COMPUTER BASED TEST (CBT) - Resonance · | JEE MAIN-2019 | DATE : 09-01-2019 (SHIFT-1) | PAPER-1 |...

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COMPUTER BASED TEST (CBT)

Memory Based Questions & Solutions

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| JEE MAIN-2019 | DATE : 09-01-2019 (SHIFT-1) | PAPER-1 | MEMORY BASED | MATHEMATICS

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PAGE # 1

PART : MATHEMATICS

Straight Objective Type

This section contains 30 multiple choice questions. Each question has 4 choices (1), (2), (3) and (4) for its

answer, out of which Only One is correct.

1. There are 5 girls and 7 boys. A team of 3 boys and 2 girls is to be formed such that no two specific boys

are in the same team. Number of way to do so

(1) 400 (2) 250 (3) 200 (4) 300

Ans. (4)

Sol. Without any restriction, total number of ways of forming team is 7C3 × 5C2 = 350 If two specific boys B1,

B2 are in same team then total number of ways of forming team equals to 5C1 × 5C2 = 50 ways

total ways = 350 – 50 = 300 ways

2. The equation x2 + 2x + 2 = 0 has roots and . Then value of 15 + 15 is

(1) 512 (2) 256 (3) –512 (4) –256

Ans. (4)

Sol. roots are –1 + i and –1 – i

Let = –1 + i and b = –1 – i

then 15 + 15 = (–1 + i)15 + (–1 – i)15 = 16 16(–1 i) ( 1 i)

3 1 i 1 i

=

168

2 2

( 1 i 1 i)2

1 i

= 128(–2) = –256

3. 3

0| cos x |

dx is equal to

(1) 4

3 (2)

2

3 (3) 0 (4)

8

3

Ans. (1)

Sol. / 2

3 3

0 0

| cos x | dx 2 | cos x | dx

= / 2

3

0

12 cos xdx

2

2

.13

= 4

3 (by walli's formula)

4. If x2 n + 1, n N then 2 2

2 2

2sin(x 1) sin2(x 1)x

2sin(x 1) sin2(x 1)

dx is equal to

(1) 2x 1

lncos2

+ c (2) 21 x 1

lncos2 2

+ c

(3) ln2x 1

sec2

+ c (4) 21 x 1

lnsec2 2

+ c

Ans. (4)








PAGE # 2

Sol. 2

2

(1 cos(x 1))x dx

(1 cos(x 1))

=

2

2

x 1sin

2x dx

x 1cos

2

= 2x 1

x.tan dx2

2x 1

t2

2xdx = 2dt

= dt

tan t2

= 1

2nsect + c =

1

2nsec

2x 1c

2

5. If A = i – j , B = i + j + k are two vectors and C is another vector such that A × C + B = 0 and A . C = 0

then | C |2 =

(1) 9 (2) 8 (3) 19

2 (4)

17

2

Ans. (3)

Sol. a × ( a × c ) + a × b = 0

( a × c ) a – ( a . a ) c + a × b = 0

4 a – 2 c + a × b = 0

a × b =

ˆ ˆ ˆi j k

1 1 0

1 1 1

= i (–1) – j (1) + k(2) = – i – j + 2 k

2 c = 4( ˆ ˆi j ) + (– i – j +2 k )

2 c = 3 i – 5 j + 2 k

2|c| = a 25 4

2|c| = 38

4|c|2 = 38 |c| = 19/2

6. f(x) =

5 ; x 1

a bx ; 1 x 3

b 5x ; 3 x 5

30 ; x 5

then

(1) f(x)is discontinuous a R, b R

(2) f(x) is continuous if a = 0 & b = 5

(3) f(x) is continuous if a = 5 & b = 0

(4) f(x) is continuous if a = – 5 & b = 10

Ans. (1)








