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CAREER POINT Ltd., CP Tower, IPIA, Road No.1, Kota (Raj.), Ph: 0744-6630500 www.careerpoint.ac.in 1 JEE Main Online Exam 2019 Questions & Solutions 12 th April 2019 | Shift - II MATHEMATICS Q.1 Let a 2 , 0 be fixed. If the integral tan x tan tan x tan dx = A(x) cos 2 + B(x) sin 2 + C, where C is a constant of integration, then the functions A(x) and B(x) are respectively : (1) x – and log e |cos(x – )| (2) x + and log e |sin(x – )| (3) x + and log e |sin(x + )| (4) x – and log e |sin(x – )| Ans. [4] Sol. dx sin x cos cos x sin sin x cos cos x sin = dx ) x sin( ) x sin( = dx ) x sin( ) 2 x sin( = dx ) x sin( 2 cos ) x sin( + dx ) x sin( 2 sin ) x cos( = (x – ) cos2 + sin 2 log |sin(x – )| + C Q.2 The general solution of the differential equation (y 2 – x 3 )dx – xydy = 0 (x 0) is : (where c is a constant of integration) (1) y 2 + 2x 2 + cx 3 = 0 (2) y 2 + 2x 3 + cx 2 = 0 (3) y 2 – 2x + cx 3 = 0 (4) y 2 – 2x 3 + cx 2 = 0 Ans. [2] Sol. (y 2 – x 3 )dx – xydy = 0 (x 0) y 2 – x 3 – xy dx dy = 0 xy dx dy – y 2 = –x 3 y dx dy x 1 y 2 = –x 2 … (i) Let y 2 = v 2y dx dy = dx dv put in eq n (i) 2 1 dx dv v x 1 = –x 2 dx dv + v x 2 = –2x 2 …(ii)
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JEE Main Online Exam 2019

Questions & Solutions 12th April 2019 | Shift - II

MATHEMATICS

Q.1 Let a

2,0 be fixed. If the integral

tan–xtantanxtan dx = A(x) cos 2 + B(x) sin 2 + C, where C is a

constant of integration, then the functions A(x) and B(x) are respectively : (1) x – and loge|cos(x – )| (2) x + and loge|sin(x – )| (3) x + and loge|sin(x + )| (4) x – and loge|sin(x – )|

Ans. [4]

Sol. dx

sinxcos–cosxsinsinxcoscosxsin

= dx

)–xsin()xsin(

= dx)–xsin(

)2–xsin(

= dx)–xsin(

2cos)–xsin(

+ dx)–xsin(

2sin)–xcos(

= (x – ) cos2 + sin 2 log |sin(x – )| + C Q.2 The general solution of the differential equation (y2 – x3)dx – xydy = 0 (x 0) is : (where c is a constant of integration)

(1) y2 + 2x2 + cx3 = 0 (2) y2 + 2x3 + cx2 = 0 (3) y2 – 2x + cx3 = 0 (4) y2 – 2x3 + cx2 = 0

Ans. [2] Sol. (y2 – x3)dx – xydy = 0 (x 0)

y2 – x3 – xydxdy = 0

xydxdy – y2 = –x3

ydxdy –

x1 y2 = –x2 … (i)

Let y2 = v 2ydxdy =

dxdv

put in eqn (i)

21

dxdv – v

x1 = –x2

dxdv + v

x2–

= –2x2 …(ii)

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I.F. = dxx2–

e = xn2–e = 2x1

solution of eqn (ii)

v × 2x1 =

dx

xx2– 2

2 – C

2xv = – 2x – C

y2 = –2x3 – cx2

y2 + 2x3 + cx2 = 0 Q.3 The equation of common tangent to the curves y2 = 16x and xy = –4, is :

(1) x – y + 4 = 0 (2) x + y + 4 = 0 (3) x – 2y + 16 = 0 (4) 2x – y + 2 = 0 Ans. [1]

