(3 digit number is a multiple of 3) =
4×3
4×3×2×1
=
1
2
.
[To be a multiple of 3, among these 4 numbers, one number chosen should be ‘3’,
and we have to choose, two more from others.]
−2 𝑤𝑖𝑡ℎ 𝑝𝑟𝑜𝑏. 1⁄5
−1 𝑤𝑖𝑡ℎ 𝑝𝑟𝑜𝑏. 1⁄5
0 𝑤𝑖𝑡ℎ 𝑝𝑟𝑜𝑏. 1⁄5
29.(b) X =
1 𝑤𝑖𝑡ℎ 𝑝𝑟𝑜𝑏. 1⁄5
1
{ 2 𝑤𝑖𝑡ℎ 𝑝𝑟𝑜𝑏. ⁄5
0 𝑤𝑖𝑡ℎ 𝑝𝑟𝑜𝑏. 1⁄3
Y = 1 𝑤𝑖𝑡ℎ 𝑝𝑟𝑜𝑏. 1⁄3
1
{2 𝑤𝑖𝑡ℎ 𝑝𝑟𝑜𝑏. ⁄3
Y = |𝑋|, So, X & Y are dependent &R = 0 , since 𝑋̅ = 0, 𝑌̅ = 1
30.(d) P (meeting) =
202 − 152
202
=
7
16
[Using Geometric Probability]
MSQE 2016 PEA (Mathematics) SOLVED PAPER
1. (c)
63
ISI MSQE ME(I) PREVIOUS YEAR’SANSWER KEY
𝑃(10) = 1000𝑎 + 100𝑏 + 10𝑐 + 𝑑 ; 𝑎, 𝑏, 𝑐, 𝑑 ∈ {1, 2, , … , 9}
𝑃(10) = 5000 + 800 + 60 + 1 = 5861.
So, 𝑎 = 5, 𝑏 = 8, 𝑐 = 6, 𝑑 = 1
So, c is 6.
2. (b)
𝜕2𝑓 ∗
(𝑥 ) ≤ 0
𝜕𝑥 2
Is a necessary condition for 𝑥 ∗ to be a point of local maximum of f on A.
3. (d) 𝑥1 < 𝑥2 < 𝑥3 < 𝑥4 < 𝑥5
Median = 𝑥3 ;
𝑥5 ↑
New median = 𝑥3 (unchanged)
New mean ↑
So, none of the above is true.
4. (a)𝑛𝐶0 + 𝑛𝐶1 + ⋯ + 𝑛𝐶𝑛 = 2𝑛 = 4096 = 212
∴ 𝑛 = 12
Total number of terms in the expression = 13.
∴ 𝑙𝑎𝑟𝑔𝑒𝑠𝑡 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑤𝑖𝑙𝑙 𝑏𝑒 = 12𝐶6 = 924.
5. (c) (Bayes theorem)
𝐸1 ∶ Picking up Green card ;𝐸2 ∶ Picking up Red card ; 𝐸3 ∶ Picking up mixed card
G
R
G
64
ISI MSQE ME(I) PREVIOUS YEAR’SANSWER KEY
G
R
R
𝑃(𝐸1 ) = 𝑃(𝐸2 ) = 𝑃(𝐸3 ) =
1
3
𝐴: Getting green card
𝑃(𝐴|𝐸1 ) = 1,
𝑃(𝐸3 |𝐴) =
𝑃(𝐴 |𝐸2 ) = 0, 𝑃(𝐴 |𝐸3 ) =
1
2
𝑃(𝐸3 ) × 𝑃(𝐴|𝐸3 )
2
= .
𝑃(𝐸1 )𝑃(𝐴|𝐸1 ) + 𝑃(𝐴2 )𝑃(𝐴|𝐸2 ) + 𝑃(𝐴3 )𝑃(𝐴|𝐸3 ) 3
6. (d)
𝜋/2
𝜋/2
𝐼 = ∫ 𝑥 sin 𝑥 𝑑𝑥 = ∫ 𝐹(𝑥)𝑑𝑥
0
0
∫ 𝐹(𝑥)𝑑𝑥 = −𝑥 cos 𝑥 + sin 𝑥 + 𝑐 (𝑈𝑠𝑖𝑛𝑔 𝐼𝑛𝑡𝑒𝑔𝑟𝑎𝑡𝑖𝑜𝑛 𝑏𝑦 𝑝𝑎𝑟𝑡𝑠)
𝜋/2
𝑆𝑜,
𝜋
∫ 𝐹(𝑥)𝑑𝑥 = 𝑓 ( ) − 𝑓(0) = 1 − 0 = 1
2
0
7. (d)
lim 𝑓(𝑥) = 0
𝑥→0−
lim+ 𝑓(𝑥) = 𝑏
} 𝑠𝑜, 𝑏 = 0 𝑓𝑜𝑟 𝑓(𝑥)𝑡𝑜 𝑏𝑒 𝑐𝑜𝑛𝑡𝑖𝑛𝑢𝑜𝑢𝑠 𝑎𝑡 𝑥 = 0
𝑥→0
𝑅𝑓 ′ (0) =
𝑑
[𝑎𝑥 + 𝑏]𝑥=0+ = 𝑎
𝑑𝑥
𝐿𝑓 ′ (0) =
𝑑
[sin 2𝑥]𝑥=0− = 2
𝑑𝑥
So, 𝑎 ≠ 2, 𝑠𝑜, 𝑎 = 1, 𝑏 = 0.
8. (b)
It’s a case of linear regression with food consumption and income as Y and X.
65
ISI MSQE ME(I) PREVIOUS YEAR’SANSWER KEY
𝑅 2will remain same in both the cases.
𝑅2 = 1 −
𝑆𝑆𝑅
.
𝑆𝑆𝑇
9. (b) For extremum,
𝑓𝑥 = 3𝑒 𝑦 − 3𝑥 2 = 0
} ⇒ 𝑥 = 1, 𝑦 = 0
𝑓𝑦 = 3𝑥𝑒 𝑦 − 3𝑒 3𝑦 = 0
𝑓𝑥𝑦 = 3𝑒 𝑦 = 𝑓𝑦𝑥
𝑓𝑥𝑥 = −6𝑥
𝑓𝑦𝑦 = 3𝑥𝑒 𝑦 − 9𝑒 3𝑦
𝑓𝑥𝑥
𝑠𝑜, |𝐷| = |
𝑓𝑦𝑥
𝑓𝑥𝑦
−6 3
|=|
| = 27 > 0 .
𝑓𝑦𝑦
3 −6
10. (a)
𝑓(𝑓(𝑥)) =
𝑥+√3
1−√3𝑥
+ √3
1 − √3.
𝑓(𝑓(𝑥)) =
(𝑥+√3)
=
𝑥 + √3 + √3 − 3𝑥
1 − √3𝑥 − √3𝑥 − 3
1−√3𝑥
2𝑥 − 2√3
2 + 2√3𝑥
=
𝑥 − √3
1 + √3𝑥
.
11. (d)
1
∫ 𝑓(𝑥)𝑑𝑥 = 1
0
𝑘
1
⇒ ∫ 𝑎𝑑𝑥 + ∫ 𝑏 𝑑𝑥 = 1 ⇒ 𝑎𝑘 + 𝑏(1 − 𝑘) = 1
0
⇒𝑘=
𝑘
1−𝑏
𝑎−𝑏
1
𝑘
1
𝑆𝑜, 𝐸(𝑋) = ∫ 𝑥𝑓(𝑥)𝑑𝑥 = ∫ 𝑥 𝑎𝑑𝑥 + ∫ 𝑥 𝑏𝑑𝑥
0
0
𝑘
66
ISI MSQE ME(I) PREVIOUS YEAR’SANSWER KEY
∴ 𝐸(𝑋) =
1−𝑏 2
)
𝑎−𝑏
𝑎(
=
𝑎𝑘 2 + 𝑏(1 − 𝑘 2 )
2
1−𝑏 2
) )
𝑎−𝑏
+ 𝑏 (1 − (
2
=
(1 − 𝑏)2 + 𝑏(𝑎 − 𝑏)2
2(𝑎 − 𝑏)
=
1 − 2𝑏 + 𝑎𝑏
.
2(𝑎 − 𝑏)
12. (b)𝑥 2 − 3|𝑥| + 2 < 0
𝑥2
𝑥≥0
𝑥<0
2
− 3𝑥 + 2 < 0 𝑥 + 3𝑥 + 2 < 0
⇒ (𝑥 − 2)(𝑥 − 1) < 0
⇒ (𝑥 + 2)(𝑥 + 1) < 0
⇒ 𝑥 ∈ (1, 2)
⇒ 𝑥 ∈ (−2, −1)
𝑆𝑜, {𝑥 ∶ −2 < 𝑥 < −1} ∪ {𝑥 ∶ 1 < 𝑥 < 2}
13. (a)
𝑥
0
4𝑑 − 1 1
1
[ 0
−1
1 ] [𝑦] = [0]
0
0
0 4𝑑 − 1 𝑧
|𝐴| = 0
⇒ (4𝑑 − 1)(1 − 4𝑑) = 0
⇒𝑑=
14. (a) Small mistake in the Question.
Right Question will be
67
1
4
ISI MSQE ME(I) PREVIOUS YEAR’SANSWER KEY
𝑎𝑥 + 𝑏
𝐹(𝑥) = { 2
𝑥 −𝑥+1
𝑖𝑓 𝑥 ≤ 𝑎
𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒
Then answer will be F is continuous in (1, 1)
15. (c)
D is the feasible region. So, all the 3 options will satisfy the given 3 constraints.
max (3𝑥𝑦 − 𝑦 3 )is possible when (x, y) = (15, 5).
