JEE Main Online Exam 2019 - Career...

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JEE Main Online Exam 2019

Questions & Solutions 12th April 2019 | Shift - I

MATHEMATICS

Q.1 The integral xx1x2

4

3 dx is equal to :

(Here C is a constant of integration)

(1) Cx

1xlog 2

3

e

(2) Cx

1xlog

21

2

3

e

(3) Cx

1xlog3

e (4) C

x)1x(log

21

3

23

e

Ans. [3]

Sol. xx1x2

4

3 dx

xx)2x2()1x4(

4

33dx

xx1x4

4

3 dx – x

12 dx

x4 + x = t (4x3 + 1) dx = dt

tdt – dx

x12

n|t| – 2nx + C

n 2

4

xxx + C n

x1x3 + C

Q.2 For x (0, 3/2), let f(x) = )x(g,x = tan x and h(x) = 2

2

x1x1

If (x) = ((hof)og)(x), then

3 is equal to :

(1) 127tan (2)

1211tan (3)

12tan (4)

125tan

Ans. [2] Sol. (x) = [(hof)og](x)

hof(x) = x1x1

(hof)g(x) = xtan1xtan1

CAREER POINT



3 =

3tan1

3tan1

3 = tan

34= tan

1211

Q.3 Let P be the point of intersection of the common tangents to the parabola y2 = 12x and the hyperbola

8x2 – y2 = 8. If S and S' denote the foci of the hyperbola where S lies on the positive x-axis then P divides SS' in a ratio :

(1) 14 : 13 (2) 13 : 11 (3) 5 : 4 (4) 2 : 1 Ans. [3]

Sol. genttancommon1

8yx:C

x12y:C2

22

21

sameBoth8)m(1mxy:C

m3mxy:CofTangent

22

1

2

m3

= m2 – 8

9 = m2 (m2–8) let m2 = t

t (t–8) = 9 t2 –8t – 9 = 0

(t–9) (t +1) = 0 m2 = 9

m = –3, 3 common tangent y = 3x + 1 y = –3x –1

P intersection of tangents P

0,31

foci of hyperbola a = 1, b = 2 2 8 = 1 (e2–1) e = 3 is S(3, 0), S(–3, 0)

x

k 1

S(3, 0) P

0,31 S(–3, 0)

31

= 1k

3k3

k = 5 : 4

CAREER POINT



Q.4 If

20

),n(mdxecxcosxcot

xcot then mn is equal to

(1) – 1 (2) 1 (3) 21

(4) 21

Ans. [1]

Sol. I =

2/

0

dxecxcosxcot

xcot

I =

2/

0xcos1

xcos = 2/

02

2

2xcos2

12xcos2

I =

2/xsec211 2

2/

0

dx

I = 2/

0)2/xtan(

22x

12

= 2

2

m = 21 , n = –2

mn = –1

Q.5 If and are the roots of the equation 375 x2 – 25x – 2 = 0, then

n

1r

rn

1r nr

nlimlim is equal to :

(1) 116

7 (2) 35829 (3)

121 (4)

34621

Ans. [3]

Sol. + = 37525

= 375

2

& (–1, 1), then

n

lim

n

1r

rr )(

+ 2 ……….infinite + + 2........ infinite

11

)1)(1(

)1()1( =

121

CAREER POINT



Q.6 The number of solutions of the equation 1 + sin4 x = cos2 3x, x

25,

25 is :

(1) 5 (2) 3 (3) 7 (4) 4 Ans. [1]

Sol. 1 + sin4x = cos2 3x ; x

25,

25

L.H.S. ≥ 1 R.H.S. ≤ 1 Both satisfy when L.H.S. = R.H.S. = 1 sin4x = 0 ; cos23x= 1 x = –2, –, 0, , 2 total 5 solution Q.7 The equation |z – i| = |z – 1|, i = ,1 represents :

(1) a circle of radius 1. (2) the line through the origin with slope – 1

(3) a circle of radius .2

1 (4) the line through the origin with slope 1.

