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JEE Main Online Paper

JEE Main Online Exam 2019

Questions & Solutions 10th April 2019 | Shift - I

Mathematics Q.1 The region represented by |y–x| 2 and | x + y| 2 is bounded by a :

(1) rhombus of area 28 sq. units (2) square of side length 22 units (3) square of area 16 sq. units (4) rhombus of side length 2 units Ans. [2]

Sol. shown figure is square with side length 22 .

(0,2)

(2,0)

(0,–2)

(–2, 0)

y

x

Q.2 All the pairs (x, y) that satisfy the inequality 5xsin2–xsin 22

ysin 24

1 1 also satisfy the equation

(1) sin x = |ysin| (2) sin x = 2 sin y (3) 2 sin x = sin y (4) |xsin|2 = 3 sin y Ans. [1]

Sol. 5xsin2–xsin 22 ysin– 2

4 1

4)1–x(sin 2

2 ysin24

4)1–x(sin 2 0 2

2 sin2 y 2

this is possible only if sin x = 1 & |sin y| = 1

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Q.3 If and are the roots of the quadratic equation, x2 + x sin – 2 sin = 0,

2,0 , then

2412–12–

1212

)–( is equal to :

(1) 6

12

)8–(sin2

(2) 12

6

)8(sin2

(3) 12

12

)8(sin2

(4) 12

12

)4–(sin2

Ans. [3]

Sol. x2 + x sin – 2 sin = 0

+ = – sin

= – 2 sin

Now, 24

1212

1212

)–(11

= 24

12

)–()(

= 122

12

4–)(

)(

= 12

2 4–)(

= 12

2 sin8sinsin2–

= 12

12

)8(sin2

Q.4 If a > 0 and z = i–a)i1( 2 , has magnitude

52 , then z is equal to :

(1) – 51 +

53 i (2) i

53–

51– (3) i

53–

51 (4) i

51–

53–

Ans. [2]

Sol. z = i–a)i1( 2 =

1a)ia(i2

2

| z | = 1a

22

= 52 a = 3

z = 10

)i–3(i2–

5

i3–1–

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Q.5 If y = y(x) is the solution of the differential equation dxdy = (tan x – y) sec2 x, x

2,

2– , such that

y (0) = 0, then

4–y is equal to :

(1) 21 – e (2) e – 2 (3)

e12 (4) 2–

e1

Ans. [2]

Sol. dxdy = (tan x – y) sec2 x

dxdy + y sec2 x = tan x sec2 x

Let tan x = t sec2 x = dxdt

dtdy = (t – y)

dtdy + y = t (Linear differential equation)

After solving we get y et = et (t – 1) + c y = (tan x – 1) + ce – tan x y (0) = 0 c = 1 y = tan x –1 + e– tan x

So, y

4– = e – 2

Q.6 If the length of the perpendicular from the point (, 0, ) ( 0) to the line, 1x =

01–y =

1–1z is

23 , then

is equal to : (1) 2 (2) 1 (3) –2 (4) –1 Ans. [4] Sol.

B(, 0, )

A(0, 1, –1) C (, 1, – –1)

1x =

01–y =

1–1z =

A point on this line is A(0, 1, –1) AC BC = 0

we get = – 21

C

21–,1,

21–

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32BC

2

22

21)1(

21

=

32

= 0, –1 = – 1 ( 0)

Q.7 If 1 = x1cos1x–sin–

cossinx

and 2 =

x12cos1x–2sin–2cos2sinx

, x 0 ; then for all

2,0 :

(1) 1 – 2 = x (cos 2 – cos 4) (2) 1 + 2 = – 2x3

(3) 1 + 2 = – 2(x3 + x –1) (4) 1 – 2 = – 2x3 Ans. [2]

Sol. 1 = x1cos1x–sin–

cossinx

= x (– x2 – 1) – sin (– x sin – cos ) + cos (– sin + x cos ) – x3

2 = x12cos1x–2sin–2cos2sinx

– x3 1 + 2 = – 2x3 Q.8 Assume that each born child is equally likely to be a boy or a girl. If two families have two children each,

then the conditional probability that all children are girls given that at least two are girls is :

(1) 101 (2)