16. (d)
Take 𝑓(𝑥) = 𝑥 ↑
𝑓 ′ (𝑥) = 1
𝑓 ′ (1) = 1
𝑔(𝑥) = 𝑥; 𝑔′ (𝑥) = 1
𝑔(1) = 1 ; 𝑔′ (1) = 1
17. (c)𝑃(𝑥) = 𝑎𝑥 2 + 𝑏𝑥 + 𝑐
68
ISI MSQE ME(I) PREVIOUS YEAR’SANSWER KEY
𝑃(1) = 𝑎 + 𝑏 + 𝑐 = −3
𝑃(−1) = 𝑎 − 𝑏 + 𝑐 = −9 }
𝑃(−2) = 4𝑎 − 2𝑏 + 𝑐 = 0
On solving 𝑎 = 4, 𝑏 = 3, 𝑐 = −10
So, 𝑃(𝑥) = 4𝑥 2 + 3𝑥 − 10
∴ 𝑃(𝑥) = (𝑥 + 2)(4𝑥 − 5)
5
∴ 𝑃( ) = 0
4
18. (b)
𝑐(𝑛, 3)
𝑐(𝑐(𝑛, 2), 3)
19. (c)𝑥 + 𝑦 ≤ 1 𝑎𝑛𝑑 4𝑥 + 𝑦 ≤ 2
𝑥
0
1
𝑥
0
0.5
𝑦
1
0
𝑦
2
0
𝑥+𝑦 =1
4𝑥 + 𝑦 = 2
⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯
1
𝑥= ,
3
𝑦=
2
3
𝐴𝑡 𝐴(0, 1); 𝑧𝑚𝑎𝑥 = 𝑏
69
ISI MSQE ME(I) PREVIOUS YEAR’SANSWER KEY
1 2
𝐴𝑡 𝐵 ( , ) ; 𝑧𝑚𝑎𝑥
3 3
1
𝐴𝑡 𝐶 ( , 0) ; 𝑧𝑚𝑎𝑥
2
So, (c) is not an optimal value.
20. (c)𝐹 ′ (𝑥) ↑
𝐹 ∶ [0, 1] → ℝ
𝑥, 𝑦 ∈ [0, 1]
𝑥>𝑦
⇒ 𝐹 ′ (𝑥) ≥
𝐹(𝑥) − 𝐹(𝑦)
𝑥−𝑦
⇒ 𝐹(𝑥) − 𝐹(𝑦) ≤ (𝑥 − 𝑦)𝐹 ′ (𝑥)
21. (d)
Total balls = 𝑁
Red balls = 𝑎
𝑃2 = 𝑅𝑅 + 𝑁𝑅
=
𝑎 𝑎−1 𝑁−𝑎
𝑎
+
+
×
𝑁 𝑁−1
𝑁
𝑁−1
=
𝑎(𝑎 − 1) + (𝑁 − 𝑎)𝑎
𝑁(𝑁 − 1)
=
𝑎
𝑁
22. (a)(𝑡 − 𝑥) = √𝑥 2 + 2𝑏𝑥 + 𝑐
⇒ 𝑥 2 + 𝑡 2 − 2𝑡𝑥 = 𝑥 2 + 2𝑏𝑥 + 𝑐
⇒ 𝑡 2 − 2𝑡𝑥 = 2𝑏𝑐 + 𝑐
Differentiating w.r.t. t,
70
ISI MSQE ME(I) PREVIOUS YEAR’SANSWER KEY
2𝑡 − 2𝑥 − 2𝑡
⟹
𝑑𝑥
𝑑𝑥
= 2𝑏
𝑑𝑡
𝑑𝑡
𝑑𝑥 𝑡 − 𝑥
=
𝑑𝑡 𝑡 + 𝑏
23. (b)
Take 𝑛 = 2; 𝐴2×2 𝑖𝑠 𝑎 2 × 2 𝑚𝑎𝑡𝑟𝑖𝑥
𝐴=[
1 1
]
1 2
|𝐴| = 1
By induction, |𝐴| = 1.
24. (a) Given 𝑥1 + 𝑥2 + 𝑥3 = 10;
total number of non−negative integer solution of this equation
= 𝑛 + 𝑟 − 1𝐶𝑟−1 = 10 + 3 − 1𝐶2
= 12𝐶2 = 66
25. (c)
2𝑏
∫
𝑏
𝑥2
𝑥
𝑑𝑥
+ 𝑏2
2𝑏
1 2𝑥
= ∫
𝑑𝑥
2
𝑧
𝑏
𝑧 = 𝑥 2 + 𝑏2
𝑑𝑧
= 2𝑥
𝑑𝑥
𝑑𝑧 = 2𝑥 𝑑𝑥
2𝑏
1 𝑑𝑧
= ∫
2
𝑧
𝑏
1
2𝑏
1
2𝑏
1
5𝑏 2
1
5
= [log 𝑧]
= [log(𝑥 2 + 𝑏 2 )]
= log ( 2 ) = log ( )
2
𝑏
2
𝑏
2
2𝑏
2
2
26. (d) (Lagrange’s Multiplier)
71
ISI MSQE ME(I) PREVIOUS YEAR’SANSWER KEY
Logic:If the curve g(x, y) = 0 is closed and bounded then absolute maxima and minima of f(x, y) exist.
Here 𝑔(𝑥, 𝑦) = 𝑥 − 𝑦 = 0 is not bounded. So, no extremum exists.
Alternate way:
𝑓𝑥 = 𝑥 2 + 2 = 0 ⇒ 𝑥 = √2
𝑓𝑦 = −3𝑦 = 0 ⇒ 𝑦 = 0
So, 𝑥 ≠ 𝑦. So, f has neither maximum nor minimum.
(1 − 𝑝)3
⏟
27. (a) Probability =
𝑒𝑙𝑖𝑚𝑖𝑛𝑎𝑡𝑒𝑑 𝑖𝑛 1𝑠𝑡 𝑟𝑜𝑢𝑛𝑑
𝑝
⏟3(𝑛−1) (1 − 𝑝)3
𝑝3 (1 − 𝑝)3
⏟
+
+ 𝑝6 (1 − 𝑝)3 + ⋯ +
𝑒𝑙𝑖𝑚𝑖𝑛𝑎𝑡𝑒𝑑 𝑖𝑛 2𝑛𝑑 𝑟𝑜𝑢𝑛𝑑
+⋯
𝑒𝑙𝑖𝑚𝑖𝑛𝑎𝑡𝑒𝑑 𝑖𝑛 𝑛𝑡ℎ 𝑟𝑜𝑢𝑛𝑑
=
(1 − 𝑝)3
; |𝑝| < 1
1 − 𝑝3
28. (a)
𝑅𝑒𝑞𝑢𝑖𝑟𝑒𝑑 𝑃𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 =
24
2
2! 2! 2! 2!
=
=
120 15
5!
[Each couples 𝑎𝑟𝑒 𝑎𝑟𝑟𝑎𝑔𝑒𝑑 𝑖𝑛 2! ways and 2! For arrangements of husband & wife]
29. (a)
Take f(x, y) = 𝑥 − 𝑦
𝑓(𝑦, 𝑧) = 𝑦 − 𝑧
𝑓(𝑧, 𝑥) = 𝑧 − 𝑥
So, 𝑓(𝑥 𝑦) + 𝑓(𝑦, 𝑧) + 𝑓(𝑧, 𝑥) = 0
So, 𝑓(𝑥, 𝑦) − 𝑓(𝑥, 0) + 𝑓(𝑦, 0) = 𝑥 − 𝑦 − 𝑥 + 𝑦 = 0
72
ISI MSQE ME(I) PREVIOUS YEAR’SANSWER KEY
30. (c)
𝑎+
1 6
{
1
≥ 2 (𝐵𝑦 𝐴𝑀 ≥ 𝐺𝑀)
𝑎
1
(𝑥 + 𝑥) − (𝑥 6 + 𝑥 6 ) − 2
1 3
1
(𝑥 + 𝑥) + (𝑥 3 + 𝑥 3 )
So, minimum value of the expansion is 6 at 𝑥 = 1.
73
} ≥
26 − 2 − 2
=6
23 + 2
ISI MSQE PEB
(Economics)
SOLVED PAPERS
Solution to Sample Questions for ME II (Economics) 2004
Q.1. Solution:
(i) Here we need to find out optimal bundle in Perfect substitutes and then to apply formula for
own price elasticity. We know in case of perfect substitutes decrease in PX by 1% decreases QX
by 1%. And formula for own price elasticity for any good x is
So, here we have
(ii) I = px0 + y ; I – C = qx0 + y ;
Optimal choice means that keeping in consideration both the markets. The choice has been made
given. Smooth convex means strictly convex, i.e., a unique optimal. So, C = (p – q) x0 .
(iii) Production function is minimum form then cost function would be in Perfect Substitute
form. Since a minimum function, so here we have
.
We know Cost Function = wL + rK = w.aQ + r.Q2b = Q (wa + 2rb)
(iv) During a period net loan from abroad of an economy is positive. Then it will imply that
change in foreign exchange reserve is positive.
(v) We know Y = C + I + G = C + I + tY »
Differentiating w.r.t. t, we have
, So, Y increases.
(vi) Interest on public debt is a part of national income. So, answer is (C).
Q.2. Solution:
(i) FALSE. The price per gallon of gasoline divided by the price per barrel of crude oil is
dependent on crude oil production since a barrel of crude oil yields a fixed number of gallons of
gasoline.
(iii) TRUE. Utility function which include homogeneous function is called homothetic. If a
utility function is homothetic rate of commodity substitute will depend upon relative rather than
absolute commodity quantities. For example, taking
is not a homogeneous
function but it is homothetic since
. Due to the homothetic form it always passes
through origin.
Q.3. Hints:
One of the firm will go out of market because of incidence of fixed cost. Find out.
Q.5. Solution:
(i) Original form of Budget Line is M = PX .X + PY .Y
Here
(
)
which is the required budget line. Now,
if you plot XE and XO in X & Y axis then we will get a curve which will be concave to the origin.
Draw the curve yourself.
(ii) Put R = 0 and I = 1, then
{
Then the optimum consumption bundle is
.
(iii) Here
------(1)
Now we need to find the maximum value of R that the company can extract from the company.
Differentiate (1) w.r.t.
minima. Then the answer will be max{R} = 1 – ½ -
. Use the process of maxima &
.
Q.6. Justification:
Even if income increases then consumption will not increase after a certain point of time. This is
because the utility function is continuously differentiable and it follows diminishing marginal
utility. Hence after a certain amount of positive consumption of x and y, total utility will not
increase.
Solution to Sample Questions for ME II (Economics) 2005
Q.1. Solution:
(a) Yes, I do agree with this claim.
Justification : Pareto optimum is a position from which it is possible to make anyone better off
without making someone else worse off. That’s why here the allocation (
) is the only
Pareto Optimum Allocation. In other cases, the given condition to be Pareto optimum will be
violated.
(b) (i) Yes, the given transaction will be included in GDP. Since, Mr. Ramgopal is hiring Mr.
Gopi for works at his home, so it is clear that he is making expenditure for consumption purpose.
