Strictly Confidential — (For Internal and Restricted Use Only)
Senior School Certificate Examination
July — 2015 (Comptt.)
Marking Scheme — Mathematics (Delhi) 65/1/1, 65/1/2, 65/1/3
General Instructions:
1.
The Marking Scheme provides general guidelines to reduce subjectivity in the marking.
The answers given in the Marking Scheme are suggested answers. The content is thus
indicative. If a student has given any other answer which is different from the one given
in the Marking Scheme, but conveys the meaning, such answers should be given full
weightage.
2.
Evaluation is to be done as per instructions provided in the marking scheme. It should
not be done according to one’s own interpretation or any other consideration — Marking
Scheme should be strictly adhered to and religiously followed.
3.
Alternative methods are accepted. Proportional marks are to be awarded.
4.
In question (s) on differential equations, constant of integration has to be written.
5.
If a candidate has attempted an extra question, marks obtained in the question attempted
first should be retained and the other answer should be scored out.
6.
A full scale of marks - 0 to 100 has to be used. Please do not hesitate to award full
marks if the answer deserves it.
7.
Separate Marking Scheme for all the three sets has been given.
8.
As per orders of the Hon’ble Supreme Court. The candidates would now be permitted
to obtain photocopy of the Answer book on request on payment of the prescribed fee.
All examiners/Head Examiners are once again reminded that they must ensure that
evaluation is carried out strictly as per value points for each answer as given in the
Marking Scheme.
1
QUESTION PAPER CODE 65/1/1
EXPECTED ANSWER/VALUE POINTS
SECTION A
Marks
1ɵ 2 ɵ
7 ɵ 14 ɵ
a=
i−
j then 7aɵ =
i−
j
1. ɵ
5
5
5
5
½+½
2. ( 2a − b) ⋅ ( 2a − b) = 1 ⇒ θ = π
4
½+½
2
2
2
2
2
2
3. cos α + cos β + cos γ = 1 ⇒ sin α + sin β + sin γ = 2
½+½
 −1
4. AB = 
4
½+½
5. 1 ⋅ y + x
5
⇒ | AB |= −28
8 
dy
dy
= −c sin x ⇒ x
+ y + xy tan x = 0
dx
dx
½+½
6. order = 2, degree = 3, sum = 2 + 3 = 5
½+½
SECTION B
7. System of equation is
3x + y + 2z = 1100, x + 2y + 3z = 1400, x + y + z = 600
1½
(i) Matrix equation is
3
1

1
1
2
1
2   x  1100 
3   y  = 1400 
1   z   600 
1
(ii) |A| = –3 ≠ 0, system of equations can be solved.
½
(iii) Any one value with reason.
1
2
8.
1
3] 
 −3
[2x
2  x 
=0
0  3
x 
4x ]   = 0
3
[ 2x − 9
1
−3
2
[2x 2 − 9x + 12x] = [0] ⇒ 2x + 3x = 0, x = 0 or 2
9.
(a + 1) (a + 2)
(a + 2) (a + 3)
a+2
a +3
1
1
(a + 3) (a + 4)
a+4
1
(a + 1) (a + 2)
a+2
1
2(a + 2)
1
0
4a + 10
2
0
=
1+1+1
R 2 → R 2 − R1 

R 3 → R 3 − R1 
1+1
= 4a + 8 – 4a – 10 = –2.
π/2
10. I =
1
dx
1 + tan x
∫
0
π/ 2
I=
∫
0
1
dx =
1 + tan(π / 2 − x)
π/2
∫
⇒ 2I =
0
⇒I=
1 + tan x
dx =
1 + tan x
2
(x − 1) (x + 2)
A=
π/ 2
∫
0
1
dx =
1 + cot x
π/2
∫
0
1 ⋅ dx =
π
2
π/ 2
∫
0
tan x
dx
1 + tan x
1½
1½
π
4
x
11.
1+1
1
=
A
B
C
+
+
2
x − 1 (x − 1)
x+2
½
2
1
2
, B = ,C = −
9
3
9
1½
3
x
2
1
2
∫ (x − 1)2 (x + 2) dx = ∫ 9(x − 1) dx + ∫ 3(x − 1)2 dx − ∫ 9(x + 2) dx
=
2
1
2
log | x − 1| −
− log | x + 2 | +C
9
3(x − 1) 9
1½
1½
12. Let X be the number of defective bulbs. Then
X = 0, 1, 2
P(X = 0) =
1
10C2
15C2
=
10C1 ⋅ 5C1 10
3
=
, P(X = 1) =
15C2
21
7
1+1
5C2
2
P(X = 2) = 15 = 21
C2
1
X
0
1
2
P(X)
3
7
10
21
2
21
OR
E1: Problem is solved by A.
E2: Problem is solved by B.
P(E1) =
1
1
1
2
, P(E 2 ) = , P(E1 ) = , P(E 2 ) =
2
3
2
3
P(E1 ∩ E 2 ) = P(E1 ) ⋅ P(E 2 ) =
1
1
6
1 2 2
P(problem is solved) = 1 − P(E1 ) ⋅ P(E 2 ) = 1 − ⋅ =
2 3 3
1½
P(one of them is solved) = P(E1 )P(E 2 ) + P(E1 )P(E 2 )
=
1 2 1 1 1
⋅ + ⋅ =
2 3 2 3 2
1½
4
uuur
13. AB = −4i$ − 6j$ − 2k$