Sol. y = mx + m4 is always tangent to y2 = 16x … (i)

if it is tangent to the xy = –4

x

m4mx = –4

m2x2 + 4x = –4m m2x2 + 4x + 4m = 0 for tangent D = 0 16 – 16m3 = 0 m = 1 put in eqn (i) y = x + 4 Q.4 A plane which bisects the angle between the two given planes 2x – y + 2z – 4 = 0 and x + 2y + 2z – 2 = 0,

passes through the point : (1) (1, –4, 1) (2) (1, 4, –1) (3) (2, 4, 1) (4) (2, –4, 1)

Ans. [4] Sol. Eqn of angle bisectors are

222 2)1(–2

4–z2y–x2

= ±

222 z21

2–z2y2x … (i)

Case I : take positive sign 2x – y + 2z – 4 = x + 2y + 2z – 2 x – 3y – 2 = 0 … (ii) Case-II : take negative sign

2x – y + 2z – 4 = –(x + 2y + 2z – 2)

2x – y + 2z – 4 = –x – 2y – 2z + 2

3x + y + 4z – 6 = 0 … (iii)

option (4) satisfy eqn (iii)

(2, –4, 1)

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Q.5 A triangle has a vertex at (1, 2) and the mid points of the two sides through it are (–1, 1) and (2, 3). Then the

centroid of this triangle is :

(1)

2,

31 (2)

35,

31 (3)

37,1 (4)

1,

31

Ans. [1] Sol.

A(1, 2)

F(2, 3)E(–1, 1)

B (x2, y2)

C (x3, y3)

21x2 = –1,

22y2 = 1

21x3 = 2 &

22y3 = 3

x2 = –3, y2 = 0 x3 = 3, y3 = 4

B(–3, 0) C(3, 4)

centroid

3

yyy,

3xxx 321321

3402,

333–1 =

2,

31

Q.6 The Boolean expression ~(p (~q)) is equivalent to :

(1) p q (2) (~p) q (3) q ~p (4) p q

Ans. [1]

Sol. – (p ~q) = p q

Q.7 Let A, B and C be sets such that A B C. Then which of the following statements is not true ?

(1) If (A – B) C, then A C (2) B C

(3) (C A) (C B) = C (4) If (A – C) B, then A B

Ans. [4] Sol. Let A = {1, 2, 3, 4} B = {3, 4, 5, 6} C = {1, 2, 3, 4, 7, 8}

Here A B = {3, 4} C

A – C = B

but A B

So not true (wrong) statement is 4th

If A – C B then A B

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Q.8 If the area (in sq. units) bounded by the parabola y2 = 4x and the line y = x, > 0, is 91 , then is equal to :

(1) 34 (2) 62 (3) 48 (4) 24 Ans. [4] Sol.

y

x

y2 = 4x & y = x 2x2 = 4x

x = 0 & x = 4

Area =

/4

0

dx)x–x4( = 91

2 ×

/4

0

2/3

23

x –

/4

0

2

2x =

91

34

× 2/3

2/32 x)2(

– 2 × 2

16

= 91

3

32 – 8 =

91

38 =

91 = 24

Q.9 If , and are three consecutive terms of a non-constant G.P. such that the equations x2 + 2x + = 0 and

x2 + x – 1 = 0 have a common root, then ( + ) is equal to : (1) (2) 0 (3) (4)

Ans. [4] Sol. ,, are in G.P. x2 + 2x + = 0 & x2 + x – 1 = 0 have a common roots. Both roots will be common

1 =

12 =

1– =

= , B = 2 , = –

( + ) =

–2

= 2

– 2 =

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Q.10 If 20C1 + (22) 20C2 + (32) 20C3 + ..... + (202)20C20 = A(2), then the ordered pair (A, ) is equal to : (1) (420, 19) (2) (420, 18) (3) (380, 18) (4) (380, 19)

