Maths IIT-JEE
MC SIR
(Paper - 1)
ANSWER KEY
1.
6.
(A)
(B)
2.
7.
(C)
(ABCD)
3.
8.
(D)
(BCD)
4.
9.
(C)
(AD)
5.
10.
(C)
(BCD)
11.
(ABCD)
12.
(BCD)
13.
(20)
14.
(5)
15.
(1)
16.
(25)
17.
(7)
18.
(5)
SOLUTIONS
1.
Sol.
(A)
For odd factors, only 3, 5, 7, 9 can be used. But for 3m + 2, only 5 and 7 are allowed
 possible factors = 5 & 35
Sum = 5 + 35
2.
(C)
Sol.
sin  cos 
sin 3  cos3 


– sin 2 = 1 +
cos  sin 
cos  sin 

1
sin 4   cos 4 
– sin2 = 1 +
sin  cos 
cos  sin 

sin 4   cos 4  – 1
 1  sin 2
sin  cos 

1 – 2 sin 2  cos 2  – 1
= 1 + sin2
sin  cos 


– 2sincos = 1 + sin2
2sin2= – 1
sin2= –

=
1
2
7
12
3.
(D)
Sol.
As  + = –
b
c
, =
a
a

 
b
–

c

1 1
b
 –
 
c
....(1)
Get 10% Instant Discount On Unacademy Plus [Use Referral Code : MCSIR] 1
Maths IIT-JEE
MC SIR
b1
c1
, –  =
a1
a1
Also,
– + = –

–
b
– 1
– 
c1

–
Now,
 + = –


1
–

C

–C–1 =
1 1 b1
 
  c1
C
1
,  =
A
A
1 1

 
b b1
–
c c1
= –

C–1 =
...(2)
(by (1) – (2))
b b1

c c1
4.
(C)
Sol.
As sinx + cosx  [– 2 ,
2]

[sinx + cosx] can be –2, –1, 0, 1
Also [–sinx] & [–cosx] can be – 1, 0, 1

(LHS)max = 1 = (RHS)min
Thus LHS = RHS iff
sinx + cosx [1, 2 ], sinx  (0, 1] & cos x (0, 1]
which is possible for infinite values of 'x' in the first quadrant.
5.
Sol.
(C)
Let sinx = t  [–1, 1]
f(x) =
( t 2  4t  4)  1
2 t 2  8t  8
1
1
=1+
( t  2) 2
t  4t  4

2f(x) = 1 +
As
t  [–1, 1]
(t + 2)2 [1, 9]
2
Get 10% Instant Discount On Unacademy Plus [Use Referral Code : MCSIR] 2
Maths IIT-JEE
1+
MC SIR
1
10
 ,
2
( t  2)
9
10

2


2 f(x)   , 2
9 

5



f(x)   , 1
9
6.
(B)
Sol.
Let xn = tan,   0,



xn+1 =


2
1
sec  – 1
2
=
(
sec

–
1
)
tan 
tan 

2 sin 2  
1 – cos 
2
=
=


sin 
2 sin cos
2
2

4

2
= tan  
 
2 
2 
As x1 =1 = tan    tan


3 
2 
x2 = tan 
  
n 1 
2 
So one till xn = tan 
7.
Sol.
(ABCD)
Domain of f(x) is x  R – I
Also ; sin([x]) = 0

F(x) = 0 ; x I
Period is 1
F(–x) = F(x) = 0  Even
Range is {0}  Singleton
–1 0
1
2
3
{x}

 {x}  
0
 – 1 , domain is x  R – I and
for Sgn  Sgn 


 {x} 




{x}

we have, Sgn (1) – 1 = 0
Hence it is identical with F(x)
Get 10% Instant Discount On Unacademy Plus [Use Referral Code : MCSIR] 3
Maths IIT-JEE
8.
Sol.
MC SIR
(BCD)
Applying AM  GM, We get
x12  4 y 22
 x12 .4 y 22
2
x 22  4 y32
 x 22 .4 y32
2
x 32  4 y12
 x 32 .4 y12
2
&
Adding
x
2
1
 4 y12
2
 2( x1 y 2  x 2 y 3  x 3 y1 )
Also (xi , yi) lies on x2 + 4y2 = 1 , thus
xi2 + 4yi2 = 1, i = 1, 2, 3