171 (3)

111 (4)

121

Ans. [3]

Sol. P (Boy) = P(girl) = 21

Required probability = girlstwoleastAt

girlsfourall

= 4

24

4

34

4

4

21C

21C

21

21

= 111

Q.9 Which one of the following Boolean expressions is a tautology ? (1) (p q) (~ p q) (2) (p q) (p q) (3) (p q) (p q) (4) (p q) (p q) Ans. [2]

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Sol. from options (p q) (~ p q) (p q) (p q) Not a tautology (p q) (p q) p (q q) tautology (p q) (p q) p (q q) Not a tautology (p q) (p q) p (q q) Not a tautology Q.10 Let f(x) = x2, x R. For any A R, define g (A) = { x R : f(x) A}. If S = [0,4], then which one of the

following statements is not true ? (1) g(f(S)) S (2) f(g(S)) S (3) f(g(S)) f(S) (4) g(f(S)) = g(S) Ans. [4] Sol. g(S) = [–2, 2] f(g(S)) = [0,4] = S f(S) = [0, 16] f(g(S)) f(S) g(f(S)) = [–4, 4] g(S) therefore , g(f(S)) S

Q.11 If 1x

lim

1–x1–x4

= kx

lim

22

33

k–xk–x , then k is :

(1) 23 (2)

38 (3)

34 (4)

83

Ans. [2]

Sol. 1x

lim

1–x1–x4

= 1x

lim

(x + 1) (x2 + 1) .....(i)

kx

lim

22

33

k–xk–x =

k2kkk 222 ....(ii)

(i) = (ii)

k = 38

Q.12 If a directrix of a hyperbola centred at the origin and passing through the point (4, –2 3 ) is 5x = 4 5 and

its eccentricity is e, then : (1) 4e4 – 24e2 + 27 = 0 (2) 4e4 – 24e2 + 35 = 0 (3) 4e4 – 12e2 – 27 = 0 (4) 4e4 + 8e2 – 35 = 0 Ans. [2]

Sol. Let hyperbola be 2

2

2

2

by–

ax = 1 & passes through (4, 32– ) therefore

22 b12–

a16 = 1 .....(i) b2 = a2 (e2 –1)

x = ea

554

a2 = 2e5

16 ....(ii)

one solving (i) & (ii) 4e4 – 24 e2 + 35 = 0

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Q.13 If f (x) =

0x,x

x–xx0x,q

0x,x

xsinx)1psin(

2/3

2

is continuous at x = 0, then the ordered pair (p, q) is equal to :

(1)

21,–

23– (2)

23,

21– (3)

21,

23– (4)

21,

25

Ans. [3]

Sol. RHL = 0x

lim 2/3

2

xx–xx

= 0x

lim x

1–x1 = 21

LHL = –0x

lim

x

xsinx)1psin(

= (p + 1) + 1 = p + 2 for function to be continuous LHL = RHL = f(0)

(p, q) =

21,

23–

Q.14 If the coefficients of x2 and x3 are both zero, in the expansion of the expression (1 + ax + bx2) (1 – 3x)15 in

powers of x, then the ordered pair (a,b) is equal to : (1) (28, 861) (2) (28, 315) (3) (–21, 714) (4) (–54, 315) Ans. [2]

Sol. coefficient of x2 = 215C × 9 – 3a )C( 1

15 + b = 0

215C × 9 – 45 a + b = 0 ...(i)

coefficient of x3 = –27 × 315C + 9a × 2

15C – 3b × 115C = 0

– 273 + 21a – b = 0 ...(ii) (i) + (ii) –24 a + 672 = 0 a = 28 b = 315 Q.15 If the circles x2 + y2 + 5Kx + 2y + K = 0 and 2(x2 + y2) + 2Kx + 3y –1 = 0, (KR), intersect at the points

P and Q, then the line 4x + 5y – K = 0 passes through P and Q, for : (1) exactly two values of K (2) no value of K. (3) exactly one value of K (4) infinitely many values of K Ans. [2] Sol. equation of common chord

4 kx + 21 y + k +

21 = 0 ...(i)

and given line 4x + 5y – k = 0 ...(ii)