uuur
$
$ $
AC = −i + (λ − 5) j + 3k
uuur
$ $ $
AD = −8i – j + 3k
−4
uuur uuur uuur
AB ⋅ (AC × AD) = −1
−8
−6
λ −5
−1
1½
−2
3 =0
3
1
−4(3λ − 12) + 6(21) − 2(8λ − 39) = 0 ⇒ λ = 9
r
r
ˆ b = $i − $j + kˆ
14. a1 = $i + 2$j + k,
1
1
r
r
$ b = 2i$ + $j + 2k$
a 2 = 2i$ – $j – k,
2
$i
r r
r r
$ b ×b = 1
a 2 – a1 = $i – 3j$ – 2k,
1
2
2
1½





1
$j
k$
−1
1
1 = −3iˆ + 3kˆ
1
r r
| b1 × b 2 | = 3 2
½+1
½
r r
r
r
(a 2 – a1 ) ⋅ (b1 – b 2 ) = −3 − 6 = −9
Shortest distance =
−9
3 2
=
2
3 2
1
π
15. tan −1 2x + tan −1 3x =
4
 5x  π
tan −1 
=
 1 − 6x 2  4
5x
1 − 6x 2
1
=1
1
5
⇒ 6x 2 + 5x − 1 = 0
⇒x=
1
1
, x = −1 (rejected)
6
1
OR
sin −1
5
−1 5
= tan
13
12
1
cos −1
3
−1 4
= tan
5
3
1
5
3
5
4
+ cos −1 = tan −1 + tan –1
13
5
12
3
−1
R.H.S. = sin
 5 4 
 12 + 3 
= tan 
5 4
 1 − ⋅ 
 12 3 
−1
= tan
16. y =
x cos −1 x
1− x2
dy
=
dx
63
16
1
− log 1 − x 2

x
1 − x 2 ⋅ 1cos −1 x –

1− x2

1− x2
1 − x 2 cos −1 x − x +
=
1
1− x
2
x 2 cos −1 x
1− x2 +
(1 − x 2 ) cos −1 x + x 2 cos −1 x
=
(1 − x 2 )3/ 2
 x cos −1 x(−2x)
 −
2 1− x2

+
=
x
1− x2
2x
2(1 − x 2 )
1+1
1
cos −1 x
(1 − x 2 )3/ 2
6
1
17. y = (sin x) x + sin −1 x
x log sin x
+ sin −1 x
⇒ y= e
1
⇒
dy
1
x log sin x
[log sin x + x cot x] +
=e
dx
2 x 1− x
1½
⇒
1
dy (sin x) x (log sin x + x cot x) +
=
2 x 1− x
dx
1½
18. x = a sec3 θ
dx
= 3a sec3 θ tan θ
dθ
½
y = a tan 3 θ
dy
= 3a tan 2 θ sec2 θ
dθ
½
3a tan 2 θ sec2 θ
dy
= sin θ
=
3a sec3 θ tan θ
dx
1
d2 y
dx 2
= cos θ ⋅
dθ
cos θ
cos 4 θ
=
=
dx 3a sec3 θ tan θ 3a tan θ
1
d2 y 
1

dx 2  θ = π = 12a
1
4
19.
∫
e x (x 2 + 1)
(x + 1) 2
dx
2
x  (x − 1) + 2 
e
dx
= ∫ 
2 
 (x + 1) 
1
2 
x  x −1
e
dx
+

= ∫
2
 x + 1 (x + 1) 
1
7
x −1 x
2
2
e x dx + ∫
e x dx
= x +1 ⋅ e − ∫
2
2
(x + 1)
(x + 1)
=
1
e x (x − 1)
+C
x +1
1
SECTION C
20. (a, b) * (c, d) = (a + c, b + d) = (c + a, d + b) = (c, d) * (a, b) ∴ * is commutative
1½
[(a, b) * (c, d)] * (e, f) = (a + c, b + d) * (e, f)
= (a + c + e, b + d + f) = (a, b) * (c + e, d + f)
1
= (a, b) * [(c, d) * (e, f) ∴ * is associate
1½
Let (e, e′) be the identity
(a, b) * (e, e′) = (a, b) ⇒ (a + e, b + e′) = (a, b) ⇒ e = 0, e′ = 0
⇒ Identity element is (0, 0)
21.
4 2 B
4
C
A(4, 4)
2
x2 + y2 = 32; y = x point of intersection is y = 4
½
Correct figure
1
4
O
Required Area = ∫ y dy +
0
4 2
∫
32 − y 2 dy
4
22. x
4 2
 y2   y
y 
32 − y 2 + 16sin −1
=   +
4 2  4
 2  0  2
1½
π 
π