So, the corresponding expenditure is consumption expenditure.
(ii) Yes, the given transaction will be included in GDP. It is clear that he is making expenditure
for consumption purpose (buying a new Maruti Esteem) . So, the corresponding expenditure is
consumption expenditure.
(iii) Yes, the given transaction will be included in GDP. It is clear that he is making expenditure
for consumption purpose (flying to see Durga Puja) . So, the corresponding expenditure is
consumption expenditure.
(iv) No, the given transaction will not be included in GDP. Mr. Ramgopal is investing his money
to buy 100 stocks of Satyam Ltd., so this is clearly an Investment Expenditure.
(v) No, the given transaction will not be included in GDP. Mr. Ramgopal is investing his
moneyto build a house, so this is clearly an Investment Expenditure.
Q.2. Solution:
Here the market demand function for roses is given by P = α – Q , where Q is the number of
roses ; P is price of a rose.
(a) Full bloom roses should be sold on the same day. The seller should sell all 1000 full bloom
roses. The price per full bloom roses is P = α – 1000.
(b) Number of seller = 10 & each finds in his garden 100 roses in full bloom ready for sale on a
day. So, total full bloom roses in the market = 1000 = Number of roses sold in a day.
So, the equilibrium price per rose is P = α – Q = 1100 – 1000 = 100, since α
(c) 1000 roses again to be sold in that market as all 1000 roses are full bloom.
.
Competitive Price :- With the given information the competitive price would be α – 1000, we
α – 1000. This will be
know Average Revenue = P in the perfect competition. AR =
the competitive price & total number of roses sold on that day will remain unchanged.
Q.3. Solution:
Consider vegetable being single good x. Now when Laxmi's income is very low she can only
afford rice not veggies. This represents the case of quasi-linear preference b/w good x & rice that
is U(x,r)= v(r) +x.
Now When she can afford veggies she will only prefer to buy either potato or cabbage depending
on its price but she will never buy both so it can't represent perfect substitutability b/w veggies,
therefore the preferences she have amongst veggies is concave i.e. x=max{c,p}
The final utility function for luxmi becomes U(r,c,p)= v(r) + max{c,p}.
Explanation:Therefore max function represents the concave preference as when u exhaust income on both of
the veggies Laxmi will get comparatively lesser units as when Luxmy buy only 1 type of veggie
that is extremes are preferred over averages & also when prices of two veggies are same the
consumer will only buy only 1 type of veggie not both.
Q.4. Solution:
Here we are given the followings: income = y = C + I ----(i)
savings = s = k.y -------(ii)
Where, k is any constant. Also given that
&
. Now, we are to compute dy,
i.e., change in the equilibrium income. So, differentiating (ii) w.r.t y, we have k = 12. Then use
this value & equation (i) &
, you will get the value of dy.
Q.5. Solution:
(i) IS Equation : Y = a + bY + I0 + G0
 Y=
LM Equation : From money market equation condition, we have Ms = Md ,
So, LM equation is
.
(ii) When price level increases then real money supply
leftward. See the graph below:
. Then LM schedule shifts
Solution to Sample Questions for ME II (Economics) 2006
Q.11. Solution:
The production function is Y = √
form will be : Y +
√(
, after changes in Income, K, L as given in question the new
)(
) => 130Y = √
.
Solution to Sample Questions for ME II (Economics) 2007
Q.3. Solution:
(b)
I = A – 50r
(
) .....because
(1 – c )y = A – 50r + G .....Since
(
)
Thus y = 2.5A – 125r + 2.5G
Since an increase in government expenditure is found to crowd out 50 units of private
investment. This implies r has gone up by 1 as a result of increase in G. Solving IS-LM for r, we
get:
Putting
. Solving IS-LM for y, we get:
The above solution is not adjusted for the money supply change to rectify the crowding out: So
we need to increase the money supply such that rate of interest again fall back to its original
value, i.e., we need to do the following: we will move along the new IS curve in such a way that
r falls by 1 from its new value. Given the slope of the IS curve
We get,
So, the total change in income due to fiscal policy and corresponding monetary policy is
250+125=375.
Q.5. Solution:
(b) Let, Price of Sugar, PS = 2 paisa / 10 gms ; Price of Tea, PT = 50 paisa / 10 gms.
Money income = 55 Rs.
So, 55 = 2S + 5T
Budget constraint is : Sugar : Tea = 5 : 2
Utility function is given by U = min {
} which corresponds to the L-Shaped I.C. and
equilibrium occurs at the kink point.
At kink point we always have
Now the budget equation is 55 = 2S + 50T = 2S + 50( ) = 22S => S =
Therefore, she wants 1gm Tea & 2.5gms Sugar per month.
Q.7. Solution:
(i) Consumption = 0.8Y ; Investment = 100 + 0.4Y ; Net Exports = 100 – 0.3Y ;
So, Y = 0.8Y + 100 + 0.4Y + 100 – 0.3Y
Solving Y = 2000, Imports = 0.3 2000 = 600.
But 600 > 450, which is the limit.
Therefore, assuming all 450 will be imported, so
Y = 0.8Y + 100 + 0.4Y + 100 – 450
 Y = 1250.
 Y = 1.2Y – 250
Solution to Sample Questions for ME II (Economics) 2008
Q.2. Solution:
(a) Let y = log Y = log (UNα) = log U + αlogN = α(u – logα) + αlogN ------(*)
Also it is given that
=> d
Y=
.
Now integrating both sides, we have Y =
Implying log Y = (log W – log P) + log N
 y = (w – p) + log N
 log N = (y – w + p)
Putting value of log N in equation (*), we have
y = α(u – logα) + α(y – w + p)
 y (1 – α) = α(u – logα) – α(w – p)
 y=
(u – logα)
(w – p) ----------(1)
(b) Substituting y = m – p => p = m – y & w = θ p = θ(m – y) in equation (i), we have
y=
(u – logα)
 y
 y*
 y=
[θ(m – y) – m + y] =
y(θ – 1) =
+=
(u – logα)
[(m – y)(θ – 1)]
(u – logα + m – mθ)
(u – logα + m – mθ)
(u – logα + m – mθ) -------------(2)
(c) If aggregate demand increases then aggregate output also increases. So,
(i) If 0 < α < 1 & 0 < θ < 1, then from equation (2), y will increase, so monetary policy will be
effective.
(ii) If θ = 1, then y is positive, i.e., monetary policy will be effective.
Q.3. Solution:
Firm 1’s objective:
Subject to
0
Differentiating the objective we get =
When
40.
is given, if
at
If
at
If
at some
then the best response of firm 1 is
then the best response of firm 1 is 0.
then the best response of firm 1 is
To summarize, best response correspondence of firm 1 is:
{
Firm 2’s objective:
Subject to
0
Similar to above, best response correspondence of firm 2 is:
{
Put a = 200 , the Nash equilibrium is: (
Put a = 100 , the Nash equilibrium is: (
(
)
Q.4. Solution:
(a) Maximize U = √
Subject to : 100 = PX .X + PY.Y
Define, α = √
+ λ [ 100
PX .X + PY.Y ] --------(i)
Now, differentiating (i) w.r.t. X, we have
√
.
Also differentiating (i) w.r.t. Y, we have
√
.
So, finally, we have PX .X= PY .Y = 50.
Now, from cost equation C = wL + rk = wL , since rk = 0 ;
Assuming 0 savings, we have C = PX .X + PY.Y = 100 = wL.
Given that 10 units of labour are available, so w = 10 is the equilibrium wage in the competitive
market.
(b)
Q.6. Solution:
(a) Product market equation is given by: Y = C + I + G






YD + T = C + I + G
YD + 200 = 200 + .5 YD + 150 – 1000 r + 250
0.5 YD = 400 – 1000 r
0.5 Y – 100 = 400 – 1000 r
Y = 1000 – 2000 r
Y = 1000 – 2000 i ----------(1)
Money market equation is given by: ( )
( )
 2 Y – 4000 i = 1600
 Y = 800 + 2000 i -------------(2)
Solving (1) and (2), we have i = 0.05 is the rate of interest. And Y = Income = 900.
(b) Here nominal money supply is increased by 2%, i.e. ,( )
Now ( )
( )
 2 Y – 4000 i = 1632
 Y = 816 + 2000 i -----------(3)
Now solving (1) & (3), we have i = 0.046 & Y = Income = 908.
Q.9. Solution:
We have to find the total types of probable deletion that may occur. And here we have 16 types
of deletion.
So, if A starts the game, obviously B will delete the last box & will become looser.
Player A’s winning strategy is to start the game himself.
Now write down 16 probable deletion from the given box of 16 squares.
Q.10. Solution:
(i) (a) 100000 = PP . P + PM . M
= (50000 – 10000) + 1.M
= 40000 + M implying M = 60000
(b) It will remain same.
(c) He would be better off if he wants. He can now enjoy cash transferred by Govt. to him by
reducing petrol consumption or lowering the expenses.
(ii)
Mr. B earns Rs. 500 today and Rs. 500 tomorrow. He can save for future by investing today
in bonds that return tomorrow the principal plus the interest. He can also borrow from his
bank paying an interest. When the interest rates on both bank loans and bonds are 15%, Mr.
B chooses neither to save nor to borrow.
Plot the two budget sets, call the old budget set B and the new budget set B′.
Here B is
c2=500+1.15(500−c1) and B′ is
Note that B′⊂ B. If (c1∗,c2∗) is the best point of B and is still available in B′, it will also be the
best point of B′.
In our case, (500,500) is the best point of B that is also available in B′.
Solution to Sample Questions for ME II (Economics) 2009
Q.1. Solution:
(i) Y = C + ̅ + ̅ = C0 + C1YD + ̅ + ̅
Y = C0 + C1(Y – T) + ̅ + ̅ = C0 + C1Y – C1(t0+t1Y) + ̅ + ̅ ----(i)
» t1 =
(
̅
̅
)
̅
Here Y > 0, C0 > 0, 0 < C1 < 1, ̅
So, t1 < 1.
(ii) From (i), we get
So, numerator is greater than 0 and less than
Y [1 – C1 (1 – t1)] = C0 – C1t0 + ̅ + ̅
» Y* = (C0 – C1t0 + ̅ + ̅ ) [1 – C1 (1 – t1)]
(iii) To get tax rate multiplier, we differentiate (i) w.r.t G, we have
(
»
(
)
)
̅
is the multiplier.