PAGE # 3

Sol. f(x) =

5 ; x 1

a bx ; 1 x 3

b 5x ; 3 x 5

30 ; x 5

for function to be continuous at x = 1

f(1) = h 0lim

f(1 + h) = h 0lim

f(1 – h)

f(1) = 5

h 0lim

f(1 + h) = h 0lim

a + b(1 + h) = a + b

h 0lim

f(1 – h) = h 0lim

5 = 5

hence a + b = 5 ..…(1)


f(3) = h 0lim

f(3 + h) = h 0lim

f(3 – h)

f(3) = b + 15

h 0lim

f(3 + h) = h 0lim

b + 5(3 + h) = b + 15

h 0lim

f(3 – h) = h 0lim

a + b(3 – h) = a + 3b

hence b + 15 = a + 3b a + 2b = 15 ………(2)


f(5) = h 0lim

f(5 + h) = h 0lim

f(5 – h)

f(5) = 30

h 0lim

f(5 + h) = 30

h 0lim

f(5 – h) = h 0lim

b + 5(5 – h) = b + 25

hence b + 25 = 30 b = 5 ………..(3)

from equation (2) a = 5

from equation (1) a = 0

hence f(x) is discontinuous for a, bR s

7. Average height & variance of 5 students in a class is 150 and 18 respectively. A new student whose

height is 156 cm is added to the group. Find new variance.

(1) 20 (2) 22 (3) 16 (4) 14

Ans. (1)

Sol. Let 5 students are x1, x2, x3, x4, x5

Given x = ix150

5

5

i

i 1

x 750

…..(1)

2

2ix(x) 18

5

22ix

(150) 185

2

ix = (22500 + 18)× 5 5

2

i

i 1

x

= 112590 …… (2)

Height of new student = 156 (Let x6)

now x1 + x2 + x3 + x4 + x5 + x6 = 750 + 156

newx = 1 2 3 4 5 6x x x x x x 906

6 6

= 151 ……..(3)








PAGE # 4

Variance (new) = 2

2ix(x)

6

= 2 2 2 2 2 2

21 2 3 4 5 6x x x x x x(151)

6

from equation (2) and (3)

var (new) = 2

2112590 (156)(151)

6

= 22 821 – 22801 = 20

8. a, b, c are in G.P. a + b + c = x b , x can not be

(1) 2 (2) –2 (3) 3 (4) 4

Ans. (1)

Sol. a + ar + ar2 = xar

since a 0 so 2r r 1

xr

; 1 +

1r x

r

r + 1

r (–, –2] [2, ) x (–, –1] [3, )

9. 4032

15

= k

15then find k.

(1) 2 (2) 8 (3) 1 (4) 4

Ans. (2)

Sol. 2032

15

8.2200 8.1650 = 8(1 + 15)50 = 8(1 + 15)

hence remainder is 8.

10. y 0lim

4

4

1 1 y 2

y

=

(1) 1

4 2 (2)

1

2 2 (3)

1

2 2(1 2) (4) does not exist

Ans. (1)

Sol. Using rationalization

y 0lim

4 4

44

1 1 y 2 1 1 y 2

y 1 1 y 2

= 4 4

4 4y 0 4

1 y 1 1 y 11lim

y 1 y 11 1 y 2

= y 0lim 44

1 1

1 y 11 1 y 2

by putting value of limit

= 1 1 1

2 2 2 4 2








PAGE # 5

11. There is a parabola having axis as x -axis, vertex is at a distance of 2 unit from origin & focus is at

(4, 0). Which of the following point does not lie on the parabola.