Ans. [2] Sol. (1 + x)20 = 20C0 + 20C1x + 20C2x2 + ........ + 20C20x20 … (i) different eqn (i) w.r.t. x 20(1 + x)19 = 20C1.1 + 2.20C2x + ......... + 2020C20x19 … (ii) Multiply eqn (ii) by x 20x(1 + x)19 = 20C1.x+ 2.20C2x2 + ........ + 2020C20x20 … (iii) diff. eqn (iii) w.r.t. x 20[(1 + x)19 + 19x(1 + x)18] = 1.20C1 + 22.20C2x + ........ + (202)20C20x19 … (iv) put x = 1 in eqn (iv) 20(219 + 19.218) = 12.20C1 + 22.20C2 + ........ + (202) 20C20

= 20 × 218(2 + 19) = 20 × 21 × 218 = 420 × 218 A = 420, = 18

Q.11 The length of the perpendicular drawn from the point (2, 1, 4) to the plane containing the lines

r = ( ji ) + ( k–j2i ) and

r = ( ji ) + µ(– k2–ji ) is :

(1) 31 (2)

31 (3) 3 (4) 3

Ans. [4] Sol. Equation of plane containing both lines is

2–11–1–21

z1–y1–x = 0

(x – 1) (–4 + 1) + (y – 1) (1 + 2) + z (1 + 2) = 0 –3(x – 1) + 3(y – 1) + 3z = 0 –x + 1 + y – 1 + z = 0 –x + y + z = 0 distance from point (2, 1, 4) is

222 111

412–

= 3

Q.12 A circle touching the x-axis at (3, 0) and making an intercept of length 8 on the y-axis passes through the

point : (1) (1, 5) (2) ( 2, 3) (3) (3, 5) (4) (3, 10)

Ans. [4] Sol. Equation of required circle will be (x – 3)2 + (y ± r)2 = r2

x2 – 6x + 9 + y2 ± 2ry + r2 = r2 x2 + y2 – 6x ± 2ry + 9 = 0 …(i)

Length of y intercept = 2 c–f 2 f = ±r

8 = 9–r2 2 16 = r2 – 9 r = 5

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So eqn of required circle will be x2 + y2 – 6x ± 10y + 9 = 0

two circles

x2 + y2 – 6x + 10y + 9 = 0 … (ii)

x2 + y2 – 6x – 10y + 9 = 0 … (iii)

option 4th (3, 10) satisfy eqn (iii)

Q.13 A person throws two fair dice. He wins Rs. 15 for throwing a doublet (same numbers on the two dice), wins Rs. 12 when the throw results in the sum of 9, and loses Rs. 6 for any other outcome on the throw. Then the expected gain/loss (in Rs.) of the person is :

(1) 41 loss (2)

21 gain (3)

21 loss (4) 2 gain

Ans. [3] Sol.

win +15 +12 –6

Prob. 366

364

3626

Probability of doublet = 366

Probability of sum of 9 = 364

Other probability = 3626

Expected gain/loss = 15 × 366 + 12 ×

364 – 6 ×

3626

= 3690 +

3648 –

36156 =

21–

21–

So, 21 loss

Q.14 Let R and the three vectors a = k3ji ,

b = k–ji2 and

c = k3j2–i . Then the set

S = { : a ,

b and

c are coplanar}

(1) contains exactly two numbers only one of which is positive (2) is singleton (3) contains exactly two positive numbers (4) is empty

Ans. [4]

Sol. [a

b

c ] = 0

32–

–1231

= 0

(3 – 2) + 1 (– 2 – 6) + 3(– 4 – ) = 0 3 – 22 – 2 – 6 – 12 – 3 = 0 –32 – 18 = 0 2 + 6 = 0 not possible for real S is empty set

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Q.15 The tangents to the curve y = (x – 2)2 – 1 at its points of intersection with the line x – y = 3, intersect at the point :

(1)

1–,

25 (2)

1–,

25– (3)

1,

25 (4)