Ans. [4] Sol. |z – i| = | z – 1| z = x + iy |x + i(y – 1)| = |(x – 1) + iy| x2 + (y – 1)2 = (x – 1)2 + y2 y = x straight line having slope (+)ve 1, pass origin (0, 0)

Q.8 If ey + xy = e, the ordered pair

,dxyd

,dxdy

2

2at x = 0 is equal to :

(1)

2e

1,e1 (2)

2e1,

e1 (3)

2e1,

e1 (4)

2e1,

e1

Ans. [1] Sol. ey + xy = e at x = 0; ey + 0 = e y = 1 y(0) = 1 diff. w.r.t. x ey. y1 + xy1 + y = 0 e(1).y1 + 0 + 1 = 0 y1 = –1/e at x = 0 diff again w.r.t. to x ey.y2 + y1 ey.y1 + xy2 + y1 + y1 = 0 at x = 0

CAREER POINT



e(1) (y2) + 2

e1

e1 + 0.y2 + 2

e1 = 0

y2 = 1/e2 x = 0

2e1,

e1

Q.9 The number of ways of choosing 10 objects out of 31 objects of which 10 are identical and the remaining 21

are distinct, is : (1) 220 – 1 (2) 220 (3) 221 (4) 220 + 1

Ans. [2] Sol. total 31

10identical 10 & distinct 21

S = 21C10 + 21C9(1) + 21C8(1) + 21C7 + 21C6 …… 21C0(1) S = 21C11 + 21C12 + 21C13 …… 21C21 2S = 221 S = 220 Q.10 For x R, let [x] denote the greatest integer < x, then the sum of the series

10099

31.....

1002

31

1001

31

31 is :

(1) – 153 (2) – 135 (3) – 133 (4) – 131 Ans. [3]

Sol.

)times67(110066

31.....

1001

31

31

+

)times33(210099

31.....

10068

31

10067

31

= – 67 – 66 = – 133 Q.11 If m is the minimum value of k for which the function f(x) = 2xkxx is increasing in the interval [0,3]

and M is the maximum value of f in [0, 3] when k = m, then the ordered pair (m, M) is equal to :

(1) (5, 63 ) (2) (4, 33 ) (3) (4, 23 ) (4) (3, 33 )

Ans. [2] Sol. f(x) : f (x) ≥ 0 ; & f(x1) ≤ f(x2), x[0, 3]

f(0) < f(3) f (x) = 2xkx2

)x2k(x

+ 2xkx

0 < 9k33 = 2

2

xkx2

)xkx(2)x2k(x

≥ 0

k > 3 = 2

2

xkx2

x4kx3

≥ 0

& kx – x2 > 0

CAREER POINT



3kx – 4x2 ≥ 0 x (3k – 4x) ≥ 0 x[0,3] x (4x – 3x) ≤ 0 k > 4 k = 4 minimum value of k m = 4

f(3) = M = 9)3(43 = 33

M = 33

(m, M) = (4, 3 3 ) Q.12 If the angle of intersection at a point where the two circles with radii 5 cm and 12 cm intersect is 90°, then

the length (in cm) of their common chord is :

(1) 5

13 (2) 1360 (3)

13120 (4)

213

Ans. [3] Sol.

D 12 5

C2 13 C1

A

B

AC1C2

tan = 5

12 sin = 1312 …..(i)

ACD:

sin = 5

2/AB …..(ii)

(i) & (ii)

5.2

AB = 1312

AB =13

120

Q.13 The coefficient of x18 in the product (1 + x) (1 – x)10 (1 + x + x2)9 is :

(1) 126 (2) – 84 (3) – 126 (4) 84 Ans. [4] Sol. coefficient of x18 (1 + x) (1 – x)10 (1 + x + x2)9 (1 + x) (1 – x) (1 – x)9 (1 + x + x2)9 (1 – x2) (1 – x3)9

CAREER POINT



(1 – x2) [9Cr (–1)r (x3r)]

9Cr (–1)r x3r – 9Cr (–1)r x3r+2

for x18

3r = 18 3r+2=18

r = 6 r not possible

then coefficient of x18 is 9C6 (–1)6 = 84

Q.14 Consider the differential equation, y2dx +

y1x dy = 0, If value of y is 1 when x = 1, then the value of x

for which y = 2, is :

(1) e

123 (2) e

23 (3)

e1

21 (4)

e1

25

Ans. [1]

Sol. dydx + 2y

x = 3y1

I.F. = dyy/1 2e = e–1/y

sol. y/1xe = y/1e (1/y3)dy

y1

= t

2y1 dy = dt

xet = dt).t(et

x.et = – tt ee.t + c

x = –t + 1 + c.e–t

x = y1 +1 + c. y/1e

give y(1) = 1

1 = 1 +1 + c.e1/1

c = –e–1

x = y1 + 1 + (–)e–1. e1/y

at y = 2

x = 23 –

e1

CAREER POINT



Q.15 If the area (in sq. units) of the region {(x, y) : y2 4x, x + y 1, x 0, y 0} is a 2 + b, then a – b is equal

to :

(1) 38 (2) 6 (3)

310 (4)

32

Ans. [2] Sol. C1 : y2 ≤ 4x C2 : x + y ≤ 1 x ≥ 0 y ≥ 0

(0,1) P A(2 2 -2)