Second Part: YES, Because C0, ̅ ̅ if change then all will increase or
accelerate the income Y.
Q.2. Solution:
(i) c0 =
; c1 =
(
)
(ii) As r increases c0 stays the same & c1 increases. This is clear from the answer (i)
(iii) If she consumes the same amount in both periods then c0 = c1
»(
)
Q.3. Solution:
(i)
(ii) Here you just need to show that min(2𝛼x1, 𝛼x1+ 𝛼x2) = 𝛼min(2x1, x1+ x2) for 𝛼 > 0.
(iii) If input prices are equal then we will get C equals to Q.
Q.4. Solution:
(i) You need to write entire Edgeworth Box.
(ii) 45 line from top left to bottom right in an edgeworth box with price ratio equal to 1.
(iii) Yes, perfectly competetive outcomes are pareto optimal. This is necessary condition to hold
pareto optimality because existance of perfect competetive market does not guarantee to hold
pareto optimality. If any other conditions to hold pareto optimality violates then pareto optimal
does not hold in all exchange economics, according to the theory of second best.
No, it does not generally hold in in all economics.
Q.5. Solution:
(i) Ans is : p =
; q1 =
; q2 =
.
The restriction is given since this is the case of dumping in market 2, the monopolist faces
perfectly competetive situation so here it is assumed that price can’t exceed more than 20.
(ii) p1 = 7.5 ; q1 = 2.5 ; p2 =
; q2 =
.
(iii) You just need to compare the profits here.
(iv) CS(i) = *(
)
(
) + ; CS(ii) = *
(
) +;
Q.6. Solution:
(i) Answer is q1 = (
)
(ii) q1 = q2 = q3 = 30.
(iii) Case 1: If Firm 2 and 3 merge. Then q1 + q2 + q3 = 40, so Firm 1 is better off and Firm 2 and
3 are worse off.
Case 2: If All 3 Firms merge. Then all three Firms are better off.
Q.8. Solution:
(i) (1+a)yi – T
2yi for j
{H,L}
(ii) High ability per go to college for T
100. Low ability per go to college for T
80.
(iii) Both High & Low ability person will attain education.
(iv) Assuming tuition fee paid by H-Type is 100 and L-Type is 80. Total subsidy is (100 –
60) 5 + (80 – 60) 5 = 300. Now for total tax receipts equal subsidy, the equation is
5(
)(
)
» x = 23.08%
Q.10. Solution:
(i) Lowest marks for which it should admit the 1st applicant = 50
(ii) Lowest marks for which it should admit the 1st applicant = 62.5
Lowest marks for which it should admit the 2nd applicant = 50
Solution to Sample Questions for ME II (Economics) 2010
Q.2. Solution:
(a) Given C(q) = mq ; P1 = a – q1 – sq2 ; P2 = a – q2 – sq1 .
Define, п1 = aq1 – q12 – sq1q2 – mq1 – mq2 –(1) & п2 = aq2 – q22 – sq1q2 – mq1 – mq2 –(2)
Differentiating (1) w.r.t. q1 & (2) w.r.t. q2 & equating with ‘0’, we have
–
–
–
–
–
–
–
–
–
–
-------(1)
-------(2)
From (1) & (2), we have respectively,
4q1 – s2q1 = (2 – s)(a – m)
&
4q2 – s2q2 = (2 – s)(a – m)
So, therefore, we have the Cournot equilibrium quantities of two firms as :
q1* = q2* =
.
Q.3. Solution:
(a) RA = PAQA = 15QA – QA2
and
RB = PBQB = 15QB – QB2
Total cost function is given by: TC = 5 + 3(QA + QB)
Therefore, Profit (п) = RA + RB – TC = 15QA – QA2 + 15QB – QB2 – 5 – 3(QA + QB)
QA* = 6
&
Now, we need to put the values of QA* & QB* in the above cases.
So, Total output = Q* = QA* + QB* = 11.5
Total Cost = TC = 5 + 3(6 + 5.5) = 39.5
Profit = п = 19.5
QB* = 5.5 ;
Q.7. Solution:
(a) TRUE. In first case, the consumer is an irrational consumer & in the second case the
consumer is a rational consumer. So, arguments are same but applied to two different types of
consumers.
(b) FALSE. When price will increases he must want to sell crops but here he is consuming the
crops when price increases. So, the crop is a Giffen good here.
(c) TRUE. Of course he will choose to work more as his wage rate will increases.
(d) FALSE. Stipends of ISI is given to students to fulfill their monthly expenses. So, this will be
included in National Income through Expenditure method but will not be included in GDP.
Solution to Sample Questions for ME II (Economics) 2011
Q.1. Solution:
Assuming zero cost of production.
(i)
(ii)
(iii)
For price of X = 3, price of Y = 4, net profit to monopolist = 12
Price of bundle = 4, net profit = 12
Price of X = 4, price of Y = 4, price of bundle = 6, net profit = 14
Thus, (iii) is the optimal pricing strategy.
Q.3. Solution:
Utility maximization problem is
Q.7. Solution:
T = 49
Solution to Sample Questions for ME II (Economics) 2012
Q.1. Solution:
In the short run, k is fixed as k* so the only decision the firm needs to make is how much labour
to employ. Whatever labour the firm employs, the production function is
and since capital is fixed as k*,
times labour which is
. Now cost in the short run is wage rate
.
In the long run, to maximize profits, set marginal derivative of profit w.r.t. labour = 0, which
gives
Similarly, set marginal derivative of profit w.r.t. to capital = 0, which gives
Derive the two equations to get K = L.
Now,
, and L = K = Y2
So, cost = wL + rK = 20 Y2
Q.2. Solution:
IS Equation is given by : Y = C + I + G + (X – M)
LM Equation is given by : (m/p) = L(Y, i).
For the given system, IS equation is :
Y = 200 + 0.5YD + 150 + 0.25Y – 1000i + 250
= 200 + 0.5(Y – 200) + 0.25Y – 1000i + 400
= 0.75Y + 500 – 1000i
» 0.25Y = 500 – 1000i
» Y = 2000 – 4000i -----------------(1)
For the given system, LM equation is:
For money market to be in equilibrium, (m/p) = (m/p)d
» 1600 = 2Y – 8000i
» Y = 800 + 4000i -----------------(2)
Solving (1) and (2), we have i =
% . Now put this value in given equations.
Equilibrium Income(Y) = 1400 , Investment(I) is 650.
Second Part: Put (m/p) = 1840 in the above equation & solve the variables again.
Third Part: Now you have an equation of investment depending upon the money supply.
Differentiate the equation w.r.t. money supply & the result. This derivative will be the short run
effect of expansionary monetary policy on investment in the short run. (Do the steps yourself)
Q.3. Solution:
Our problem is to maximize utility function subject to the budget constraint.
s.t. M = Px + Py
Here price of both goods are equal = P, and 𝛼> 0.
Using Lagrange’s Method of Maximization as follows:
L=
[
]
, where
is Lagrange multiplier.
For points of extremum Lx = 𝛼
Ly = 𝛼
L =
Differentiate partially again we will get the component of the matrix D.
Now note that D = [
] , and see ∣D∣> 0 (check yourself)
From (i) and (ii) we get x = y . Substituting the value in (iii) we get y* =
= x*
Hence we get demand function of x and y.
Second Part : Assume Px = 2Py ,
Where Px = price of x & Py = price of y.
Rest is similar as done in above.
Q.5. Solution:
Here is the plot of the data provided in the problem.
The answer to the questions is yes. There exist preferences that will choose exactly what A
has chosen from A's budget set and what B has chosen from B's budget set. Here are couple
of examples to demonstrate that:
Q.8. Solution:
Solution to Sample Questions for ME II (Economics) 2013
Q.1. Solution:
Agent’s utility maximization problem is the following :
s.t. (
)
(
)
&
(a) Solving the above problem we get :
(
Hence, saving = w -
)
(
(
)
)
.
(b) Clearly, Savings doesn’t change in rate of interest rate r .
Q.2. Solution:
(a) If the price of a MD is Rs. 20 and the marginal cost is Rs. 15 per MD , vendor’s profit
maximization problem is the following :
s.t.
Thus, each vendor would want to sell 100 MD a day .
(b) Given competitive behavior, free entry-exit from the industry and constant returns to scale
technology , we have zero profit condition.
That is , price equals marginal cost. Thus, demand is
d(15) = 4400 – 120(15) = 2600
Since each vendor sells 100 units and demand is 2600 units , there are 26 vendors selling MD in
the market .
(c) If number of vendors are 20 and each vendor produces 100 units , price is given by
p=
(d) The maximum price that a vendor is willing to pay for the permit is equal to the profit that a
vendor gets if he operates i.e. 5×100=500.
Q.3. Solution:
The production possibility frontier of the two inputs is given by
Since final product can be sold at the end of the day at a per unit price of Rs. 1.
The firm’s profit maximization problem is :
√
s.t.
&
Thus, firm will hire the worker, produces
√ .
Q.4. Solution:
Since monopolist’s sale to the government is positive, his marginal revenue at the point of sale in
the private market must be Rs. 100 . Now price in the private market is Rs. 150. We can compute
the price elasticity of demand in the following way :
TR(x) = p(x) .x
Differentiating TR(x) w.r.t. x , we get ,
MR(x) = p(x) + x
( )
= p(x) + ( )
( )
( )
= p(x)(
)
Now substituting p(x) = 150 and MR(x) = 100 in the above we get elasticity, η= - 3 .
Q.5. Solution:
(a) Let f (K,L) denotes the production function.
(
)
( ) ( )
(
(
).
Thus, production function satisfies constant returns to scale.
(b) Profit maximization problem of the competitive producer is
L≥0,K ≥0
s.t.
The above problem is equivalent to
(
s.t.
Clearly, when 0 ≤ w ≤
)
(
)
L≥0,K ≥0
, demand function for labor is not defined.
(c) Also, the demand function for capital is defined when price of capital service is zero
provided w >
Q.6. Solution:
Suppose the production function has changed from
where a > 0. Labor demand curve is, therefore,
* +
( )
Where
{
is the real wage and it changes to
( )
* +
{
( )
( )
(
)
Let us assume that Labor supply is exogenously given and is equal to ̅ . Solving for the equilibrium in
labor market we get that the real wage has changed from a to 2a but the equilibrium employment is ̅ in
̅
̅ . Given any
both cases. Therefore, Aggregate Supply curve has shifted from
( ) it is easy to see that the new equilibrium level of output in the
aggregate demand curve,
model will be twice as much as it was earlier.