1,

25–

Ans. [1] Sol. x – y – 3 = 0 … (i) will be chord of contact of parabola y = x2 – 4x + 3 Let the required point is P(x1, y1) chord of contact for point P is

2

yy 1 = xx1 – 42

)xx( 1 + 3

y + y1 = 2x1x – 4x – 4x1 + 6 (2x1 – 4)x – y + (–4x1 – y1 + 6) = 0 … (ii) eqn (i) & (ii) are same line

1

4–x2 1 =1–1– =

3–6y–x4– 11

2x1 – 4 = 1 – 4x1 – y1 + 6 = –3

x1 = 25 –10 – y1 + 9 = 0

y1 = –1

Ans.

1–,

25

Q.16 If [x] denotes the greatest integer x, then the system of linear equations [sin ]x + [–cos]y = 0, [cot]x + y = 0

(1) has a unique solution if

32,

2 and have infinitely many solutions if

67,

(2) have infinitely many solutions if

32,

2 and has a unique solution if

67,

(3) have infinitely many solutions if

32,

2

67,

(4) has a unique solution if

32,

2

67,

Ans. [2] Sol. [sin ]x + [–cos]y = 0 … (i) [cot]x + y = 0 … (ii) Case-I :

When

32,

2 sin

1,

23

cos

0,

21– – cos

21,0

cot

0,

31–

[sin ] = 0, [– cos] = 0, [cot] = –1 eqn (i) & (ii) will

0yx–0y0x0

system will have infinitely many solution

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Case-II :

When

67, sin

0,

21–

cos

23–,1–

cot ),3(

[sin] = –1, [cos] = –1 [cot] = {1,2,3,.....} –x – y = 0 Ix + y = 0 I = {1, 2, .....} It will have unique solution is all cases x = 0, y = 0 Q.17 A group of students comprises of 5 boys and n girls. If the number of ways, in which a team of 3 students can

randomly be selected from this group such that there is at least one boy and at least one girl in each team, is 1750, then n is equal to :

(1) 24 (2) 25 (3) 27 (4) 28 Ans. [2] Sol. Given 5 boys and n girls total ways of forming team of 3 member under given condition = 5C1. nC2 + 5C2. nC1 5C1 . nC2 + 5C2 . nC1 = 1750

2

)1–n(n5 + 10n = 1750

2

)1–n(n + 2n = 350

n2 + 3n = 700 n2 + 3n – 700 = 0 n = 25 Q.18 An ellipse, with foci at (0, 2) and (0, –2) and minor axis of length 4, passes through which of the following

points ?

(1) (2, 2 ) (2) (2, 22 ) (3) ( 2 , 2) (4) (1, 22 ) Ans. [3] Sol. Given 2a = 4 and 2be = 4 a = 2, be = 2 b2e2 = 4 b2 – a2 = 4 b2 = 8 equation of ellipse

4

x2 +

8y2

= 1

Clearly option (3) satisfy the given curve.

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Q.19 For an initial screening of an admission test, a candidate is given fifty problems to solve. If the probability

that the candidate solve any problem is 54 , then the probability that he is unable to solve less than two

problems is :

(1) 48

51

25164

(2)

48

54

25316

(3)

49

51

5201

(4)

49

54

554

Ans. [4] Sol. Total problems = 50

P(Solving) = 54

P(Not solving) = 51

P(unable to solve less than two problems) = P(not solving one problem) + P(not solving zero problem)

= 50C0 0

51

50

54

+ 50C1

1

51

49

54

= 50

50

54 + 50. 49

49

5.54

= 50

54

+ 10.

49

54

= 49

54

+

10

54

= 5

54 . 49

54

Q.20 A straight line L at a distance of 4 units from the origin makes positive intercepts on the coordinate axes and

the perpendicular from the origin to this line makes an angle of 60º with the line x + y = 0. Then an equation of the line L is :

(1) x + 3 y = 8 (2) 3 x + y = 8

(3) ( 3 + 1)x + ( 3 – 1)y = 28 (4) ( 3 – 1)x + ( 3 + 1)y = 28 Ans. [3,4] Sol.