(1, 0) B (0, 0) O

y2 = 4x ; y2 = 4 (1–y) y2 + 4y + 4 = 0

y = 2 2 –2 , – 2 2 –2

Area : shaded region of curve OAB

A = Area of OBP – Area of region OAP

OBP = 21 × 1 × 1 =

21

Area of OAP = 222

0

2

4y dy +

1

222

)y1( dy

= 121 222

03y

+ 1

222

2

2yy

= 121 [(2 2 –2)3] +

2)222()222(

211

2

= 623 – 2

38

A = 21 –

623 +

328

a = 38 , b = –

620

a – b = 6

CAREER POINT



Q.16 If the data x1, x2, …… x10 is such that the mean of first four of these is 11, the mean of the remaining six is 16 and the sum of squares of all of these is 2,000 ; then the standard deviation of this data is :

(1) 2 (2) 2 (3) 22 (4) 4 Ans. [2] Sol. x1 + x2 + x3 + x4 = 44 x5 + x6 ….. x10 = 96 xi = 140

40

140x = 14

2x = 2000

SD = 22

nx

nx

=2

10140

102000

= 196200 = 2

Q.17 If A is a symmetric matrix and B is a skew-symmetric matrix such that A + B =

1532

, then AB is equal

to :

(1)

4124

(2)

4124

(3)

4124

(4)

4124

Ans. [4] Sol. A is symmetric AT = A B is skew 8 symmetry BT = – B

A + B =

1532

… (i)

Transpose

AT + BT =

1352

A – B =

1352

….(ii)

From (i) + (ii)

A =

1442

From (i) – (ii)

B =

0110

then AB =

4124

CAREER POINT



Q.18 If the normal to the ellipse 3x2 + 4y2 = 12 at a point P on it is parallel to the line, 2x + y = 4 and the tangent to the ellipse at P passes through Q(4,4) then PQ is equal to :

(1) 261 (2)

2221 (3)

2157 (4)

255

Ans. [4]

Sol. Ellipse : 9y

4x 22

= 1

Normal at P : (2cos, 3 sin) is 2xsec – 3 cosec y = 4 = 3 = 1 Normal at P: 2x sec– 3 y cosec

Slope of normal =

eccos3sec2 =

32

cossin

Normal parallel to 2x + y = 4

then 3

2 tan = –2; tan = – 3

32

Point P

23,1 , Q(4, 4)

PQ = 2

2234)14(

= 4

125 = 255

Q.19 Let a = k2j2i3 and b

k2j2i be two vectors. If a vector perpendicular to both the vectors

ba

and ba

has the magnitude 12 then one such vector is : (1) )kj2i2(4 (2) )kj2i2(4 (3) )kj2i2(4 (4) )kj2i2(4

Ans. [1] Sol. k2j2i3a

k2j2ib

j4i4ba

k4i2ba

A vector r perpendicular to )ba(&)ba(

& magnitude 12 n.12r

n = |)ba()ba(|

)ba()ba(

)ba()ba(

= 402044kji

= )8(k)16(j)16(i = ]kj2i2[8 | )ba()ba(

| = 8.3

n = 3

)kj2i2(

r = 4. )kj2i2(

CAREER POINT



Q.20 If B =

13120125

is the inverse of a 3 × 3 matrix A, then the sum of all values of for which

det(A) + 1 = 0, is : (1) 2 (2) – 1 (3) 0 (4) 1

Ans. [4] Sol. If B is inverse of A then

AB = I det(AB) = det(I)

det(A).det(B) = 1 Given det (A) = –1 then det (B) = –1

13

120125

= –1

5(–2 –3) – 2[0 – ] + 1 [–2] = –1 22 – 2 – 25 = – 1 22 – 2 – 24 = 0 ( – 4)( + 3) = 0 ; = 4, – 3 Sum of value of = 4 – 3 = 1 Q.21 A 2 m ladder leans against a vertical wall. If the top of the ladder begins to slide down the wall at the rate

25 cm/sec, then the rate (in cm/sec.) at which the bottom of the ladder slides away from the wall on the horizontal ground when the top of the ladder is 1 m above the ground is :

(1) 3

25 (2) 25 (3) 325 (4) 3

25

Ans. [4] Sol. x2 + y2 = 4 Diff. w.r.t. t

2x dtdx + 2y

dtdy = 0

2m

x

dtdx =2

s/cm25dtdy

y

At y = 1; x = 3 then

sec/cm3

25dtdx

CAREER POINT



Q.22 The equation y = sinx sin (x + 2) – sin2 (x + 1) represents a straight line lying in : (1) first, second and fourth quadrants (2) first, third and fourth quadrants (3) second and third quadrants only (4) third and fourth quadrants only