Q.7. Solution:
E(Planned Expenditure) = C* + c(Y – T) – (I* – bi) + G* + NX(other than investment) .
Let NX = Export – Import = Export – mY , where m = marginal propensity to import.
E = (C* + G* – I* – cT + Export) + cY – mY
In equilibrium E = Y
Here Y( 1 – (c – m) ) = Autonomous component of expenditure
(
(
))
Assuming interest rate to be constant so when I* is positive output will decrease by
(
)) , where
= Change in imports.
(
So, the given statement is TRUE.
Alternative Answer: In the sample Keynesian Model, we consider a static set up where Goods market
equilibrium condition in the open economy is
( )
If the entire investment good is supplied from import the net demand is 0 and therefore the multiplier is
0. Therefore, a positive investment multiplier does not exist in an open economy simple Keynesian
model when the entire amount of investment good is supplied from import.
Q.8. Solution:
Given that income elasticity of demand for all goods is positive, i.e. (
that (
)
Hence, goods are normal.
)( )
. This implies
Q.9. Solution:
Given that the money income is
demand is
. Demand for X when price
is
is 14. Demand for X when price is
is 16. In order to find the Substitution
effect and Income Effect, we first need to find that what must be his income so that he can
afford the original equilibrium at the new price. Original equilibrium was (14, 78) where 78
( )). At the new price we need to
is his remaining income after spending on (
( )
give him income
so that he can exactly afford his original
consumption bundle at the new prices. We will now find the demand for X in this situation
when
and price is 2, and we get
effect is
and income effect is
. Therefore, Substitution
.
Q.10. Solution:
(a) In the Solow model, where the entire income is consumed, there will be no capital
formation and in the presence of depreciation capital depletes over time. If the population
is either fixed or grows over time then there will be fall in capital by labor ratio over time.
Therefore wage rate would fall and rental rate would increase over time.
(b) Yes, this economy will attain a steady state at
which is disappointing but steady.
Solution to Sample Questions for ME II (Economics) 2014
Q.1. Solution:
(a) Firm choose quantity to sell domestically,
profit maximization problem:
(
, such that they solve the following
)
(
)
Let us denote the profits by . To solve the above problem, we will first differentiate the objective ( )
with respect to
, to get expression for marginal profits:
Since marginal profits from exports are always positive, the profit maximizing monopolist will always
exhaust his capacity in the optimum. Therefore, the firm will choose
in such a way that
Therefore,
.
(b) If marginal cost is 6, the profit maximization problem is
(
)
(
)
Let us denote the profits by . To solve the above problem, we will first differentiate the objective ( )
with respect to
to get expression for the marginal profits.
Since marginal profits from exports are always negative, the profit maximizing monopolist will not
export anything. The quantity that this firm sells in the domestic market is determined by equating its
marginal profits to zero. Therefore,
.
Q.2. Solution:
(a) (i) Here is the utility maximization problem
,(
)
(
) -
Since prices are equal, we can just compare the marginal utilities and solve for the equilibrium.
)
When we equate the marginal utilities to 0, we get the unconstrained optimum (
this bundle also satisfy our budget constraint, therefore this is the equilibrium bundle.
(ii)
,(
)
(
(
). Since
) -
Since prices are equal, we can just compare the marginal utilities and solve for the equilibrium.
In this case, the bliss point (10, 10) is not affordable. Since
the equilibrium bundle satisfy
Therefore,
.
for all a<10 and likewise for
,
(b) (i) To find the deadweight loss due to monopoly, we will first find the competitive equilibrium and
monopoly equilibrium. Competitive equilibrium can be obtained by
condition which gives us
. Monopoly equilibrium can be obtained by
condition which gives us
. In case of competition, total surplus in the economy (Area of
in the graph
below) is the sum of consumer surplus (Area of
in the graph below) and producer surplus (Area of
in the graph below) when
. In case of monopoly, total surplus in the
economy (Area of the trapezium AEDC in the graph below) is the sum of consumer surplus (Area of
in the graph below) and producer surplus (Area of trapezium AEDG in the graph below) when
and
. Deadweight loss is the loss in surplus due to monopoly which is the Area of
in the graph below. Its value is 100.
(ii) Government can achieve minimum deadweight loss by setting a price ceiling of 30. This will result in
the equilibrium which is same as the competitive equilibrium and hence zero deadweight loss.
Here is the graphical demonstration of monopoly equilibrium.
Fig: (i) Monopoly
Fig: (ii) Monopoly with price ceiling
Q.3. Solution:
(a) (i) Profit maximization problem of the cinema hall is
(
(
Or equivalently,
)
)
(
(
)
)
(
)
(
)
(ii) Profit maximization problem of the cinema hall with the preference swap is:
(
)
(
)
(b) Trader utility function is given by
* + indicates consumers chicken or
not and
is his money balance. If the price of chicken is p, then individual will choose to buy one unit
of chicken if his utility from consuming chicken exceeds his utility from not consuming it i.e.,
(
)
(
)
This is equivalent to saying that consumes chicken if his valuation for chicken exceeds the price i.e.,
. Also, it is given that total supply of chicken is 6 units. To check for demand side, we will plot
trader’s valuation.
Let us consider the price of chicken
. In the graph below only trader 1 has valuation higher that
9.5 and therefore, at this price the demand is just one unit. So, this cannot be an equilibrium price
because supply exceeds demand.
When we reduce the price further to
equilibrium price.
, the demand−supply match. Therefore, this is an
Note that this is not the only equilibrium price. Any price
Let us now consider a lower price, say
demand is still short of supply.
When we reduce the price further to
equilibrium price.
,
- can be the equilibrium price.
. At this price, there are three buyers and therefore
, the demand−supply match. Therefore, this is an
Q.4. Answer Key:
(a) (i) K=24,So q=8 and p=12; (ii) K =18, So q=6 and p=14
Q.5. Solution:
(a) Equilibrium in this market consists of (
firm 1 given the firm 2’s choice of output :
) such that
(
)
solves the profit maximization problem of
solves the profit maximization problem of firm 2 given the firm 1’s choice of output
(
:
)
To find the equilibrium we will first solve the following problem of firm 1 at every possible choice for
firm 2’s output :
(
)
Solution to the above problem will be a function of q2 and is also known as best response function. The
standard methodology gives us the following as the best response function of firm 1:
By symmetry, we will get the best response of firm 2 as:
Solving the above system of equations, we will get the equilibrium as (
(b) Equilibrium in this market consists of (
firm 1 given the firm 2’s choice of output :
) such that
(
)
)
(
).
solves the profit maximization problem of
solves the profit maximization problem of firm 2 given the firm 1’s choice of output
(
:
)
We will first find the best response function of firm 1 by solving the following problem at every level of
output of firm 2, :
(
)
We will get
{
By symmetry, we will get the best response of firm 2 as
{
Solving the above system of equations, we will get the equilibrium as (
)
(
Q.6. Answer Key:
(a) (ii) Yes
(iii) If it changes then less housing and more of other goods will be consumed.
(b) (i) profit maximizing q=25, p=75
(ii) industry output =90, price =10 and number of firms =18.
Explanation:
).
Q.7. Solution:
(a)
Let us represent per−capita capital at time t by the variable
we have
, and per capita output at time t by
, so
To solve for the steady state, we will divide the law of motion of capital by N to get
(
)
This gives us law of motion of capital in per capita terms,
(
Using
)
, we can re−write the law of motion of capital as
(
)
Solving for the steady state per capita capital, we get
. /
And the per capita output in the steady state will be
. /
(b)
To find the extent of change in steady state output per worker, we will differentiate y with respect to s,
. /
(c) To find the savings rate that will maximize the steady state consumption per worker, we will
maximize the following:
(
Solving it, we will get the optimal savings rate is
)
. /
.
Q. 10. Solution:
(a)
(
(
)
(
̅)
)
̅ ( )
̅
The first –order conditions that the solution will satisfy are
(
(
)
̅)
(
(
̅)
̅ ( )
(
̅)
̅
)
,
-
̅ ( ) ,
-
,
(b) We can conclude from the first order condition ̅ ( )
wage and decreases with the interest rate.
-
(
̅ ) that optimal e increases with
ISI MSQE PEB (Economics) 2015 Solutions
Q.1. Solution:
) is a concave function and hence is also quasi−concave, therefore
(a) Check yourself that (
solution to the above problem can be obtained through the standard slope analysis.
And the budget line is
. Solution is not at the corner
because at this corner,
Therefore consumer will benefit from spending some money on . Solution is not at the other corner
because
at this consumption level, therefore, it is beneficial to spend less on .
Hence, the solution set is in the interior and satisfy
, and we get
.
(b) For the new problem, solution is not at the corner
because
Therefore, it pays to move some money to . Solution is not at the other corner
at this consumption level, therefore it pays to move money to
Therefore, the solution satisfy
And we get
and
.
.
(c) In order to find the extra income, we just need to find the x1 at which the individual will attain the
same level of satisfaction as in (a) at the prices specified in part (b). Given that the price ratio is ¼,
. To find x1, we will solve the following for x1:
( )
(
)
(
)
( )
(
)
(we equate the satisfaction level in (a) to the satisfaction level from a bundle in which
). Thus,
. Now we find the income needed to afford
at prices (1, 4) and we get
33.5 which is 13.5 higher than his current income. So, the compensation needed is 13.5.
Q.2. Solution:
(a) Equilibrium bribe rate per connection will be 0.6 and the net social surplus will be the consumer
surplus on consumption of 0.4 units plus the bribe minus the marginal cost that equals 0.08 + (0.6 – 0.5)
0.4 0.12.
(b) With privatization, equilibrium quantity will be ¼ and the equilibrium price will be ¾. Social surplus,
monopoly profits plus consumer surplus, is equal to 3/32.
(c) For 0 < c < ½, monopoly quantity is (1−c)/2 and the monopoly price is (1+c)/2. Consumer surplus (in
case of monopoly) as a function of c equals (
) . Range of values of c for which privatization
increases consumers’ surplus satisfy
(
)
> 0.08 that gives us 0 < c < 0.2.
Q.3. Solution:
(a) An individual located at distance
from inside and outside if (
̂ from the center of B will be indifferent between buying G
̂ or equivalently, consumer located ̂
)
distance inside
the boundary is indifferent between buying from inside and outside the region. Therefore, proportion of
people who will buy from outside the circular region equals
̂)
(
̂)
(
(b) Given the tax rate t, social welfare of B is given by the sum of welfare of people buying from outside
the state plus the welfare of the people buying G from inside the state and the Government tax revenue.