3
 2 (x1y2 + x2y3 + x3y1)
2

x1y2 + x2y3 + x3y1 
3
4
9.
Sol.
(AD)
x2 – 6x + 10 = (x – 3)2 + 1 > 0  x R

f(x) = sinx + tanx + 1
Period is LCM(2, ) = 2
Hence , 4is also a period
10.
(BCD)
Sol.
f(x + 4) = sin  [ x  4] 

2



2


= sin  [ x ]  2 

2


= sin  [ x ]  = f(x)

Also

periodic with period '4'


3
[x] = 0,
, ,
 x [0, 4)
2
2
2

2


sin  [ x ]   0,1,0,–1
Hence, into
Also, f(x) = 0  x [0, 1)
Hence, many one
Get 10% Instant Discount On Unacademy Plus [Use Referral Code : MCSIR] 4
Maths IIT-JEE
11.
Sol.
MC SIR
(ABCD)
General term is
n
= a b c (x2)a (2x)b (2)c
G.T.
n
= a b cx
(A)
0 1 n –1
x1 · 2 n
= (2n · n)x
2a + b = 2
a, b, c = (0, 2, n – 2) or (1, 0 , n – 1)
 Coeff. of x2 is
n
0 2 n–2
(C)
2bc
where a + b + c = n
2a + b = 1
 a=0,b=1,c=n–1
 term of 'x' is
n
(B)
2a b
·2 n 
n
·2 n –1
1 0 n –1
= 2n–1 (n(n – 1) + n)
= n2 · 2n–1
2a + b = 3
a, b, c = 0, 3, n – 3 or 1, 1, n – 2
 Coeff. of x3 is
n
0 3 n –3
·2 n 
n
1 1 n–2
·2 n –1
2

= 2n–1  n (n – 1)(n – 2)  n (n – 1)
6

n – 2 
1
= 2n–1 · n(n – 1) 
 3

n ( n – 1)(n  1)
= 2n × n+1C3
3
2a + b = 4
a, b, c (0, 4, n – 4) or (1, 2, n – 3) or (2, 0, n – 2)
Hence , Coeff. is
= 2n–1
(D)
n
0 4 n–4
·2 n 
= 2n–2 nC2 ·
n
1 2 n–3
·2 n –1 
n
2 0 n–2
·2 n – 2
(n 2  n – 3)
3
Get 10% Instant Discount On Unacademy Plus [Use Referral Code : MCSIR] 5
Maths IIT-JEE
12.
MC SIR
(BCD)
P
C
a
Sol.
b
Q
30º
s/2
C3
R
s/2
For PQR, circumcentre and incentre coincide, hence it is equilateral

b
= sin(30º) 
a
a = 2b
&
b
 tan(30º )  s = 2 3b
(s / 2)
a
3 2
3
s =
(12b 2 ) = 3 3 b 2 = 3 3  
Hence, area (PQR) =
4
4
2
13.
20
Sol.
f(x) =
2
=
3 3 2
3 3
a =
ab
4
2
3  2 sin x

x
x
2  cos  sin 
2
2

2
x
x

1  2 sin  cos 
2
2

f(x) =

x
x
2  sin  cos 
2
2

Case-I
f(x) =
x
 I Quadrant
2
1
 2t
2t
Now, t = sin
=
x
x
 cos
2
2
 x 
2 sin     [1,
2 4
2 ] as
 
x
 0, 
2
 2
2 t  [ 2 , 2]
Get 10% Instant Discount On Unacademy Plus [Use Referral Code : MCSIR] 6
Maths IIT-JEE
MC SIR