= 8 +  0 + 16 ⋅  −  8 + 16 ⋅  = 4π
2 
2

1½
dy
dy  1

+ y − x + xy cot x = 0 ⇒
+  + cot x  y = 1
dx
dx  x

1
1½
4
½

I.F. = ∫  x + cot x  dx = x sin x
e
1
8
Solution: y ⋅ x sin x = ∫ 1 ⋅ x sin x dx
1½
⇒ yx sin x = –x cos x + sin x + C
1
when
x=
π
, y = 0, we have C = −1
2
1
yx sin x + x cos x – sin x = 1
1
OR
x dy + (xy + y )dx = 0 ⇒
2
2
Put y = vx ⇒
⇒ v+x
1
dy
dv
= v+x
dx
dx
1
dv
dx
dv
2
=−
= −(v + v ) ⇒ 2
dx
dx
v + 2v
dv
⇒
∫ (v + 1)2 − (1)2 − ∫
⇒
C
=
x
dx
1
v
⇒ log
= − log x + log C
x
2
v+2
y
y+x
If x = 1, y = 1, then C =
⇒
dy −(xy + y 2 )
=
dx
x2
1
3x =
1
1
1
3
1
y
y+x
1
23. Plane passing through the intersection of given planes:
(x + y + z – 1) + λ(2x + 3y + 4z – 5) = 0
1
(1 + 2λ)x + (1 + 3λ)y + (1 + 4λ)z +(–1 – 5λ) = 0
1½
Now (1 + 2λ) 1 + (1 + 3λ) (–1) + (1 + 4λ)1 = 0
1½
9
⇒λ= −
1
3
1
Equation of required plane is
⇒x–z+2=0
1
24. E1: First bag is selected.
E2: Second bag is selected.
A: both balls are red.
P(E1) =





1
1
1  A  12  A  2
, P(E 2 ) = , P   = , P 
=
2
2  E1  56  E 2  56
½+½+1+1
A
1 12
P(E1 ) P  
×
E
 E1 
6
 1
2 56
P  =
=
=
1
12
1
2
A
7
A
 A 
 
P(E1 ) P   + P(E 2 ) P 
 2 × 56 + 2 ⋅ 56
 E1 
 E2 
25.
½ + 1½
Let x and y be the number of takes. Then
Maximise:
z=x+y
1
Subject to:
200 x + 100y ≤ 5000
25x + 50y ≤ 1000
x ≥ 0, y ≥ 0
Y





Correct figure
50
at (20, 10), z = 20 + 10 = 30 is maximum.
20
at (25, 0), z = 25 + 0 = 25
B(20, 10)
A
O
2
25
40
X
at (0, 20), z = 20
10
2





1
26. l × b × 3 = 75 ⇒ l × b = 25
1
Let C be the cost. Then
C = 100 (l × b) + 100 h(b + l)
1
 25 
 25 
C = 100  l ×  + 300  + l 
l 

 l

1
dC
 −25 
= 0 + 300  2 + 1
dl
 l

dC
=0⇒l=5
dl
d 2C
dl 2
1
> 0 ⇒ C is maximum when l = 5 ⇒ b = 5
1
C = 100 (25) + 300(10)) = Rs. 5500
1
OR
A
θ
b
B
D
a
θ
Correct figure
1
AD = b sec θ, DC = a cosec θ
1
L = AC = b sec θ + a cosec θ
1
dL
= b sec θ tan θ – acosec θ cot θ
dθ
1
dL
a
= 0 ⇒ tan3 θ =
dθ
b
1
C
d2L
dθ2
L=
> 0 ⇒ minima
b ⋅ a 2 / 3 + b2 / 3
b1/ 3
⇒ L = (a2/3 + b2/3)2/3
11
+
a a 2 / 3 + b2 / 3
a1/ 3