(1) (6, 8) (2) (5, 2 6 ) (3) (8, 4 3 ) (4) (4, –4)

Ans. (1)

Sol. As per the given data diagram of parabola must be as below

(2, 0)

(4, 0)

then equation of parabola will be

(y – 0)2 = 4.2(x – 2)

y2 = 8(x – 2)

now check all options (C)

12. Find sum of all possible values of in the interval ,2

for which 3 2isin

1 2isin

is purely imaginary

(1) 3

(2) (3)

2

3

(4)

2

Ans. (3)

Sol. Let z = 3 2isin

1 2isin

,

2

for z to be purely imaginary

z + z = 0

3 2isin 3 2isin

01 2isin 1 2isin

2 2

2

3 6isin 2isin 4sin 3 6isin 2isin 4sin

1 4sin

3 – 4sin2 = 0 sin2 = 3

4

= n ± 3

Given ,2

hence = 2

, ,3 3 3

sum of all possible value of = 2

3








PAGE # 6

13. Let A = cos sin

sin cos

Find the value of A–50 at = 12

(1)

3 1

2 2

1 3

2 2

(2)

3 1

2 2

1 3

2 2

(3)

3 1

2 2

1 3

2 2

(4)

1 3

2 2

3 1

2 2

Ans. (2)

Sol. A = cos sin

sin cos

A–50 = cos( 50 ) sin( 50 )

sin( 50 ) cos( 50 )

A–50 at = 12

is

25 25cos sin

6 6

25 25sin cos

6 6

=

cos sin6 6

sin cos6 6

=

3 1

2 2

1 3

2 2

14. If (A B) ( A B) = A B what should be proper symbol in place of and to hold the equation

(1) and (2) and (3) and (4) and

Ans. (1)

Sol. Check all option repeatedly

(i) (A B) (~A B) A(B (~ A B))

A (B) A B

(i) is correct

(ii) (A B) (~A B) (A~A) B

fB f

(iii) (A B) (~A B) B

(iv) (A B) (~A B)

B (A ~A) = B f f

only (1) is correct








PAGE # 7

15. If y(x) is solution of x dy

dx + 2y = x2 , y(1) = 1 then value of y

1

2

=

(1) –49

16 (2)

49

16 (3)

45

8 (4)

45

8

Ans. (2)

Sol. xdy

dx+2y = x2

dy

dx+

2

xy = x

This is linear differential equation

I.F. = 2

dxxe = 2 nxe = x2

Solution of differential equation is

y . x2 = x . x2 dx + C

yx2 = 4x

4+ C

y = 4

2

x C

4 x

given y (1) = 1

1 = 1

4+ C C =

3

4

Hence y(x) = 2x

4+

2

3

4x

y1

2

= 1

16+

3

14

4

y1

2

= 49

16

16. From a well shuffled deck of cards, 2 cards are drawn with replacement. If x represent numbers of

times ace coming, then value of P(x = 1) + P (x = 2) is

(1) 25

169 (2)

24

169 (3)

49

169 (4)

23

169

Ans. (1)

Sol. P(x = 1) = 4

52 ×

48

52× 2 =

24

169

P(x = 2) = 4

52 ×

4

52=

1

169

P(x = 1) + P (x = 2) = 25

169

17. If eccentricity of the hyperbola 2

2

x

cos –

2

2

y

sin = 1 is more than 2 when 0,

2

Find the possible

values of length of latus rectum

(1) (3, ) (2) (1, 3/2) (3) (2, 3) (4) (–3, –2)

Ans. (1)








PAGE # 8

Sol. 2

2

x

cos –

2

2

y

sin = 1

e > 2 (given)

e2 > 4 1 + 2

2

sin

cos

> 4

1 + tan2 > 4 tan2 > 3

0,2

hence ,3 2

Latus rectum = 22sin

cos

= 2 tan sin

for ,3 2

, 2 tan sin is

An increasing function

Hence latus rectum (3, ) Ans(1)

18. If slant height of a right circular cone is 3 cm then the maximum volume of cone is

(1) 2 3 cm3 (2) 4 3 cm3 (3) (2 + 3 ) cm3 (4) (2 – 3 ) cm3

Ans. (1)