̂
∫
(
)
(
)
(
̂)
(
)
̂
Differentiating it with respect to t, we get the first order condition
(
̂)
(
̂)
(
̂)
(
̂
)
The above holds when
̂
(
)
̂
(
)
̂
Therefore, the social welfare maximizing tax rate is
.
Now to find the tax revenue maximizing tax rate, we will first write the expression for Tax−Revenue:
(
̂)
(
̂
̂)
[
̂
]
Finding the revenue maximizing t is equivalent to finding the revenue maximizing ̂ . Maximizing above,
we get
̂
And hence the revenue maximizing tax rate
Therefore, necessary and sufficient condition for the revenue maximizing tax rate to be the same as
welfare maximizing tax rate is
.
(c) From (b), we know that revenue maximizing tax rate is
And the revenue is
(
* +)
Therefore, the optimal revenue is
Therefore, the elasticity of tax revenue with respect to p is 0.
Q.4. Solution:
(a) Given the inverse demand functions:
We can solve for the best response function of the firm 1 by maximizing:
And of firm 2 by maximizing,
So we get
( )
{
}
( )
{
}
If both firms produce quantities then the output of the two firms solves
We solve the above system to get
(b)
in equilibrium if and only if
(c) When
model.
, then this model reduce to the standard Courhot
In the standard case, the two firms produce homogeneous and indistinguishable goods, which
would mean the inverse demand function should be same.
and
Q.5. Solution:
Firm solves the following profit maximization problem:
) since it is a monopolist as well as monopsonist. By eliminating Q
The firm choose(
and P using the demand constraints and the production function, the above problem can be rewritten
as:
)(
(
)
Now we will use the labor supply equations to write the above problem just in terms of input prices:
(
)(
)
FOCs:
(
)
(
)
(
)
(
(
)
(
)
)
Above can be rewritten as
(
(
)
)
(
(
)
)
(
)
(
)
Above can be rewritten as
(
)
(
)
(
)
(
)
(
)
(
)
Again, above can be rewritten as
(
)
(
)
(
)
(
)
(
)
(
)
Dividing them we get
(
(
Using
)
)
, we get
Q.6. Solution:
Supply of output (Y) and demand for input (L):
(a) Firm choose (Y, L) by maximizing profit
PY – WL
Subject to the constraint Y
( )
Private demand for output (c), demand for money balances (M) and supply of input (L);
Household choose (C, M, L) by maximizing
U
– ( )
Subject to the constraint
PC + M
̅–
Government’s demand for the final good is given by (G), and it must satisfy the budget
constraint;
̅
M
–
Private demand for the final good + Govt. demand for it
Y
supply of the final good;
C+G
Demand Supply in input market and money market; also, market clearing conditions in labor
market and money market tells us that the equilibrium labor employment must be the solution of
both the household’s problem and the firm’s problem and equilibrium level of money balances
must solve the household’s utility maximization problem and satisfy the Govt’s Budget
constraint.
Therefore, given the exogenous variables (G, ̅ ), equilibrium of the above economy consists of
prices (W, P) and the endogenous variables (Y, C, M, T, , L) such that the above mentioned
holds, i.e., (Y, L) solves the firm’s problem given (W, P), (C, M, L) solves the household’s
problem given (W, P) and T must satisfy the budget constraint of the government.
Finally, (W, P) must be such that demand equals supply holds in all the markets.
Now, we will write the conditions that the equilibrium prices (W, P) and the equilibrium vector
(Y, C, M, T, , L) must satisfy :
From firm’s profit maximization problem:
Y F(L)
F’ (L)
PY – WL
From household’s utility maximization problem:
PC + M
M
WL +
̅ – PT
PC
d’(L)
And we have the government’s budget constraint:
M
̅
PG – PT
Finally, the market clearing condition Y
C+G
Market clearing conditions for the money market and labor market are implicit in above since we
denoted labor demand and labor supply by the same variable L and money demand and money
supply by the same variable M.
We will reduce the above system of conditions by using the household’s optimization condition
& substituting M PC everywhere in the system:
Y F(L)
F’(L)
PY – WL
FC + PC
WL +
̅ – PT
d’(L)
PC
̅
Y
C+G
– PT
Now, we will eliminate T from the system by substituting PT
govt.’s budget constraint)
Y F(L)
F’(L)
–
WL + – PG
PC
d’(L)
Y
C+G
Next, we will eliminate
–
Y F(L)
F’(L)
d’(L)
by substituting it with
everywhere, we will then reduce the system to
PC + ̅ + PC (using the
Y
C+G
Now we eliminate
Y
F’(L) everywhere
F(L)
( )
d’(L)
Y
by substituting
C+G
(b) Differentiating the above system with respect to G ,
( )
( )
d’’(L)
( )
–
+1
Eliminating
, we get
( )
( )
( )
Now solving for
(
(
( ))
, we get
( ))
( )
( )
(0, 1)
The above follows from F’(L) > 0,
F’’(L) < 0 and d’’(L) >0. Since
– 1 and we get
Q.7. Solution:
Fundamental differential equation of Solow model :
( )
( + n)k
–
( )
(
)
In the steady state
0. If Sf(K) > ( + n)k, then k increases.
Let us plot Sf(k) – (
)k when
(
f(k)
So, we will plot
(
)(
)(
)(
)
We can easily see that one steady state is
The other steady states (where k
g(k) :
(
)(
and observing where g(k)
)(
)(
(
).
) .
0.
0) can be obtained by plotting
)
(
)
0
Figure: Plot of g
The above graph shows that there are 3 more steady states.
For n and
sufficiently small, there are
and
. For K <
, g(K) < 0
And therefore K < 0, thus Sf(K) < ( + n)K, and K decreases. Therefore
is not locally stable.
Similarly, we check for
and
and we will find that
is locally stable but
is not. Note
that
is locally stable steady state because when K is +ve and close to 0, then < 0.
CONCLUSION: Four steady states: :
. Two of them are locally stable:
and
.
Q.8. Solution:
(a) In this model, the law of motion of capital is given by
(
( )
( )
( )
Using ( )
motion of capital as follows:
( )
)
(
) ( )
(
( )
( )
) ( )
)
) ( )
) ( )
) ( )
(
(
(
( )
(
(
In the steady state,
)
(
) ( ), we can rewrite the law of
( )
( )
) ( )
(
(
(
( )
( )
) ) ( )
(
) ) ( ( ))
( ). Therefore, K* satisfy
(
) )
(
)
on K*, we will differentiate the above expression with respect to :
(
Assuming (
(
(
(
To find the effect of higher
( )
(
) )
(
)
(
)
) )
(b) Yes, a higher lead to a lower value of the capital stock in every period (i.e., along the entire
transition path). Let
the law of motion of capital is
(
)
(
) ( )
(
(
) ) ( ( ))
)
(
(
) ) (
And for ’ the law of motion of capital is
(
Starting from ( )
K’(1).
)
(
( )
( ), we need to show that K(t) > K’(t)
( )
( ))
. We will first show that K(1) >
(
) ( )
(
(
) ) ( ( ))
(
) ( )
(
(
) ) ( ( ))
(
)
(
(
) ) (
( )
( )
( ))
Suppose, by the induction procedure, K(t) > K’(t) we will show that K(t+1) > K’(t+1)
(
)
(
) ( )
(
(
) ) ( ( ))
(
) ( )
(
(
) ) ( ( ))
(
)
(
(
) ) (
(
Therefore, a higher
( )
)
lead to a lower value of the capital stock in every period.
( ))
ISI MSQE 2016 PEB SOLUTION
Q.2. Solution:
Monopolist solves the following profit maximization problem:
(
Solution (
)
(
)
) to the above problem satisfy the following condition:
(
)
So,
Here is the graphical demonstration of equilibrium and surplus:
Q.3. Solution:
For (1) Budget is
Here is the plot of the budget:
and the optimal consumption is (E,B)=(0.5,0.5). Budget is blue and indifference curves (IC)
are red in color. The highest possible IC is through (E,B)=(0.5,0.5) on the budget line. See
graph below:
Q.4. Solution:
Q.5. Solution:
Q.7. Solution:
Q.8. Solution:
Q.9. Solution:
(i) (
(ii) (
)
(
)
)
(
(iii) Constant Return to Scale.
*
+)
Q.10. Solution:
2017
Booklet No.
Questions: 30
TEST CODE: PEA
Forenoon
Time: 2 hours
• On the answer booklet write your Name, Registration number,
Test Centre, Test Code and the Number of this Booklet in the
appropriate places on the Answer-sheet.
• This test has 30 questions. ANSWER ALL QUESTIONS.
All questions carry equal (4) marks.
• For each of the 30 questions, there are four suggested answers.
Only one of the suggested answers is correct. You will have to
identify the correct answer to get full credit for that question.
Indicate your choice of the correct answer by darkening the appropriate oval completely on the answer-sheet.
• You will get:
4 marks for each correctly answered question,
0 marks for each incorrectly answered question, and
1 mark for each unanswered question.