1
QUESTION PAPER CODE 65/1/2
EXPECTED ANSWER/VALUE POINTS
SECTION A
Marks
1. 1 ⋅ y + x
dy
dy
= −c sin x ⇒ x
+ y + xy tan x = 0
dx
dx
½+½
2. order = 2, degree = 3, sum = 2 + 3 = 5
½+½
1 $ 2 $
7 $ 14 $
i−
j then 7a$ =
i−
j
3. a$ =
5
5
5
5
½+½
r r
r r
4. ( 2a − b) ⋅ ( 2a − b) = 1 ⇒ θ = π
4
½+½
2
2
2
2
2
2
5. cos α + cos β + cos γ = 1 ⇒ sin α + sin β + sin γ = 2
½+½
 −1
6. AB = 
4
½+½
5
⇒ | AB |= −28
8
SECTION B
7. y =
x cos −1 x
1− x
2
− log 1 − x 2

x
1 − x ⋅ 1cos −1 x –

1− x2

1− x2
2
dy
=
dx
x 2 cos −1 x
1 − x cos x − x +
1− x2 + x
1− x2
1− x2
2
=
2x
2(1 − x 2 )
1+1
−1
(1 − x 2 ) cos −1 x + x 2 cos −1 x
=
 x cos −1 x(−2x)
 −
2 1− x2

+
(1 − x 2 )3/ 2
=
1
cos −1 x
(1 − x 2 )3/ 2
12
1
8. y = (sin x) x + sin −1 x
x log sin x
+ sin −1 x
⇒ y= e
1
⇒
dy
1
x log sin x
[log sin x + x cot x] +
=e
dx
2 x 1− x
1½
⇒
1
dy (sin x) x (log sin x + x cot x) +
=
2 x 1− x
dx
1½
9. x = a sec3 θ
dx
= 3a sec3 θ tan θ
dθ
½
y = a tan 3 θ
dy
= 3a tan 2 θ sec2 θ
dθ
½
3a tan 2 θ sec2 θ
dy
= sin θ
=
3a sec3 θ tan θ
dx
1
d2 y
dx 2
= cos θ ⋅
dθ
cos θ
cos 4 θ
=
=
dx 3a sec3 θ tan θ 3a tan θ
1
d2 y 
1

dx 2  θ = π = 12a
1
4
10.
∫
e x (x 2 + 1)
(x + 1) 2
dx
2
x  (x − 1) + 2 
e
dx
= ∫ 
2 
 (x + 1) 
1
 x −1
2 
dx
+
= ∫ ex 
2
 x + 1 (x + 1) 
1
13
x −1 x
2
2
e x dx + ∫
e x dx
= x +1 ⋅ e − ∫
2
2
(x + 1)
(x + 1)
=
1
e x (x − 1)
+C
x +1
1
11. System of equation is
3x + y + 2z = 1100, x + 2y + 3z = 1400, x + y + z = 600
1½
(i) Matrix equation is
3
1

1
12.
1
2
1
2   x  1100 
3   y  = 1400 
1   z   600 
(ii) |A| = –3 ≠ 0, system of equations can be solved.
½
(iii) Any one value with reason.
1
[2x
[ 2x − 9
1
3] 
 −3
2  x 
=0
0  3
x 
4x ]   = 0
3
1
−3
2
[2x 2 − 9x + 12x] = [0] ⇒ 2x + 3x = 0, x = 0 or 2
13.
1
(a + 1) (a + 2)
a+2
1
(a + 2) (a + 3)
(a + 3) (a + 4)
a +3
a+4
1
1
(a + 1) (a + 2)
2(a + 2)
=
4a + 10
a+2
1
2
1
R → R 2 − R1
0 2
R → R 3 − R1
0 3
1+1+1
1+1
= 4a + 8 – 4a – 10 = –2.
1+1
14
π/2
14. I =
1
dx
1 + tan x
∫
0
π/ 2
I=
∫
0
1
dx =
1 + tan(π / 2 − x)
π/2
∫
⇒ 2I =
0
⇒I=
1 + tan x
dx =
1 + tan x
0
π/2
∫
0
1 ⋅ dx =
π/ 2
∫
0
tan x
dx
1 + tan x
π
2
(x − 1) (x + 2)
1½
=
A
B
C
+
+
2
x − 1 (x − 1)
x+2
½
2
1
2
, B = ,C = −
9
3
9
x
1½
2
1
2
∫ (x − 1)2 (x + 2) dx = ∫ 9(x − 1) dx + ∫ 3(x − 1)2 dx − ∫ 9(x + 2) dx
=
1½
1
2
A=
∫
1
dx =
1 + cot x
π
4
x
15.
π/ 2
2
1
2
log | x − 1| −
− log | x + 2 | +C
9
3(x − 1) 9
1½
1½
16. Let X be the number of defective bulbs. Then
X = 0, 1, 2
P(X = 0) =
1
10C2
15C2
=
10C1 ⋅ 5C1 10
3
=
, P(X = 1) =
15C2
21
7
1+1
5C2
2
P(X = 2) = 15 = 21
C2
1
X
0
1
2
P(X)
3
7
10
21
2
21
15
OR
E1: Problem is solved by A.
E2: Problem is solved by B.
P(E1) =
1
1
1
2
, P(E 2 ) = , P(E1 ) = , P(E 2 ) =
2
3
2
3
P(E1 ∩ E 2 ) = P(E1 ) ⋅ P(E 2 ) =
1
1
6
1 2 2
P(problem is solved) = 1 − P(E1 ) ⋅ P(E 2 ) = 1 − ⋅ =
2 3 3
1½
P(one of them is solved) = P(E1 )P(E 2 ) + P(E1 )P(E 2 )
=
1 2 1 1 1
⋅ + ⋅ =
2 3 2 3 2
1½
ɵ− 2k
ɵ
17. AB = −4iɵ− 6j