A

P

O

B

4

OP = 4 Given OP makes 60º with x + y = 0

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let slope of OP = m

tan60º = m–11m

1–m1m = 3 or – 3

m + 1= m3 – 3 or m + 1 = 3 – 3 m

m( 3 – 1) = 3 + 1 or m( 1 + 3 ) = 3 – 1

m = 1–313 or m =

131–3

tan = 1–313 or tan =

131–3

eqn of line x cos + y sin = P ( 3 + 1)x + ( 3 – 1)y = 28 or ( 3 – 1)x + ( 3 + 1)y = 28

Q.21 A value of such that

1

)1x)(x(dx = loge

89 is :

(1) 2 (2) –2 (3) 21 (4)

21–

Ans. [2]

Sol.

1

)1x)(x(dx = loge

89

1

dx)1x)(x()x(–)1x( = loge

89

1

xdx –

1

1xdx = loge

89

loge

1

1xx

= loge

89

loge

2212 – log

122 = loge

89

log

212

2212 = loge

89

)1(4

)12( 2

=

89

8[42 + 4 + 1] = 9[42 + 4] 322 + 32 + 8 = 362 + 36 42 + 4 – 8 = 0 2 + – 2 = 0 ( + 2) ( – 1) = 0 = 1, –2

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Q.22 1x–xsin–1xsin2x

xsin2xlim220x

is :

(1) 6 (2) 1 (3) 3 (4) 2 Ans. [4]

Sol. 1x–xsin–1xsin2x

xsin2xlim220x

= 1–xxsin–1xsin2x

xsin2xlim 220x

× )1x–xsin1xsin2x( 22

= xxsin–xsin2x

xsin2xlim 220x

× (2)

Applying L’H Rule

= 1xcosxsin2–xcos2x2

)xcos21(2lim0x

= 12)3(2

= 2

Q.23 The derivative of tan–1

xcosxsinxcos–xsin , with respect to

2x , where

2

,0x is :

(1) 32 (2) 1 (3) 2 (4)

21

Ans. [3]

Sol. Given y = tan–1

xcosxsinxcos–xsin

y = tan–1

1xtan1–xtan

y = –tan–1

xtan1xtan–1

y = –tan–1

x–

4tan

0 < x < 2

–2 < –x < 0

–4 <

4 – x < 0

y = –

x–

4

2

,2

–xxxtantan 1–

y = x –4

)2/x(ddy =

)2/1(1 = 2

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Q.24 Let S be the set of all R such that the equation, cos2x + sinx = 2– 7 has a solution. Then S is equal to : (1) [2, 6] (2) [3, 7] (3) [1, 4] (4) R

Ans. [1] Sol. Given cos2x + sinx = 2 – 7 1 – 2sin2x + sinx = 2 – 7 2sin2x – sinx + 2 – 8 = 0

sin x =4

)8–2(8–2

sin x =4

)8–(

sin x = 4

8– , 4

8–

sin x = 2 (Not possible) for solution

–1 4

8–2 1

–4 2a – 8 4 4 2 12 [2, 6]

Q.25 Let z C with Im(z) = 10 and it satisfies nz2n–z2

= 2i – 1 for some natural number n. Then :

(1) n = 20 and Re(z) = –10 (2) n = 40 and Re(z) = 10 (3) n = 40 and Re(z) = –10 (4) n = 20 and Re(z) = 10

Ans. [3] Sol. Let z = x + 10i

given nz2n–z2

= 2i – 1

n)i10x(2n–)i10x(2

= 2i – 1

(2x – n) + 20i = (2i – 1) [(2x + n) + 20i] Comparing real and imaginary part 2x – n = 2(–20) – (2x + n) and 20 = 2(2x + n) – 20 2x – n = –40 – 2x – n and 20 = 4x + 2n – 20 4x = –40 and 4x + 2n = 40 x = –10 and –40 + 2n = 40 n = + 40 n = 40 and Re(z) = –10 Q.26 Let f(x) = 5 – |x – 2| and g(x) = |x + 1|, x R. If f(x) attains minimum value at , then