Sol. V = 1

3 r2 h , r2 + h2 = 9

V = 1

3 h (9 – h2)

dv

dh=

1

3 (9 – 3h2) = 0

9 – 3h2 = 0

h2 = 3 , h = 3

V = 1

3() (6) 3 = 2 3

19. If cos–1 2

3x

+ cos–1 3

4x

= 2

, x >

3

4 then x =

(1) 145

11 (2)

145

12 (3)

146

10 (4)

146

11

Ans. (2)

Sol. cos–1 12 3cos

3x 4x 2

3x

4

cos–1 2 2

2 3 4 91 1

3x 4x 29x 16x

2

1

2x =

2 2

2

9x 4 16x 9

12x

6 = 2 29x 4 16x 9








PAGE # 9

square both side

36 = 144x4 – 81x2 – 64x2 + 36

144x4 = 145x2

x4 = 2145x

144 x = ±

145, 0

12

x > 3

4 hence x =

145

12

20. If px + qx + r = 0 represent a family of straight lines such that 3p + 2q + 4r = 0 then

(1) All lines are parallel (2) All lines are inconsistence

(3) All lines are concurrent at 3 1

,4 2

(4) All lines are concurrent at 3, 2

Ans. (3)

Sol. Given relation is 3p + 2q + 4r = 0

3

4p +

q

2+ r = 0 ...........(ii)

comparing (1) and (2) we get x = 3

4, y =

1

2

hence we can say that these lines are concurrent at 3 1

,4 2

21. Consider the system of equations x + y + z = 1, 2x + 3y + 2z = 1, 2x + 3y + (a2 – 1) z = a + 1 then

(1) system has a unique solution for |a| = 3

(2) system is inconsistence for |a| = 3

(3) system is inconsistence for a = 4

(4) system is inconsistence for a = 3

Ans. (2)

Sol. D = 2

1 1 1

2 3 2

2 3 a 1

= 3( a2 –1) – 6 – 2(a2 – 1) + 4

= a2 – 1–2 = a2 – 3

If |a| 3 system has unique solution

if |a| = 3

x y z 1

2x 3y 2z 1

2x 3y 2z 3 1

Hence system is inconsistent for |a| = 3

22. The value of 3(cos – sin)4 + 6(sin + cos)2 + 4sin6 is where ,4 2

(1) 13 – 4 cos4 (2) 13 – 4 cos6

(3) 13 – 4 cos6 + 2sin4 cos2 (4) 13 – 4 cos4 + 2sin4 cos2

Ans. (2)








PAGE # 10

Sol. 3(sin – cos)4 + 6(sin + cos)2 + 4sin6

= 3(1 – sin2)2 + 6(1 + sin2) + 4(1 – cos2)3

= 3(1 – 2sin2 + sin22) + 6(1 + sin2) + 4(1 – 3cos2 + 3cos4 – cos6)

= 13 + 3sin22 + 12cos2(cos2 – 1) – 4cos6

= 13 + 12sin2 cos2 – 12sin2 cos2 – 4cos6

= 13 – 4cos6

23. 3 circles of radii a, b, c (a < b < c) touch each other externally and have x-axis as a common tangent

then

(1) a, b, c are in A.P. (2) 1

b=

1

a+

1

c

(3) a , b , c are in A.P. (4) 1

c +

1

b =

1

a

Ans. (4)

Sol.

a c b

2 2(a b) – (a –b) + 2 2(a c) – (a – c) = 2 2(b c) – (b – c)

ab + ac = bc

1

c +

1

b =

1

a

24. If f(x) = 1

x, f2 (x) = 1 – x, f3 (x) =

1

1 xthen find J(x) such that f2 o J o f1 (x) = f3 (x)

(1) f1 (x) (2) 1

xf3 (x) (3) f3 (x) (4) f2(x)

Ans. (3)

Sol. f2 (Jo(f1(x))) = f3(x)

1 – J (f1(x)) = f3 (x)