ɵ
ɵ
ɵ
AC = −i + (λ − 5) j + 3k 


uuur
$
$
$

AD = −8i – j + 3k
−4
uuur uuur uuur
AB ⋅ (AC × AD) = −1
−8
−6
λ −5
−1
1½
−2
3 =0
1
3
−4(3λ − 12) + 6(21) − 2(8λ − 39) = 0 ⇒ λ = 9
1½
r
r
ˆ b = $i − $j + kˆ
18. a1 = $i + 2$j + k,
1
1




r
r
$ b = 2i$ + $j + 2k$ 
a 2 = 2i$ – $j – k,
2
$i
r
r
r r
$ b ×b = 1
a 2 – a1 = $i – 3j$ – 2k,
1
2
2
1
$j
−1
1
k$
1 = −3iˆ + 3kˆ
1
16
½+1
r r
| b1 × b 2 | = 3 2
½
r r
r
r
(a 2 – a1 ) ⋅ (b1 – b 2 ) = −3 − 6 = −9
Shortest distance =
3 2
−9
=
2
3 2
−1
−1
19. tan 2x + tan 3x =
1
π
4
 5x  π
tan −1 
=
 1 − 6x 2  4
5x
1
=1
1
⇒ 6x 2 + 5x − 1 = 0
1
1 − 6x 2
⇒x=
1
, x = −1 (rejected)
6
1
OR
sin −1
5
−1 5
= tan
13
12
1
3
−1 4
= tan
5
3
1
cos −1
−1
R.H.S. = sin
5
3
5
4
+ cos −1 = tan −1 + tan –1
13
5
12
3
 5 4 
+ 

= tan  12 3 
 1 − 5 ⋅ 4 
 12 3 
−1
= tan
1
63
16
1
17
SECTION C
20.
Let x and y be the number of takes. Then
Maximise:
z=x+y
1
Subject to:
200 x + 100y ≤ 5000
25x + 50y ≤ 1000
Y
x ≥ 0, y ≥ 0





2
50
Correct figure
at (20, 10), z = 20 + 10 = 30 is maximum.
20
B(20, 10)
A
O
25
40
at (25, 0), z = 25 + 0 = 25
X
at (0, 20), z = 20
21. l × b × 3 = 75 ⇒ l × b = 25
2





1
1
Let C be the cost. Then
C = 100 (l × b) + 100 h(b + l)
1
 25 
 25 
C = 100  l ×  + 300  + l 
l 

 l

1
dC
 −25 
= 0 + 300  2 + 1
dl
 l

dC
=0⇒l=5
dl
d 2C
dl 2
1
> 0 ⇒ C is maximum when l = 5 ⇒ b = 5
C = 100 (25) + 300(10)) = Rs. 5500
1
1
18
OR
A
θ
Correct figure
1
AD = b sec θ, DC = a cosec θ
1
L = AC = b sec θ + a cosec θ
1
dL
= b sec θ tan θ – acosec θ cot θ
dθ
1
dL
a
= 0 ⇒ tan3 θ =
dθ
b
1
D
b
a
θ
B
C
d2L
dθ2
L=
> 0 ⇒ minima
b ⋅ a 2 / 3 + b2 / 3
1/ 3
b
+
a a 2 / 3 + b2 / 3
1/ 3
a
⇒ L = (a2/3 + b2/3)2/3






22. (a, b) * (c, d) = (a + c, b + d) = (c + a, d + b) = (c, d) * (a, b) ∴ * is commutative
1
1½
[(a, b) * (c, d)] * (e, f) = (a + c, b + d) * (e, f)
= (a + c + e, b + d + f) = (a, b) * (c + e, d + f)
1
= (a, b) * [(c, d) * (e, f) ∴ * is associate
1½
Let (e, e′) be the identity
(a, b) * (e, e′) = (a, b) ⇒ (a + e, b + e′) = (a, b) ⇒ e = 0, e′ = 0
⇒ Identity element is (0, 0)
23.
4 2 B
4
C
A(4, 4)
2
x2 + y2 = 32; y = x point of intersection is y = 4
½
Correct figure
1
O
4
Required Area =
∫ y dy + ∫
0
19
4 2
4
32 − y 2 dy
1½
4 2
4
24. x
 y2   y
y 
32 − y 2 + 16sin −1
=   +
4 2  4
 2  0  2
1½
π 
π