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on comparing (i) & (ii) we get

k = k–

2/1k101

No. real value of k exist

Q.16 The sum 222

333

22

33

2

3

321)321(7

21)21(5

113

+......... upto 10th term is :

(1) 660 (2) 680 (3) 600 (4) 620 Ans. [1]

Sol. Tn = )n........21(

)n......21)(2)1–n(3(222

333

=23 n(n + 1)

Sn = nT

= )1n(n23

on solving Sn = 2

)2n)(1n(n S10 = 660

Q.17 Let f : R R be differentiable at c R and f(c) = 0. If g(x) = )x(f , then at x = c, g is :

(1) differentiable if f ' (c) = 0 (2) differentiable if f ' (c) 0 (3) not differentiable (4) not differentiable if f ' (c) 0 Ans. [1]

Sol. g'(c) = 0h

lim

h

|)c(f|–|)hc(f|

= 0h

lim

h

|)hc(f| ( f (c) = 0)

= 0h

lim

h

|h|h

)c(f–)hc(f

= 0h

lim

|f '(c)| h

|h| = 0 if f ' (c) = 0

i.e. g(x) is differentiable at x = c if f '(c) = 0 Q.18 If Q(0, –1 –3) is the image of the point P in the plane 3x – y + 4z = 2 and R is the point (3, –1, –2), then the

area (in sq. units) of PQR is :

(1) 265 (2) 132 (3)

291 (4)

491

Ans. [3]

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Sol.

P

M

R

Q

MQ = 1619

|2–12–1|

= 26

13

= 2

13

PM = 26

RQ = 1019

RM = 27

213–10

Ar (PQR) = 21 × 26 ×

27 =

291

Q.19 If 22 )10x2–x(dx = A

10x2–x)x(f

31–xtan 2

1– + C where C is a constant of integration then :

(1) A =541 and f(x) = 9(x–1)2 (2) A =

541 and f(x) = 3(x–1)

(3) A =811 and f(x) = 3(x–1) (4) A =

271 and f(x) = 9(x–1)

Ans. [2]

Sol. 22 )10x2–x(dx =

22 9)1–x(

dx

Let x – 1 = 3 tan dx = 3 sec2 d

dcos271 2 =

541 d2cos1 =

541

22sin

= 541 C

10x2–x)1–x(3

31–xtan 2

1–

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Q.20 The line x = y touches a circle at the point (1,1). If the circle also passes through the point (1, – 3), then its radius is :

(1) 3 (2) 2 (3) 2 2 (4) 23 Ans. [3] Sol.

(1,1)

Equation of circle is given as S + L = 0 (x – 1)2 + (y – 1)2 + (x – y) = 0 passes through(1, – 3) 16 + × 4 = 0 = – 4 (x – 1)2 + (y – 1)2 – 4 (x –y) = 0 r = 22

Q.21 The value of dx)]x3cos1(x2[sin2

0

, where [t] denotes the greatest integer function is :

(1) 2 (2) (3) –2 (4) – Ans. [4]

Sol. I = dx)]x3cos1(x2[sin2

0

2I =

2

0

dx]x3cosx2sin–x2sin[–)]x3cos1(x2[sin

2I = 2

0

dx–

2I =

0

dx–2

I =

0

dx– –

Q.22 Let A (3,0, –1), B(2, 10, 6) and C(1, 2, 1) be the vertices of a triangle and M be the midpoint of AC. If G divides BM in the ratio, 2 : 1, then cos (GOA (O being the origin) is equal to :

(1) 151 (2)

1061 (3)

301 (4)

1521

Ans. [1] Sol. G will be centroid of ABC G (2, 4, 2) OG = k2j4i2

OA = k–i3

cos (GOA) = |OA||OG|

OAOG = 151

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Q.23 If the system of linear equations x + y + z = 5 x + 2y + 2z = 6 x + 3y + z = , (, R), has infinitely many solutions, then the value of + is : (1) 10 (2) 9 (3) 12 (4) 7 Ans. [1] Sol. x + 3y + z – = a (x + y + z – 5) + b (x + 2y + 2z – 6) comparing coefficients we get a + b = 1 and a + 2b = 3 (a, b) (–1, 2) So, x + 3y + z – = x + 3y + 3z – = 7, = 3 Q.24 The number of 6 digit numbers that can be formed using the digits 0, 1, 2, 5, 7 and 9 which are divisible by