hence f(x)   2 


Similarly for
1
1
,2  
2
2
x
 III quadrant
2
 
 
Range f(x)   –  2 
Case II
1
f(x)   ,
2
x
x
– cos =
2
2
x

2 sin  2 – 4   [1,
2]
Similarly for IV Quadrant


14.
Sol.
2
3 
 has integer {1}
2 
f(x)   –


 No integer
5
– 2t
2t
where t = sin

1 
1 

, –  2 
2 
2 
x
 II Quadrant
2
x
x

5 – 2 sin – cos 
2
2

f(x) =
x
x


2  sin – cos 
2
2

f(x) =
 No integer
3
1
, –   has integer {–1}
2 2
f(x) has two integers in range {1, –1}
N=2
5
x6 + x + = (x – 2)2 (x – x1) (x – x2) (x – x3)(x – x4)

x 6  x  
(x – x1)(x – x2) (x – x3)(x – x4) =
( x – 2) 2

x 6  x  
(2 – x1) (2 – x2) (2 – x3) (2 – x4) = lim
x 2
( x – 2) 2
Apply LH rule twice,
30x 4
RHS = lim
x 2
2
= 15 × 24
= 240
Get 10% Instant Discount On Unacademy Plus [Use Referral Code : MCSIR] 7
Maths IIT-JEE
MC SIR
15.
1
Sol.
C
BAE =
2
C
as BE subtends equal angles at C & A
In ACD, by sine rule
b
C/2
C/2
CD
AD

sin A sin(C / 2)
D
A C/2
In ADE,
a
B
B
AD
DE

sin B sin(C / 2)
E
Equating AD, we get,
CD. sin(C / 2) DE. sin B

sin A
sin(C / 2)

CD sin A sin B

DE sin 2 (C / 2)
Add 1,
CD
sin A sin B  sin 2 (C / 2)
1 
DE
sin 2 (C / 2)
CE 2 sin A sin B  2 sin 2 (C / 2)


DE
2 sin 2 (C / 2)
=
cos(A – B) – cos(A  B)  1 – cos C
2 sin 2 (C / 2)
=
1  cos(A – B)
2 sin 2 (C / 2)
=
 A – B
2 cos 2 

 2 
2 sin 2 (C / 2)
=
  A – B 
 cos 2  

 
 sin  C  
  

2 
{ cos(A + B) = cos( – C) = – cos C}
2
Get 10% Instant Discount On Unacademy Plus [Use Referral Code : MCSIR] 8
Maths IIT-JEE
MC SIR
2
=

 A – B   C 
 2 cos 2  cos 2  

  

 2 sin C  cos C  
    

2 2 
=

 A – BC
A –B–C
 cos
  cos

2
2







sin C




=
 sin B  sin A 


sin C


=
ba


 c 
2
2
2
{using sine sule}
 K=1
16.
25
(4, 5)
5

5
(1, 1)
5
Area
5
1

= 2  .5.5 sin  
2

= 25 sin
 (Area)max = 25 (when sin= 1)
17.
7
Sol.
ab
A &
2
ab  G
 A = 2G

ab
 2 ab
2
or
ab 2

2 ab 1
Componendo & Dividendo,
a  b  2 ab
2 1
=
2 –1
a  b – 2 ab
Get 10% Instant Discount On Unacademy Plus [Use Referral Code : MCSIR] 9
Maths IIT-JEE
 a b


 a– b



MC SIR
2
=3
a b
3

1
a– b
Componendo & Dividendo,

2 a
3 1

 2 3
2 b
3 –1
Squaring , we get
a
 43 4 3
b
=7+ 4 3
 k=7
18.
Sol.
5
For n  
n
1
1
1

n –

.......
 1
2
3
1   enln(1+1/n) = e  n 2 n 3 n 
 n
 given limit is
 2 n  1n  13n 3  15n 5 ..... 2  2