= 8 +  0 + 16 ⋅  −  8 + 16 ⋅  = 4π
2 
2

1½
dy
dy  1

+ y − x + xy cot x = 0 ⇒
+  + cot x  y = 1
dx
dx  x

1
½

I.F. = ∫  x + cot x  dx = x sin x
e
1
Solution: y ⋅ x sin x = ∫ 1 ⋅ x sin x dx
1½
⇒ yx sin x = –x cos x + sin x + C
1
when
x=
π
, y = 0, we have C = −1
2
1
yx sin x + x cos x – sin x = 1
1
OR
x dy + (xy + y )dx = 0 ⇒
2
2
Put y = vx ⇒
⇒ v+x
dy −(xy + y 2 )
=
dx
x2
1
dy
dv
= v+x
dx
dx
1
dv
dx
dv
2
=−
= −(v + v ) ⇒ 2
dx
dx
v + 2v
dv
⇒
∫ (v + 1)2 − (1)2 − ∫
⇒
C
=
x
dx
1
v
⇒ log
= − log x + log C
x
2
v+2
y
y+x
1
1
20
If x = 1, y = 1, then C =
⇒
1
3x =
1
3
1
y
y+x
1
25. Plane passing through the intersection of given planes:
(x + y + z – 1) + λ(2x + 3y + 4z – 5) = 0
1
(1 + 2λ)x + (1 + 3λ)y + (1 + 4λ)z +(–1 – 5λ) = 0
1½
Now (1 + 2λ) 1 + (1 + 3λ) (–1) + (1 + 4λ)1 = 0
1½
⇒λ= −
1
3
1
Equation of required plane is
⇒x–z+2=0
26.
1
: First bag is selected.


E2: Second bag is selected. 


A: both balls are red.
E
1
P(E1) =
1
1
1  A  12  A  2
, P(E 2 ) = , P   = , P 
=
2
2  E1  56  E 2  56
A
1 12
P(E1 ) P  
×
E1 
 E1 
6

2
56
P  =
=
=
A
 A  1 × 12 + 1 ⋅ 2 7
A
P(E1 ) P   + P(E 2 ) P 

 E1 
 E 2  2 56 2 56
21
½+½+1+1
½ + 1½
QUESTION PAPER CODE 65/1/3
EXPECTED ANSWER/VALUE POINTS
SECTION A
Marks
2
2
2
2
2
2
1. cos α + cos β + cos γ = 1 ⇒ sin α + sin β + sin γ = 2
½+½
 −1
2. AB = 
4
½+½
3. 1 ⋅ y + x
5
⇒ | AB |= −28
8 
dy
dy
= −c sin x ⇒ x
+ y + xy tan x = 0
dx
dx
½+½
4. order = 2, degree = 3, sum = 2 + 3 = 5
½+½
1 $ 2 $
7 $ 14 $
i−
j then 7a$ =
i−
j
5. a$ =
5
5
5
5
½+½
r r
r r
6. ( 2a − b) ⋅ ( 2a − b) = 1 ⇒ θ = π
4
½+½
SECTION B
x
7.
(x − 1)2 (x + 2)
A=
=
A
B
C
+
+
x − 1 (x − 1)2 x + 2
½
2
1
2
, B = ,C = −
9
3
9
x
1½
2
1
2
∫ (x − 1)2 (x + 2) dx = ∫ 9(x − 1) dx + ∫ 3(x − 1)2 dx − ∫ 9(x + 2) dx
=
2
1
2
log | x − 1| −
− log | x + 2 | +C
9
3(x − 1) 9
1½
1½
8. Let X be the number of defective bulbs. Then
X = 0, 1, 2
1
22
P(X = 0) =
10C2
15C2
=
10C1 ⋅ 5C1 10
3
=
, P(X = 1) =
15C2
21
7
1+1
5C2
2
P(X = 2) = 15 = 21
C2
1
X
0
1
2
P(X)
3
7
10
21
2
21
OR
E1: Problem is solved by A.
E2: Problem is solved by B.
P(E1) =
1
1
1
2
, P(E 2 ) = , P(E1 ) = , P(E 2 ) =
2
3
2
3
P(E1 ∩ E 2 ) = P(E1 ) ⋅ P(E 2 ) =
1
1
6
1 2 2
P(problem is solved) = 1 − P(E1 ) ⋅ P(E 2 ) = 1 − ⋅ =
2 3 3
1½
P(one of them is solved) = P(E1 )P(E 2 ) + P(E1 )P(E 2 )
=
1 2 1 1 1
⋅ + ⋅ =
2 3 2 3 2
1½
uuur
9. AB = −4i$ − 6j$ − 2k$



uuur
$i + (λ − 5) $j + 3k$ 
−
=
AC


uuur
$ – $j + 3k$

=
−
8i
AD
−4
uuur uuur uuur
AB ⋅ (AC × AD) = −1
−8
−6
λ −5
−1
1½
−2
3 =0
3
1
23
−4(3λ − 12) + 6(21) − 2(8λ − 39) = 0 ⇒ λ = 9
r
r
ˆ b = $i − $j + kˆ
10. a1 = $i + 2$j + k,
1
1
1½