Ans. [4] Sol. y = sinx sin(x + 2) – sin2(x + 1)

y = )]2xsin(xsin2[21

– )]1x(sin2[21 2

y = )]2x2cos()2[cos(21

– )]1x(2cos1[21

y = )]12[cos21

y = (–) 1sin)2(21 2

y = – sin21

y = – sin2l Graph IIIrd & IVth Q.23 If the truth value of the statement p (~q r) is false (F), then the truth values of the statements p, q, r are

respectively : (1) T, F, T (2) F, T, T (3) T, T, F (4) T, F, F

Ans. [3] Sol. p (~ q r) is false It is true when p T & (~ q r) = false It will true : ~ q false & r false ~q F | r F q T Truth value of p, q, r T, T, F

Q.24 Let a random variable X have a binomial distribution with mean 8 and variance 4. If P(X < 2) = ,2k16 then k

is equal to : (1) 17 (2) 1 (3) 137 (4) 121

Ans. [3] Sol. Given : np = 8 ; nqp = 4

q = 21 p =

21 n = 16

P(x 2) = P(x = 0) + P(x = 1) + (x = 2)

= 16C0 016

21

21

+ 16C1

15

21

1

21

+ 16C2

14

21

2

21

=

16

21

[1 + 16 + 120] = 162

137

k = 137

CAREER POINT



Q.25 Let Sn denote the sum of the first n terms of an A.P. If S4 = 16 and S6 = – 48, then S10 is equal to : (1) – 320 (2) – 380 (3) – 410 (4) – 260

Ans. [1] Sol. Sn = Sum of n terms of an A.P. S4 = 16 = a + 3d …..(i) S6 = –48 = a + 5d …..(ii) From (i) & (ii) d = – 32 a = 112

S10 = 2

10 [2.(112) + (10 – 1)(–32)]= 5[–64]

S10 = – 320

Q.26 Let f : R R be a continuously differentiable function such that f(2) = 6 and f(2) = .481 If

f(x)

63 ),x(g)2x(dtt4 then )x(gLim

2xis equal to :

(1) 18 (2) 36 (3) 12 (4) 24 Ans. [1]

Sol. )x(f

6

3 dt.t4 (x – 2)g(x) ; f(2) = 6 ; f '(2) = 481

g(x) = )2x(

dtt4)x(f

6

3

2x

lim

)2x(

dtt4)x(f

6

3

at x = 2

00 form

L-hospital rule

2x

lim

1

)x('f.)]x(f[4 3

At x = 2, 4[f(2)]3.f '(2)

4(6)3

481 = 18

Q.27 The value of sin–1

1312

53sin 1 is equal to :

(1)

6563sin 1 (2)

659cos

21 (3)

6533cos 1 (4)

6556sin

21

Ans. [4]

Sol. sin–1

1312 – sin–1

53

= sin–1

1691441

53

2591

1312

CAREER POINT



= sin–1

6533

= cos–1

6556

= 2 – sin–1

6556

Q.28 If the volume of parallelopiped formed by the vectors kj,kji and ki is minimum, then is

equal to :

(1) 3

1 (2) 3 (3)

31 (4) 3

Ans. [3] Sol. Volume of parallelopiped

10

1011

= 1[1 – 0] – [– 2] + 1 [–]

V = |1 + 3 – |

13ddV

0ddV

= 3

1 , –3

1

6d

Vd2

2 for minimum

2

2

dVd

> 0 at = 3

1

Q.29 If three of the six vertices of a regular hexagon are chosen at random, then the probability that the triangle

formed with these chosen vertices is equilateral is :

(1) 101 (2)

103 (3)

203 (4)

51

Ans. [1] Sol. Number of total triangle = 6C3

Equilateral = 2

Prob. = 3

6C2 =

101

CAREER POINT



Q.30 If the line 11z

21y

32x

intersects the plane 2x + 3y – z+13 = 0 at a point P and the plane

3x + y + 4z = 16 at a point Q, then PQ is equal to :

(1) 72 (2) 14 (3) 142 (4) 14

Ans. [3]

Sol.

11z

g1y

32x

General point (3 + 2, 2 – 1, – + 1) = intersect plane 2x + 3y – z + 13 = 0 at P then 2(3+ 2) + 3 (2 – 1) – (– + 1) + 13 = 0 = – 1 P(– 1, – 3, 2) Line intersect plane : 3x + y + 4z = 16 at Q then 3(3 + 2) + 2 – 1 + 4(– +1) = 16 = 1

Q(5, 1, 0) then PQ = 222 )20()31()15(

PQ = 2 14

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