11 and no digit is repeated is : (1) 36 (2) 60 (3) 72 (4) 48 Ans. [2] Sol. Let the six digit number be abcdef for this number to be divisible by 11, |(a + c + e) – (b + d + f)| must be

multiple of 11 possibility is a + c + e = b + d + f = 12 Case : 1 {a, c, e} = {7, 5, 0} & {b, d, f} = {9, 2, 1} So, number of numbers = 2 × 2! × 3! = 24 Case : 2 {a, c, e} = {9, 2, 1} & {b, d, f} = {7, 5, 0} So, number of numbers = 3! × 3! = 36 total 24 + 36 = 60 Q.25 If for some x R, the frequency distribution of the marks obtained by 20 students in a test is :

Marks 2 3 5 7 Frequency (x + 1)2 2x – 5 x2 – 3x x

then the mean of the marks is (1) 3.0 (2) 2.8 (3) 2.5 (4) 3.2 Ans. [2]

Sol. Mean x =

i

ii

f

fx

if = (x + 1)2 + (2x –5) + (x2 – 3x) + x = 20 x = 3, – 4(rejected)

x =

i

ii

f

fx = 2.8

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Q.26 ABC is a triangular park with AB = AC = 100 metres. A vertical tower is situated at the mid-point of BC. If the angles of elevation of the top of the tower at A and B are cot–1 (3 2 ) and cosec–1 (2 2 ) respectively, then the height of the tower (in metres) is :

(1) 33

100 (2) 25 (3) 20 (4) 510

Ans. [3] Sol.

Q

h

P

100C A

100

B

x

cosec = 22 cot = 23

hx = 23 ...(i)

So 24 x–10

h = 7

1 .....(ii)

from (i) & (ii) h = 20 Q.27 If a1, a2, a3, ............... an are in A.P. and a1 + a4 + a7 + ........... + a16 = 114, then a1 + a6 + a11 + a16 is equal to : (1) 38 (2) 98 (3) 76 (4) 64 Ans. [3] Sol. a1 + a4 + a7 + a10 + a13 + a16 = 114

26 (a1 + a16) = 114

a1 + a16 = 38

So, a1 + a6 + a11 + a16 = 24 (a1 + a16)

= 2 × 38 76 Q.28 Let f(x) = ex – x and g(x) = x2 – x, x R. Then the set of all x R, where the function h(x) = (fog) (x) is

increasing, is :

(1) [0, ) (2)

21–,1–

,

21 (3) ),1[0,

21–

(4)

21,0 [1, )

Ans. [4]

CAREER POINT



Sol. h(x) = f(g(x)) h'(x) = f '(g(x)) g'(x) and f '(x) = ex –1 h'(x) = (eg(x) – 1) g'(x)

h'(x) = 0)1–x2()1–e( x–x 2

Case :1 1e x–x2 and 2x –1 0

x

21,0 .....(i)

Case : 2 1e x–x 2 and 2x –1 0

x [1, ) ....(ii) from (i) & (ii)

x

21,0 [1,)

Q.29 n

lim

3/4

3/1

3/4

3/1

3/4

3/1

n)n2(.......

n)2n(

n)1n( is equal to :

(1) 34 (2)3/4 (2)

43 (2)4/3 –

43 (3)

34 (2)4/3 (4)

34–)2(

43 3/4

Ans. [2]

Sol. n

lim3/1n

1r nrn

n1

= 3/11

0

)x1( dx = 43 (24/3 –1)s

Q.30 If the line x – 2y = 12 is tangent to the ellipse 2

2

ax + 2

2

by = 1 at the point

29–,3 , then the length of the latus

rectum of the ellipse is :

(1) 8 3 (2) 212 (3) 5 (4) 9

Ans. [4] Sol. Tangent at (3, – 9/2)

2ax3 – 2b2

y9 = 1

comparing with x – 2y = 12

2a3 = 2b4

9 = 121

a = 6 & b = 33

length of latus rectum = ab2 2

= 9

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