r
r
$ b = 2i$ + $j + 2k$
a 2 = 2i$ – $j – k,
2
$i
r
r
r r
$ b ×b = 1
a 2 – a1 = $i – 3j$ – 2k,
1
2
2
1
$j
k$
−1
1 = −3iˆ + 3kˆ
1
1
r r
| b1 × b 2 | = 3 2
½+1
½
r r
r
r
(a 2 – a1 ) ⋅ (b1 – b 2 ) = −3 − 6 = −9
Shortest distance =
3 2
−9
=
2
3 2
11. tan −1 2x + tan −1 3x =
1
π
4
 5x  π
tan −1 
=
 1 − 6x 2  4
5x
1
=1
1
⇒ 6x 2 + 5x − 1 = 0
1
1 − 6x 2
⇒x=
1
, x = −1 (rejected)
6
1
OR
sin −1
5
−1 5
= tan
13
12
1
cos −1
3
−1 4
= tan
5
3
1
24
5
3
5
4
+ cos −1 = tan −1 + tan –1
13
5
12
3
−1
R.H.S. = sin
 5 4 
 12 + 3 
= tan 

 1 − 5 ⋅ 4 
 12 3 
−1
= tan
12. y =
x cos −1 x
1− x2
dy
=
dx
1
63
16
1
− log 1 − x 2

x
1 − x 2 ⋅ 1cos −1 x –

1− x2

1− x2
1 − x 2 cos −1 x − x +
=
1− x
2
x 2 cos −1 x
1− x2 +
(1 − x 2 ) cos −1 x + x 2 cos −1 x
=
(1 − x 2 )3/ 2
 x cos −1 x(−2x)
 −
2 1− x2

+
=
x
1− x2
2x
2(1 − x 2 )
1+1
1
cos −1 x
(1 − x 2 )3/ 2
1
13. y = (sin x) x + sin −1 x
x log sin x
+ sin −1 x
⇒y= e
1
⇒
dy
1
x log sin x
[log sin x + x cot x] +
=e
dx
2 x 1− x
1½
⇒
1
dy (sin x) x (log sin x + x cot x) +
=
2 x 1− x
dx
1½
14. x = a sec3 θ
dx
= 3a sec3 θ tan θ
dθ
½
25
y = a tan 3 θ
dy
= 3a tan 2 θ sec2 θ
dθ
½
3a tan 2 θ sec2 θ
dy
= sin θ
=
3a sec3 θ tan θ
dx
1
d2 y
dx 2
= cos θ ⋅
dθ
cos θ
cos 4 θ
=
=
dx 3a sec3 θ tan θ 3a tan θ
1
d2 y 
1

dx 2  θ = π = 12a
1
4
15.
∫
e x (x 2 + 1)
(x + 1) 2
dx
 (x 2 − 1) + 2 
dx
= ∫e 
2 
 (x + 1) 
1
2 
x  x −1
 dx
= ∫ e  x +1 +
(x + 1) 2 

1
x −1 x
2
2
x
e
dx
+
e x dx
∫
= x +1 ⋅ e − ∫
2
2
(x + 1)
(x + 1)
1
x
=
e x (x − 1)
+C
x +1
1
16. System of equation is
3x + y + 2z = 1100, x + 2y + 3z = 1400, x + y + z = 600
1½
(i) Matrix equation is
3
1

1
1
2
1
2   x  1100 
3   y  = 1400 
1   z   600 
26
1
17.
(ii) |A| = –3 ≠ 0, system of equations can be solved.
½
(iii) Any one value with reason.
1
1
3] 
 −3
[2x
2  x 
=0
0  3
x 
4x ]   = 0
3
[ 2x − 9
1
2
[2x 2 − 9x + 12x] = [0] ⇒ 2x + 3x = 0, x = 0 or
18.
(a + 1) (a + 2)
a+2
1
(a + 2) (a + 3)
a +3
1
(a + 3) (a + 4)
a+4
1
(a + 1) (a + 2)
a+2
1
2(a + 2)
1
0
4a + 10
2
0
=
−3
2
1+1+1
R 2 → R 2 − R1
1+1
R 3 → R 3 − R1
= 4a + 8 – 4a – 10 = –2.
π/2
19. I =
∫
0
π/ 2
I=
∫
0
1
dx
1 + tan x
1
dx =
1 + tan(π / 2 − x)
π/2
⇒ 2I =
∫
0
⇒I=
1+1
1 + tan x
dx =
1 + tan x
π/2
π/ 2
∫
0
1
dx =
1 + cot x
π
∫ 1 ⋅ dx = 2
π/ 2
∫
0
tan x
dx
1 + tan x
1½
1½
0
π
4
1
27
SECTION C
20. Plane passing through the intersection of given planes:
(x + y + z – 1) + λ(2x + 3y + 4z – 5) = 0
1
(1 + 2λ)x + (1 + 3λ)y + (1 + 4λ)z +(–1 – 5λ) = 0
1½
Now (1 + 2λ) 1 + (1 + 3λ) (–1) + (1 + 4λ)1 = 0
1½
⇒λ= −
1
3
1
Equation of required plane is
⇒x–z+2=0
1
21. E1: First bag is selected.