lim e 
– e n
n 


  2  2 ......  2
2   3n 2 5 n 4

lim
e
– 1n
= n e


  2 2  2 4 .....   2
 3n 5n

– 1n
e

2 

2
 2

lim e

 ....... 
n 
2
 2
  3n 2 5n 4

 2  4  .....
 3n 5n

= e2 × 1 ×
2
2e2

3
3
(a + b)min = 5
Get 10% Instant Discount On Unacademy Plus [Use Referral Code : MCSIR] 10
Maths IIT-JEE
MC SIR
(Paper - 2)
ANSWER KEY
1.
6.
11.
(A)
(A)
(A,D)
2.
(B)
7. (B,C,D)
12.
(A,B)
16.
(27)
17.
3.
(C)
8. (A,B,D)
13.
(2)
(4)
18.
4.
(B)
9. (A,B,C)
14.
(19)
5.
(A)
10. (A,C,D)
15.
(3)
(1)
SOLUTIONS
1.
Sol.
(A)
Note that
5a + 2b +c = a + 4a + 2b + c = a + f(2)
If a > 0 f(2) > 0
1
2
2
If a < 0 f(2) < 0
1
 a(a + f(2)) > 0 for both cases
2.
(B)
A
N
M
H
I

Sol.
C
B
90º–C
45º–
C
2
45º–
B
2
90º–B
As BMC is right angled
MBC = 90º – C
Similarly NCB = 90º – B
Also , BI is angle bisector of MBC
B
C 

BC
  = 180º –  45º–  45º–  = +
2
2 2
2

Get 10% Instant Discount On Unacademy Plus [Use Referral Code : MCSIR] 1
Maths IIT-JEE
3.
Sol.
MC SIR
(C)
As a, b, c, d are in HP
b=
2ac
2bd
& c=
a c
bd
Appling AM > HM, we have (no equality because a, b, c, d are distinct)
a  c 2ac

b
2
ac
or a + c > 2b
Also,
......(1)
b  d 2bd

c
2
bd
 b + d > 2c
(1) + (2) gives
a + b + c + d > 2b + 2c
......(2)
or
a+d>b+c
4.
Sol.
(B)
Let the number formed is a1a2a3 ........ a8 (where ai represents the digit)
Now if product of digits at 5-consecutive positions is divisible by 5 then digit a4 or a5 must be 5
hence ways = 2C1 × 1 × 7! = 2(7!)
5.
Sol.
(A)
As f(x) is odd, f(–x) = – f(x)
Put x = 0 , f(0) = – f(0)
 f(0) = 0
Also, f(x) is periodic with period 2, thus
f(x + 2) = f(x)
i.e. f(0) = f(2) = f(4) = .......  f(4) = f(0) = 0
6.
(A)
Q
2
P
Sol.

2

2
4
6
R
S
Get 10% Instant Discount On Unacademy Plus [Use Referral Code : MCSIR] 2
Maths IIT-JEE
cos=
MC SIR
4
= 60º
8
 PR = 2 (2sin60º) = 2 3
& QS = 2(6sin60º) = 6 3
Distance between parallel sides = 2 cos60º + 2 + 6 – 6 cos60º
 Area of trapezium PQRS =
7.
Sol.
1
(2 3  6 3 ) (6)
2
=6
= 24 3
(BCD)
As A > G > H > 0 for positive unequal numbers, we have
f(x) = Ax2 – Gx – 2H
f(0) = – 2H < 0
f(–1) = A + G – 2H
= (A – H) + (G – H) > 0
f(2) = 2[2A – G – H]
= 2 [(A – G) + (A – H)] > 0
0
–1
2
Hence, one root  (–1, 0) & other  (0, 2)
{other root may also be '1'}
8.
(A, B, D)
Sol.
|B – A| = 2cot–12 –
1 
1 
–1 
–1 
 – cot 2  + 3cot–1 3 –
 – cot 3 
2 2
3 2