1. The dimension of the space spanned by the vectors (−1, 0, 1, 2),
(−2, −1, 0, 1), (−3, 2, 0, 1) and (0, 0, −1, 1) is
A. 1
B. 2
C. 3
D. 4.
2. How many onto functions are there from a set A with m > 2 elements to a set B with 2 elements?
A. 2m
B. 2m − 1
C. 2m−1 − 2
D. 2m − 2.
3. The function f : R2+ → R given by f (x, y) = xy is
A. quasiconcave and concave
B. concave but not quasiconcave
C. quasiconcave but not concave
D. none of the above.
4. The function f : R2+ → R given by f (x, y) = xy is
A. homogeneous of degree 0
B. homogeneous of degree 1
C. homogeneous of degree 2
D. not homothetic.
5. You have n observations on rainfall in centimeters (cm) at a certain
location, denoted by x, and you calculate the standard deviation,
variance, and coefficient of variation (CV). Now, if instead, you
were given the same observations measured in millimeters (mm),
then
1
A. the standard deviation and CV would increase by a factor
of 10, and the variance by a factor of 100
B. the standard deviation would increase by a factor of 10,
the variance by a factor of 100, and the CV would be
unchanged
C. the standard deviation would increase by a factor of 10,
and the variance and CV by a factor of 100
D. none of the above.
6. You have n observations on rainfall in centimeters (cm) at two locations, denoted by x and y respectively, and you calculate the
covariance, correlation coefficient r, and the slope coefficient b of
the regression of y on x. Now, if instead, you were given the same
observations measured in millimeters (mm), then
A. the covariance would increase by a factor of 10, b by a
factor of 100, and r would be unchanged
B. the covariance and b would increase by a factor of 100,
and r would be unchanged
C. the covariance would increase by a factor of 100, and b
and r would be unchanged
D. none of the above.
7. Let 0 < p < 100. Any solution (x∗ , y ∗ ) of the constrained maximization problem
)
(
−1
+y
max
x,y
x
subject to
px + y ≤ 10,
x, y ≥ 0,
must satisfy
A. y ∗ = 10 − p
B. x∗ = 10/p
√
C. x∗ = 1/ p
2
D. none of the above.
8. Suppose the matrix equation Ax = b has no solution, where A is a
3 × 3 non-zero matrix of real numbers and b is an 3 × 1 vector of
real numbers. Then,
A. The set of vectors x for which Ax = 0 is a plane.
B. The set of vectors x for which Ax = 0 is a line.
C. The rank of A is 3.
D. Ax = 0 has a non-zero solution.
9. k people get off a plane and walk into a hall where they are assigned to at most n queues. The number of ways in which this can
be done is
A. Ckn
B. Pkn
C. nk k!
D. n(n + 1) . . . (n + k − 1).
10. If P r(A) = P r(B) = p, then P r(A ∩ B) must be
A. greater than p2
B. equal to p2
C. less than or equal to p2
D. none of the above.
11. If P r(Ac ) = α and P r(B c ) = β, (where Ac denotes the event ‘not
A’), then P r(A ∩ B) must be
A. 1 − αβ,
B. (1 − α)(1 − β)
C. greater than or equal to 1 − α − β
D. none of the above.
12. The density function of a normal distribution with mean µ and
standard deviation σ has inflection points at
A. µ
B. µ − σ, µ + σ
3
C. µ − 2σ, µ + 2σ
D. nowhere.
13. In how many ways can five objects be placed in a row if two of
them cannot be placed next to each other?
A. 36
B. 60
C. 72
D. 24.
14. Suppose x = 0 is the only solution to the matrix equation Ax = 0
where A is m × n, x is n × 1, and 0 is m × 1. Then, of the two
statements (i) The rank of A is n, and (ii) m ≥ n,
A. Only (i) must be true
B. Only (ii) must be true
C. Both (i) and (ii) must be true
D. Neither (i) nor (ii) has to be true.
15. Mr A is selling raffle tickets which cost 1 rupee per ticket. In the
queue for tickets, there are n people. One of them has only a 2rupee coin while all the rest have 1-rupee coins. Each person in the
queue wants to buy exactly one ticket and each arrangement in the
queue is equally likely to occur. Initially, Mr A has no coins and
enough tickets for everyone in the queue. He stops selling tickets as
soon as he is unable to give the required change. The probability
that he can sell tickets to all people in the queue is:
A.
B.
C.
D.
n−2
n
1
n
n−1
.
n
n−1
.
n+1
16. Out of 800 families with five children each, how many families would
you expect to have either 2 or 3 boys? Assume equal probabilities
for boys and girls.
A. 400
B. 450
4
C. 500
D. 550
17. The function f : R → R given by
{
x
,
f (x) = |x|
1,
if x ̸= 0,
if x = 0.
is
A. concave
B. convex
C. neither concave nor convex
D. both concave and convex
2
n +1
18. As n → ∞, the sequence { 2n
2 +3 }
A. diverges
B. converges to 1/3
C. converges to 1/2
D. neither converges nor diverges.
19. The function x1/3 is
A. differentiable at x = 0
B. continuous at x = 0
C. concave
D. none of the above.
20. The function sin(log x), where x > 0
A. is increasing
B. is bounded and converges to a real number as x → ∞
C. is bounded but does not converge as x → ∞
D. none of the above.
21. For any two functions f1 : [0, 1] → R and f2 : [0, 1] → R, define the
function g : [0, 1] → R as g(x) = max(f1 (x), f2 (x)) for all x ∈ [0, 1].
A. If f1 and f2 are linear, then g is linear
5
B. If f1 and f2 are differentiable, then g is differentiable
C. If f1 and f2 are convex, then g is convex
D. None of the above
22. Let f : R → R be the function
f (x) = x3 − 3x ∀ x ∈ R.
Find the maximum value of f (x) on the set of real numbers x
satisfying x4 + 36 ≤ 13x2 .
A.
B.
C.
D.
18
−2
2
52
23. A monkey is sitting on 0 on the real line in period 0. In every
period t ∈ {0, 1, 2, . . .}, it moves 1 to the right with probability p
and 1 to the left with probability 1 − p, where p ∈ [ 12 , 1]. Let πk
denote the probability that the monkey will reach positive integer
k in some period t > 0. The value of πk for any positive integer k
is
A. pk
B. 1
pk
C. (1−p)
k
D.
p
.
k
24. Refer to the previous question. Suppose p = 12 and πk now denotes
the probability that the monkey will reach any integer k in some
period t > 0. The value of π0 is
A. 0
B. 21k
C. 12
D. 1
25. Suppose f : R → R is a differentiable function with f ′ (x) > 0 for
all x ∈ R and satisfying the property
lim f (x) ≥ 0.
x→−∞
Which of the following must be true?
6
A. f (1) < 0
B. f (1) > 0
C. f (1) = 0
D. None of the above
26. For what values of x is
x2 − 3x − 2 < 10 − 2x
A. 4 < x < 9
B. x < 0
C. −3 < x < 4
D. None of the above
27.
∫ e2
e
1
dx
x(log x)3
=
A. 3/8
B. 5/8
C. 6/5
D. −4/5
28. The solution of the system of equations
x − 2y + z = 7
2x − y + 4z = 17
3x − 2y + 2z = 14
is
A. x = 4, y = −1, z = 3
B. x = 2, y = 4, z = 3
C. x = 2, y = −1, z = 5
D. none of the above.
29. Let f : R2 → R be a twice-differentiable function with non-zero
second partial derivatives. Suppose that for every x ∈ R, there is
a unique value of y, say y ∗ (x), that solves the problem
max f (x, y).
y∈R
Then y ∗ is increasing in x if
7
A. f is strictly concave
B. f is strictly convex
C.
∂2f
∂x∂y
>0
D.
∂2f
∂x∂y
< 0.
∫
30.
√
3
A.
B.
√
3
C.
√
3
dx =
√
2x + 1
+
+c
ln 3
ln 3
√
3
2x+1
2x+1
2x+1
√
√
2x+1
√
√
2x + 1 3 2x+1
−
+c
ln 3
(ln 3)2
2x + 1 3 2x+1
−
+c
(ln 3)2
ln 3
D. none of the above.
8
2017
Booklet No.
Questions: 30
TEST CODE: PEB
Afternoon
Time: 2 hours
• On the answer booklet write your Name, Registration number,
Test Centre, Test Code and the Number of this Booklet in the
appropriate places on the Answer-sheet.
• This test has 30 questions. ANSWER ALL QUESTIONS.
All questions carry equal (4) marks.
• For each of the 30 questions, there are four suggested answers.
Only one of the suggested answers is correct. You will have to
identify the correct answer to get full credit for that question.
Indicate your choice of the correct answer by darkening the appropriate oval completely on the answer-sheet.
• You will get:
4 marks for each correctly answered question,
0 marks for each incorrectly answered question, and
1 mark for each unanswered question.
1. A researcher has 100 hours of work which have to be allocated
between two research assistants, Aditya and Gaurav. If Aditya is
allocated x hours of work, his utility is, −(x − 20)2 . If Gaurav is
allocated x hours of work, his utility is, −(x − 30)2 . The researcher
is considering two proposals: (I) Aditya does 60 hours and Gaurav
40 hours (II) Aditya does 90 hours and Gaurav 10 hours. Which
of the following statements is correct.
A. Proposal I is Pareto-efficient but Proposal II is not.
B. Proposal II is Pareto-efficient but Proposal I is not.
C. Both proposals are Pareto-efficient.
D. Neither proposal is Pareto-efficient.
2. The industry demand curve for tea is: Q = 1800 − 200P. The industry exhibits constant long run average cost (ATC) at all levels
of output at Rs 1.50 per unit of output. Which market form(s)
– perfect competition, pure monopoly and first-degree price discrimination – has the highest total market (that is, producer +
consumer) surplus?
A. perfect competition
B. pure monopoly
C. first degree price discrimination
D. perfect competition and first degree price discrimination
3. The following information will be used in the next question also.
OIL Inc. is a monopoly in the local oil refinement market. The
demand for refined oil is
Q = 75 − P
where P is the price in rupees and Q is the quantity, while the
marginal cost of production is
M C = 0.5Q.
The fixed cost is zero. Pollution is emitted in the refinement of
oil which generates a marginal external cost (MEC) equal to 31
Rs/unit. What is the level of Q that maximizes social surplus?
1
A. 50
B. 29 31
C. 17.6
D. 44
4. Refer to the previous question. Suppose the government decides to
impose a per unit pollution fee on OIL Inc. At what level should
the fee (in Rs/unit) be set to produce the level of output that
maximizes social surplus? You may use the fact that the marginal
revenue is given by: M R = 75 − 2Q.
A. 1/3
B. 2
C. 3/4
D. 5/3
5. Mr. X has an exogenous income W, and his utility from consumption is given by U (c). With probability p, an accident can occur.
If it occurs, the monetary equivalent of the damage is T. Mr. X
can however affect the accident probability, p, by taking prevention effort, e . In particular, e can take two values: 0, and a > 0.
Assume that p(0) > p(a). Let us also assume that the utility cost
of effort is Ae2 . Calculate the value of A below which effort will be
undertaken.
A.
B.
[p(a)−p(0)][u(W −T )−U (W )]
a2
p(a)−p(0)
u(W −T )−U (W )
C.
p(a)p(0)a2
u(W −T )−u(W )
D.
p(a)/p(0)
a2
u(W −T )/u(W )
6. Suppose Mr. X maximizes inter-temporal utility for 2 periods. His
total utility is given by
log(c1 ) + β log(c2 )
where β ∈ (0, 1) and c1 and c2 are his consumption in period 1
and period 2, respectively. Suppose he earns a wage only in period
2
1 and it is given by W. He saves for the second period on which
he enjoys a gross return of (1 + r) where r > 0 is the net interest
rate. Suppose the government implements a scheme where T ≥ 0
is collected from agents (thus also from Mr. X) in the first year,
and gives the same amount, T , back in the second period. What
is the optimum T for which his total utility is maximized?