E2: Second bag is selected. 


A: both balls are red.
P(E1) =
1
1
1  A  12  A  2
, P(E 2 ) = , P   = , P 
=
2
2  E1  56  E 2  56
½+½+1+1
A
1 12
P(E1 ) P  
×
E
 E1 
6


1
2
56
P  =
=
=
A
 A  1 × 12 + 1 ⋅ 2 7
A
P(E1 ) P   + P(E 2 ) P 

 E1 
 E 2  2 56 2 56
22.
½ + 1½
Let x and y be the number of takes. Then
Maximise:
z=x+y
1
Subject to:
200 x + 100y ≤ 5000
25x + 50y ≤ 1000
x ≥ 0, y ≥ 0
28





2
Correct figure
Y
50
at (20, 10), z = 20 + 10 = 30 is maximum.
at (25, 0), z = 25 + 0 = 25
20
at (0, 20), z = 20
B(20, 10)
A
25
O
2





1
X
40
23. l × b × 3 = 75 ⇒ l × b = 25
1
Let C be the cost. Then
C = 100 (l × b) + 100 h(b + l)
1
 25 
 25 
C = 100  l ×  + 300  + l 
l 

 l

1
dC
 −25 
= 0 + 300  2 + 1
dl
 l

dC
=0⇒l=5
dl
d 2C
dl 2
1
> 0 ⇒ C is maximum when l = 5 ⇒ b = 5
C = 100 (25) + 300(10)) = Rs. 5500
1
1
OR
A
θ
b
B
D
Correct figure
1
AD = b sec θ, DC = a cosec θ
1
L = AC = b sec θ + a cosec θ
1
dL
= b sec θ tan θ – acosec θ cot θ
dθ
1
dL
a
= 0 ⇒ tan3 θ =
dθ
b
1
a
θ
C
29
d2L
dθ2
L=
> 0 ⇒ minima
b ⋅ a 2 / 3 + b2 / 3
b1/ 3
+
a a 2 / 3 + b2 / 3
a1/ 3
⇒ L = (a2/3 + b2/3)2/3






24. (a, b) * (c, d) = (a + c, b + d) = (c + a, d + b) = (c, d) * (a, b) ∴ * is commutative
1
1½
[(a, b) * (c, d)] * (e, f) = (a + c, b + d) * (e, f)
= (a + c + e, b + d + f) = (a, b) * (c + e, d + f)
1
= (a, b) * [(c, d) * (e, f) ∴ * is associate
1½
Let (e, e′) be the identity
(a, b) * (e, e′) = (a, b) ⇒ (a + e, b + e′) = (a, b) ⇒ e = 0, e′ = 0
⇒ Identity element is (0, 0)
25.
4 2 B
4
C
2
x2 + y2 = 32; y = x point of intersection is y = 4
½
Correct figure
1
A(4, 4)
4
Required Area =
∫ y dy + ∫
0
O
4 2
32 − y 2 dy
4
4
26. x
4 2
 y2   y
y 
32 − y 2 + 16sin −1
=   +
4 2  4
 2  0  2
1½
π 
π

= 8 +  0 + 16 ⋅  −  8 + 16 ⋅  = 4π
2 
2

1½
dy
dy  1

+ y − x + xy cot x = 0 ⇒
+  + cot x  y = 1
dx
dx  x

1
1½
½

I.F. = ∫  x + cot x  dx = x sin x
e
1
30
Solution: y ⋅ x sin x = ∫ 1 ⋅ x sin x dx
1½
⇒ yx sin x = –x cos x + sin x + C
1
when
π
, y = 0, we have C = −1
2
1
yx sin x + x cos x – sin x = 1
1
x=
OR
x dy + (xy + y )dx = 0 ⇒
2
2
Put y = vx ⇒
⇒ v+x
1
dy
dv
= v+x
dx
dx
1
dv
dx
dv
2
=−
= −(v + v ) ⇒ 2
dx
dx
v + 2v
dv
⇒
∫ (v + 1)2 − (1)2 − ∫
⇒
C
=
x
dx
1
v
⇒ log
= − log x + log C
x
2
v+2
y
y+x
If x = 1, y = 1, then C =
⇒
dy −(xy + y 2 )
=
dx
x2
1
3x =
1
1
1
3
1
y
y+x
1
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