=
5
10 –1
5
cot–12 +
cot 3 –
2
3
12
=
1 5
5  –1 1
5
 tan –1  + cot –1 3 –
tan
2
3 6
2 
12
=
5 5 5
5    5 –1
–1
  + cot 3 – 12  24  6 cot 3
2 4 6
a = 5, b = 24, c = 5, d = 6
Get 10% Instant Discount On Unacademy Plus [Use Referral Code : MCSIR] 3
Maths IIT-JEE
9.
Sol.
(A)
MC SIR
(A, B, C)
Line passing through P will be
(A1x + B1y + C1) +  (A2x + B2y + C2) = 0
 it passes through origin
.......(1)
C1
 C1 + C2 = 0 = – C
2
from (1)
(B)
C2 (A1x + B1y + C1) = C1 (A2x + B2y + C2)
C2 (A1x + B1y) = C1 (A2x + B2y)
Slope of line will be m hence
slope =
– ( A1   A 2 )
m
B1  B2
(B1m + A1) +  (B2m + A2) = 0
from (1)
(A1x  B1y  C2 ) A 2 x  B2 y  C 2

A1  mB1
A 2  mB2
m

(C)
it can be evaluated from (B) after replacing m by
10.
Sol.
(ACD)
(A) True by extreme value theorem
(B) As f(x) is continuous in [a, b] and has extreme values, it must be bounded
(C) True by intermediate value theorem.
f(x)
M
A
m
x
x0
(D) True because the function will retain its sign between two roots as it is continuous.
x1
x2
x3
x4
x
Get 10% Instant Discount On Unacademy Plus [Use Referral Code : MCSIR] 4
Maths IIT-JEE
MC SIR
11.
(A, D)
Sol.
  
(A) Let sin–1 x = ,   – , 
 2 2

x = sin

y = cos–1
cos2 
= cos–1 |cos|
= cos–1 (cos)

  
 –  ,    – 2 , 0



= 
  ,   0,  
 2 

 – sin –1 x ; x  [–1, 0]
= 
 sin –1 x ; x  [0,1]
(B)

Let x = – t t  (1, )
tan–1x + tan–1 (1)
= tan–1 1 – tan–1 t
1– t 

= tan–1 
1 t 
1 x 

= tan–1 
1– x 
(C)

Let cos–1 x = x = cos 
y = cos–1 (cos3)
 
 1
If x  0,  , = cos–1 x   , 
3 2
 2


 3 
3 , 
 2
y = cos–1 (cos3) = –3+ 2 = 2– 3cos–1x
cos–1 (cosx)
0

2
x
Get 10% Instant Discount On Unacademy Plus [Use Referral Code : MCSIR] 5
Maths IIT-JEE
(D)


MC SIR
Let tan–1 x = 
x = tan
y = tan–1 (tan3)
 1 1 
,

If x   –
3 3




12.
Sol.
  
 = tan–1 x   – , 
 6 6
  
3  – , 
 2 2
y = 3= 3tan–1 x
(A,B)
40(x – {x}) = 4x2 + 51
40[x] = 4x2 + 51
x=
if
40[ x ] – 51
40[ x ] – 51

4
2
[x] = 2 then x =
29
2
[x] = 3 then x =
69
2
[x] = 4 then x =
109
2
[x] = 5 then x =
149
2
[x] = 6 then x =
189
2
[x] = 7 then x =
229
2
[x] = 8 then x =
269
2
[x] = 9 then x =
309
2
hence x =
Also
29
,
2
189
,
2
5
29
>
> 2.3
2
2
229
,
2
&
269
2
16.5
269
<
< 8.25
2
2
Get 10% Instant Discount On Unacademy Plus [Use Referral Code : MCSIR] 6
Maths IIT-JEE
13.
(2)
Sol.
Let L =
L = cos
MC SIR
1 1 1 1