1 – J 1

x

= 1

1 x

J 1

x

= 1 – 1

1 x= –

x

1 x

J(x) =

1

x1

1x

= 1

1 x J(x) = f3 (x)








PAGE # 11

25. Find the equation of line through (–4, 1, 3) & parallel to the plane x + y + z = 3 while the line intersects

another line whose equation is x + y – z = x + 2y – 3z + 5

(1) x 4 y 1 z 3

3 2 1

(2)

x 4

1

=

y 1

2

=

z 3

3

(3) x 4 y 1 z 3

3 2 1

(4)

x 4

1

=

y 1

2

=

z 3

3

Ans. (3)

Sol. Family of plan containing the line of intersection of planes of x + y – z = 0 = x + 2y – 3z + 5 is

(x + y – z) + (x + 2y – 3z + 5) = 0

If this plane passes through (–4, 1, 3), we get (–4 + 1 – 3) + (–4 + 2 – 9 + 5) = 0 –6 + (–6)

= –1

Hence equation of this plane is (x + y – z) – (x + 2y – 3z + 5) = 0

now required line lies in this plane and is parallel to x + y + z = 5

Hence vector, v (say), along the line is v =

ˆ ˆ ˆi j k

ˆ ˆ ˆ1 1 1 3i 2 j k

0 1 2

so required line is x 4 y 1 z 3

3 2 1

26. Consider the curves y = x2 + 2 and y = 10– x2. Let be the angle between both the curves at point of

intersection, then find |tan|

(1) 8

15 (2)

5

17 (3)

3

17 (4)

8

17

Ans. (1)

Sol. y = x2 + 2 & y = 10 – x2 meet at (±2, 6)

dy

2xdx

for first curve and dy

2xdx

for second.

Hence slope are 4 and –4

so |tan| = 4 ( 4) 8

1 4( 4) 15

27. A plane parallel to y-axis passing through line of intersection of planes x + y + z = 1 & 2x + 3y – z – 4 =

0 which of the point lie on the plane

(1) (3, 2, 1) (2) (–3, 0, 1) (3) (–3, 1, 1) (4) (3, –1, 1)

Ans. (4)

Sol. Equation of required plane is

(x + y + z – 1) + (2x + 3y – z – 4) = 0 (1 + 2)x + (1 + 3)y +(1 – ) = 0

since given plane is parallel to y-axis 3 + 1 = 0 = 1

3

Hence equation of plane is x 4z 1

03 3 3 x + 4z + 1 = 0








PAGE # 12

28. Find common tangent of the two curve y2 = 4x and x2 + y2 – 6x = 0

(1) y = x

33 (2) y =

x3

3

(3) y =

x3

3 (4) y =

x3

3

Ans. (4)

Sol. Tangent to y2 = 4x is y = mx + 1

m

If it is tangent to given circle it's distance from (3, 0) is equal to 3

Hence 2

13m

m

1 m

= 3

|3m2 + 1| = 3m 21 m

squaring both sides, we get

9m4 + 6m2 + 1 = 9m2 + 9m4 m = 1

3

Hence common tangents are y = x

3+ 3 or y = –

x3

3

29. If the area bounded by the curve y = x2 – 1 tangent to it at (2, 3) and y-axis is

(1) 2

3 (2)

4

3 (3)

8

3 (4) 1

Ans. (3)

Sol.

3

(2, 3)

–1

–5

area =3 3

5 1

xdy xdy

= 3 3

5 1

y 5y 1 dy

4

=

32

3

3 / 2

1

5

y5y

22 (y 1)4 3

=

9 2515 25

2 2

4

= |16

3 | =

8

3



COMPUTER BASED TEST (CBT) - Resonance · | JEE MAIN-2019 | DATE : 09-01-2019 (SHIFT-1) | PAPER-1 |...

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Transcript of COMPUTER BASED TEST (CBT) - Resonance · | JEE MAIN-2019 | DATE : 09-01-2019 (SHIFT-1) | PAPER-1 |...