–xlim

8x6–x)6x5–x)(1–x(

2

2

is equal to :

(1) 21 (2)

21– (3)

23 (4) –

23

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Ans. [1] Sol. f(x) = 5 – |x – 2| f(x) attains maximum value when |x – 2| = 0 x = 2 = g(x) = |x + 1| g(x) attains minimum value of x = –1 =

–x

lim8x6–x

)6x5–x)(1–x(2

2

= )4–x)(2–x(

)3–x)(2–x)(1–x(lim2x

= )4–2(

)3–2)(1–2( = 21

Q.27 The angle of elevation of the top of a vertical tower standing on a horizontal plane is observed to be 45º from

a point A on the plane. Let B be the point 30 m vertically above the point A. If the angle of elevation of the top of the tower from B be 30º, then the distance (in m) of the foot of the tower from the point A is :

(1) 15(1 + 3 ) (2) 15(3 – 3 ) (3) 15(3 + 3 ) (4) 15(5 – 3 ) Ans. [3] Sol.

P

A45º

30º B

M

N

AB = 30 m = NP In ANM

tan45º = ANMN = 1

MN = AN PM = MN – 30 = AN – 30 In BPM

tan30º = PBPM =

AN30–AN

3

1 = AN

30–AN

AN = 3 AN – 330

AN = 1–3

330 = 2

)13(330 = 15 ( 33 )

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Q.28 A value of

3,0 , for which

6cos41sincos6cos4sin1cos6cos4sincos1

22

22

22

= 0, is :

(1) 18 (2)

9 (3)

247 (4)

367

Ans. [2]

Sol.

3,0

6cos41sincos6cos4sin1cos6cos4sincos1

22

22

22

= 0

R2 R2 – R1, R3 R3 – R1

101–011–

6cos4sincos1 22 = 0

C1 C1 + C2

101–010

6cos4sin2 2 = 0

Expanding along first column 2[1 – 0] – 1[– 4cos6] = 0 2 + 4cos6 = 0

cos6 = –21

6 = 3

2 = 9

Q.29 If a1, a2, a3, ..... are in A.P. such that a1 + a7 + a16 = 40, then the sum of the first 15 terms of this A.P. is :

(1) 120 (2) 150 (3) 280 (4) 200 Ans. [4] Sol. a1, a2, a3, ......, an are in A.P. a1 + a7 + a16 = 40 a + a + 6d + a + 15d = 40 3a + 21d = 40

a + 7d = 3

40

S15 =2

15 [2a + 14d]

= 15[a + 7d]

= 15 × 3

40

= 200

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Q.30 The term independent of x in the expansion of

81x–

601 8

.6

22

x3–x2

is equal to :

(1) –36 (2) –108 (3) 36 (4) –72 Ans. [1]

Sol.

81x–

601 8

.6

22

x3–x2

term independent of x will be

601 × term independent of x in

6

22

x3–x2

81 × term of z–8 in

6

22

x3–x2

Tr+1 in 6

22

x3–x2

will be

Tr+1 = 6Cr(2x2)6 – rr

2x3–

= 6Cr 26 – r (–1)r × 3r × x12 – 2r – 2r Case-I : For term independent of x is 12 – 4r = 0 r = 3 T4 = – 6C3 × 23 × 33x6 = –20 × 23 × 33 Case-II : For term of x–8 12 – 4r = –8 4r = 20 r = 5 T6 = 6C5. 21. (–1). 35 . x–8

Required ans. =601 × (–20)23 × 33 –

811 × 6 × 2 × (–1) × 35

= –72 + 36 = –36