2 2 2 2

4
1 1 1 1 1


........
2 2 2 2 2
1 1

 cos 
2 2
4
1 1 1 1
 

 cos  .......
2 2 2 2
4
sin(  / 2)
2



= cos . cos . cos ....... =
=
/ 2

4
8
 16 
14.
(19)
Sol.
1

 {( n  1) – x}{x – n} , n  x  n  2
f(x) = 
1
(n  1) – x
,
n

 x  n 1

2
 f(x) = 0  x  I
with lim f(x) = f(n) = 0
xn
1 1

& lim1 f(x) = f  n   =
2 2
x n 

2
hence f(x) is continuous  x  R
{(2n  1) – 2 x}

 2 {(n  1) – x}{x – n}
Also, f '(x) = 

–1

, nxn
, n
1
2
1
 x  n 1
2
LHD at x  n is – 1

  f(x) is N.D. at x = n
RHD at x  n is infinity
Also at x = n +
1
,
2
LHD = 0 , RHD = – 1
 points of N.D of f(x) in x  (–5, 5)  x = n, x = n +
15.
Sol.
1
 19 points
2
(3)
Let g(x) = x2 – 2mx – 4(m2 + 1) ; D1 = 4 (5m2 + 4)
h(x) = x2 – 4x – 2m (m2 + 1) ; D2 = 8(2 + m3 + m)
For f(x) to have exactly three distinct linear factors, g(x) & h(x) must have total 3 distinct zeros.
I case D2 = 0 m3 + m + 2 = 0
 (m + 1)(m2 – m + 2) = 0 m = – 1
Get 10% Instant Discount On Unacademy Plus [Use Referral Code : MCSIR] 7
Maths IIT-JEE
MC SIR
for m = – 1
h(x) = x2 – 4x + 4 (x – 2)2
g(x) = x2 + 2x – 8 (x + 4) (x – 2)
both have total two zeros – 4, 2
Case II If g(x), h(x) have exactly one common root (let )
g() – h() = {2 – 2m– 4m2 – 4} – (2 – 4– 2m3 – 2m) = 0
 2(2 – m) + 2m2 (m – 2) + 2 (m – 2) = 0
 2(2 – m) { – m2 – 1} = 0 = m2 + 1 or m = 2
 f() = 0 f(m2 + 1) = 0
(m2 + 1)2 – 2m (m2 + 1) – 4(m2 + 1) = 0
(m2 + 1) {m2 + 1 – 2m – 4} = 0
(m2 + 1) (m2 – 2m – 3) = 0 (m2 + 1) (m – 3) (m + 1) = 0
m = 3, – 1
hence finally m can be 3 only as for m = 2, both roots are common.
16.
Sol.
(27)
Let the number N = 10x + 6 (also let it be an 'n' digit number). If we erase digit 6 then number will
convert into x hence by the given information
6.10n–1 + x = 4 (10x + 6)
6.
10n–1 =
6(10n –1 – 4)
39x + 24 x =
39
2(10n –1 – 4)
it will be smallest if n – 1 = 5 n = 6
13
and x = 15384

N = 153846
x=
17.
Sol.
(4)
ab + bc + ca = (a + c)b + ac
= (a + c)
{4 – (a  c)}
+ ac
2
(a  c) 2
= 2(a + c) –
+ ac
2
= 2(a + c) –
1 2
(a + c2)
2
= –
1
{a2 – 4a + c2 – 4c}
2
= –
1
[(a–2)2 + (c – 2)2 – 8]
2
(ab + bc + ca)max = –
1
(–8) = 4
2
Get 10% Instant Discount On Unacademy Plus [Use Referral Code : MCSIR] 8
Maths IIT-JEE
18.
Sol.
MC SIR
(1)
2cosx 3siny = (z – 1)2 + 6
RHS has minimum value 6 when z = 1
also LHS can be maximum 6 when cosx = 1 & siny = 1
hence x = 2n
and
y = 2m +

, m , n  I and z = 1
2
Get 10% Instant Discount On Unacademy Plus [Use Referral Code : MCSIR] 9