A. T = 0
B. T =
W
2β
C. T =
βW
2(1−β)
D. T =
W
2(1−β)
7. Suppose there is one company in an economy which has a fixed
supply of shares in the short run. Suppose there is new information
that causes expectations of lower future profits. How does this new
stock market equilibrium affect final output and the final price
level of the economy if you assume that autonomous consumption
spending and household wealth are positively related?
A. real GDP increases; price decreases
B. real GDP decreases; price increases
C. real GDP decreases; price decreases
D. real GDP increases; price stays constant.
8. A monopolist faces a demand function, p = 10 − q. It has two
plants at its disposal. The cost of producing q1 in the first plant is
300 + q12 , if q1 > 0, and 0 otherwise. The cost of producing q2 in
the second plant is 200 + q22 , if q2 > 0, and 0 otherwise. What are
the optimal production levels in the two plants?
A. 10 units in both plants,
B. 20 units in the first plant and 10 units in the second plant
C. 0 units in the first plant and 15 units in the second plant
D. None of the above.
3
9. Consider a firm facing three consumers, 1, 2 and 3, with the following valuations for two goods, X and Y (All consumers consume at
most 1 unit of X and 1 unit of Y .)
Consumers X Y
1
7 1
2
4 5
3
1 6
The firm can produce both goods at a cost of zero. Suppose the
firm can supply both goods at a constant per unit price of pX for
X, and py for Y. It can also supply the two goods as a bundle, for
a price of pXY . The optimal vector of prices (pX , py , pXY ) is given
by
A. (7,6,9).
B. (4,1,4).
C. (7,7,7).
D. None of the above.
10. Two individuals, Bishal (B) and Julie (J), discover a stream of
mountain spring water. They each separately decide to bottle some
of this water and sell it. For simplicity, presume that the cost of
production is zero. The market demand for bottled water is given
by P = 90−0.25Q, where P is price per bottle and Q is the number
of bottles. What would Bishal’s output QB , Julie’s output QJ ,
and the market price be if the two individuals behaved as Cournot
duopolists?
A. QB = 120; QJ = 120; P = 42
B. QB = 90; QJ = 90; P = 30
C. QB = 120; QJ = 120; P = 30
D. QB = 100; QJ = 120; P = 30
11. The next three questions (11, 12, 13) are to be answered together.
Consider the following model of a closed economy
4
△Y
△C
△Yd
△T
=
=
=
=
△C + △I + △G
c△Yd
△Y − △T
t△Y + △T0
where △Y = change in GDP, △C = change in consumption, △I =
change in private investment, △G = change in government spending, △Yd = change in disposable income (i.e., after tax income),
△T = the change in total tax collections, t is the tax rate between
(0, 1), and △T0 = the change in that portion of tax collections that
can be altered by government fiscal policy measures. The value of
the balanced budget multiplier (in terms of G and T0 ) is given by:
A.
1
1−c(1−t)
B.
−c
1−c(1−t)
C.
1−c
1−c(1−t)
D. none of the above.
12. Refer to the previous question. Suppose the marginal propensity
to consume, c = .8, and t = .375. The value of the government
expenditure multiplier is
A. 2,
B. -1.6
C. .4
D. .5
13. Refer to the previous two questions. Suppose the marginal propensity to consume, c = .8, and t = .375. The value of the tax multiplier
(with respect to T0 ) is
A. -1.6
B. 2
C. .4
D. .3
5
14. In the IS-LM model, a policy plan to increase national savings
(public and private) without changing the level of GDP, using any
combination of fiscal and monetary policy involves
A. contractionary fiscal policy, contractionary monetary policy
B. expansionary fiscal policy, contractionary monetary policy
C. contractionary fiscal policy, expansionary monetary policy
D. expansionary fiscal policy, expansionary monetary policy
15. Consider the IS-LM-BP model with flexible exchange rates but with
no capital mobility. Consider an increase in the money supply. At
the new equilibrium, the interest rate is
, the exchange rate
, and the level of GDP is
, respectively.
is
A. higher, lower, higher
B. lower, higher, higher
C. lower, higher, lower
D. higher, lower, lower
16. Consider a Solow model of an economy that is characterized by the
following parameters: population growth, n; the depreciation rate,
δ; the level of technology, A; and the share of capital in output,
α. Per-capita consumption is given by c = (1 − s)y where s is the
exogenous savings rate, and y = Ak α , where y denotes output percapita, and k denotes the per-capita capital stock. The economy’s
golden-rule capital stock is determined by which of the following
conditions?
A.
B.
C.
∂c
∂k
∂c
∂k
∂c
∂k
= Ak α − (n + δ)k = 0
= αAk α−1 − (n + δ) = 0
= (n + δ)k − sAk α = 0
D. none of the above.
6
17. In the Ramsey model, also known as the optimal growth model,
with population growth, n, and an exogenous rate of growth of
technological progress, g, the steady-state growth rates of aggregate
output, Y, aggregate capital, K, and aggregate consumption, C, are
A. 0, 0, 0
B. n + g, n + g, n + g
C. g, n + g, n
D. n + g, n + g, g
18. Consider the standard formulation of the Phillips Curve,
πt − πte = −α(ut − un )
where πt is the current inflation rate, πte is the expected inflation
rate, α is a parameter, and un is the natural rate of unemployment.
Suppose the economy has two types of labour contracts: a proportion, λ, that are indexed to actual inflation, πt , and a proportion,
1 − λ, that are not indexed and simply respond to last year’s inflation, πt−1 . Wage indexation (relative to no indexation) will
the effect of unemployment on inflation.
A. strongly decrease
B. increase
C. not change
D. mildly decrease
19. Consider a Harrod-Domar style growth model with a (i) Leontieff
aggregate production function, (ii) no technological progress, and
(iii) a constant savings rate. Let K and L denote the level of
capital and labor employed in the economy. Output, Y, is produced
according to
Y = min{AK, BL}
where A and B are positive constants. Let L̄ be the full employment
level. Under what condition will there be positive unemployment?
A. AK > B L̄
B. AK < B L̄
7
C. AK = B L̄
D. none of the above.
20. The next two questions (20 and 21) are to be answered together.
People in a certain city get utility from driving their cars but each
car releases k units of pollution per km driven. The net utility of
each person is his or her utility from driving, v, minus the total
pollution generated by everyone else. Person i’s net utility is given
by
n
∑
Ui (x1 , ..., xn ) = v(xi ) −
kxj
j=1
j̸=i
where xj is km driven by person j, n is the city population, and
the utility of driving v has an inverted U-shape with v(0) = 0,
limx→0+ v ′ (x) = ∞, v ′′ (x) < 0, and v(x̄) = 0 for some x̄ > 0. In an
unregulated city, an increase in population will
A. increase the km driven per person
B. decrease the km driven per person
C. leave the km driven per person unchanged
D. may or may not increase the km driven per person.
21. Refer to the information given in the previous question. A city
planner decides to impose a tax per km driven and sets the tax
rate in order to maximize the total net utility of the residents.
Then, if the population increases, the optimal tax will
A. increase
B. decrease
C. stay unchanged
D. may or may not increase.
22. The production function
F (L, K) = (L + 10)1/2 K 1/2
has
8
A. increasing returns to scale
B. constant returns to scale
C. decreasing returns to scale
D. none of the above.
23. Consider the production functions
F (L, K) = L1/2 K 2/3 and G(L, K) = LK.
where L denotes labour and K denotes capital.
A. F is consistent with the law of diminishing returns to
capital but G is not.
B. G is consistent with the law of diminishing returns to
capital but F is not.
C. Both F and G are consistent with the law of diminishing
returns to capital
D. Neither F nor G is consistent with the law of diminishing
returns to capital
24. A public good is one that is non-rivalrous and non-excludable. Consider a cable TV channel and a congested city street.
A. A cable TV channel is a public good but a congested city
street is not
B. A congested city street is a public good but a cable TV
channel is not
C. Neither is a public good
D. Both are public goods.
25. Firm A’s cost of producing output level y > 0 is, cA (y) = 1 + y
while Firm B’s cost of producing output level y is, cB (y) = y(1−y)2
A. A can operate in a perfectly competitive industry but B
cannot
B. B can operate in a perfectly competitive industry but A
cannot
9
C. Neither could operate in a perfectly competitive industry
D. Either could operate in a perfectly competitive industry.
26. Suppose we generically refer to a New Keynesian model as a model
with a non vertical aggregate supply (AS) curve. Under sticky
prices, the AS curve will be
, and under sticky wages, the AS
curve will be
, respectively.
A. horizontal, upward sloping
B. upward sloping, upward sloping
C. downward sloping, horizontal
D. upward sloping, horizontal
27. With perfect capital mobility, and
at influencing output.
, monetary policy is
A. fixed exchange rates, effective
B. fixed exchange rates, ineffective
C. flexible exchange rates, ineffective
D. none of the above are correct
28. The next three questions (28, 29 and 30) use the following information. Consider an economy with two goods, x and y, and two
consumers, A and B, with endowments (x, y) given by (1, 0) and
(0, 1) respectively. A’s utility is
UA (x, y) = x + 2y
while B’s utility is
UB (x, y) = 2x + y.
Using an Edgeworth box with x measured on the horizontal axis
and y measured on the vertical axis, with A’s origin in the bottomleft corner and B’s origin in the top-right corner, the set of Paretooptimal allocations is
A. a straight line segment
B. the bottom and right edges of the box
10
C. the left and top edges of the box
D. none of the above.
29. Referring to the information given in the previous question, the
following allocations are the ones that may be achieved in some
competitive equilibrium.
A. (0,1)
B. The line segment joining (0, 1/2) to (0, 1) and the line
segment joining (0, 1) to (1/2, 1)
C. The line segment joining (1/2, 0) to (1, 0) and the line
segment joining (1, 0) to 1, 1/2)
D. (1,0)
30. Referring to the information given in the previous two questions, if
the price of y is 1, then the price of x in a competitive equilibrium
A. must be 1/2
B. must be 1
C. must be 2
D. could be any of the above.
11
Scanned by CamScanner