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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 9 th April 2019 | Shift - I MATHEMATICS Q.1 Let = i ˆ 3 + j ˆ and = i ˆ 2 j ˆ + k ˆ 3 . If = 1 2 , where 1 is parallel to and 2 is perpendicular to , then 1 × 2 is equal to : (1) – i ˆ 3 + j ˆ 9 + k ˆ 5 (2) 2 1 (– i ˆ 3 + j ˆ 9 + k ˆ 5 ) (3) i ˆ 3 j ˆ 9 k ˆ 5 (4) 2 1 ( i ˆ 3 j ˆ 9 + k ˆ 5 ) Ans. [2] Sol. 1 is parallel to 1 = 1 ) j ˆ i ˆ 3 ( Given that 2 1 1 2 ) k ˆ 3 j ˆ i ˆ 2 ( ) j ˆ i ˆ 3 ( 2 k ˆ 3 ) 1 ( j ˆ ) 2 3 ( i ˆ 2 Also given that 2 is perpendicular to 2 3(3 – 2) + ( + 1) = 0 2 1 So, j ˆ 2 1 i ˆ 2 3 1 and k ˆ 3 j ˆ 2 3 i ˆ 2 1 2 ) k ˆ 5 j ˆ 9 i ˆ 3 ( 2 1 3 2 / 3 2 / 1 0 2 / 1 2 / 3 k ˆ j ˆ i ˆ 2 1 Q.2 Let S be the set of all values of x for which the tangent to the curve y = f(x) = x 3 – x 2 – 2x at (x, y) is parallel to the line segment joining the points (1, f(1)) and (–1, f(–1)), then S is equal to : (1) 1 , 3 1 (2) 1 , 3 1 (3) 1 , 3 1 (4) 1 , 3 1

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### Transcript of JEE Main Online Exam 2019 - Career Pointcareerpoint.ac.in/studentparentzone/2019/jee-main/JEE... ·...

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JEE Main Online Exam 2019

Questions & Solutions 9th April 2019 | Shift - I

MATHEMATICS

Q.1 Let = i3 + j and

= i2 – j + k3 . If

= 1

– 2

, where 1

is parallel to

and 2

is perpendicular to

, then 1

× 2

is equal to :

(1) – i3 + j9 + k5 (2) 21 (– i3 + j9 + k5 ) (3) i3 – j9 – k5 (4)

21 ( i3 – j9 + k5 )

Ans. [2]

Sol. 1 is parallel to

1 =

1 )ji3(

Given that 21

12

)k3ji2()ji3(2

k3)1(j)23(i2

Also given that 2 is perpendicular to

2

3(3 – 2) + ( + 1) = 0

21

So, j21i

23

1 and k3j23i

21

2

)k5j9i3(21

32/32/102/12/3kji

21

Q.2 Let S be the set of all values of x for which the tangent to the curve y = f(x) = x3 – x2 – 2x at (x, y) is parallel

to the line segment joining the points (1, f(1)) and (–1, f(–1)), then S is equal to :

(1)

1,

31 (2)

1–,

31 (3)

1–,

31– (4)

1,

31–

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Ans. [4] Sol. y = f(x) = x3 – x2 – 2x

slope of tangent 2x2x3)x('fdxdy 2

This tangent is parallel to line segment joining points (1, f(1)) and (–1, f(–1)) m1 = m2

11

)1(f)1(f2x2x3 2

2

)211()211(2x2x3 2

3x2 – 2x – 2 = – 1 3x2 – 2x – 1 = 0 (3x + 1) (x – 1) = 0

1,31x

Q.3 Let S = {[–2, 2] : 2cos2 + 3 sin = 0}. Then the sum of the elements of S is :

(1) (2) 6

13 (3) 2 (4) 3

5

Ans. [3] Sol. 2 cos2 + 3 sin = 0 2(1 – sin2) + 3 sin = 0 2 sin2 – 3 sin – 2 = 0 (2 sin + 1) (sin – 2) = 0

21sin

65

O X

Y

6

6

7

611

in [–2, 2]

6

11,6

7,6

,6

5

Sum of all roots

26

1175

Q.4 For any two statements p and q, the negation of the expression p (~ p q) is : (1) ~ p ~ q (2) p q (3) p q (4) ~ p ~ q Ans. [4] Sol. ~ (p p q) = ~ p (p ~ q) = (~ p p) (~ p ~ q) = c (~ p ~ q) = ~ p ~ q

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Q.5 The solution of the differential equation xdxdy + 2y = x2 (x 0) with y(1) = 1, is :

(1) y = 5x3

+ 2x51 (2) y = 3x

54 + 2x5

1 (3) y = 2x43 + 2x4

1 (4) y = 4x2

+ 2x43

Ans. [4]

Sol. 2xy2dxdyx (x 0)

xxy2

dxdy

I.F. = 2xlogdxx2

xee2

e

Solution is yx2 dxxx2

yx2 = C4x4

at y(1) = 1 C411

43C

43

4xyx

42

2

2

x43

4xy

Q.6 If a tangent to the circle x2 + y2 = 1 intersects the coordinate axes at distinct points P and Q, then the locus of

the mid-point of PQ is : (1) x2 + y2 – 2x2y2 = 0 (2) x2 + y2 – 4x2y2 = 0 (3) x2 + y2 – 16x2y2 = 0 (4) x2 + y2 – 2xy = 0 Ans. [2] Sol. Let the equation of tangent is x cos + y sin = 1 co-ordinates of P and Q are

0,

cos1P and

sin

1,0Q

Q

P

(cos , sin )

Let mid point of P and Q is (h, k)

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so, 2

0cos

1

h

and 2sin

10k

h2

1cos and k21sin

squaring and adding we get

1k41

h41

22

locus 1y41

x41

22

x2 + y2 – 4x2 y2 = 0 Q.7 Slope of a line passing through P(2, 3) and intersecting the line, x + y = 7 at a distance of 4 units from P, is :

(1) 151–5

(2)

171–7

(3)

717–1

(4)

515–1

Ans. [3] Sol. Let any point on the line is P(2 4 cos , 3 4 sin ) it also lie on line x + y = 7

P(2,3)

4 (2 + 4 cos , 3 + 4 sin )

Y

x + y = 7

(2 4 cos ) + (3 4 sin ) = 7

(sin + cos ) = 21

(sin + cos )2 = 41

412sin1

432sin

43

tan1tan2

2

3 tan2 + 8 tan + 3 = 0

tan 6

7286

728

71)71( 2

7171

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Q.8 If f(x) is a non-zero polynomial of degree four, having local extreme points at x = – 1, 0, 1; then the set S = {x R : f(x) = f(0)} contains exactly : (1) four rational numbers. (2) two irrational and two rational numbers. (3) two irrational and one rational number. (4) four irrational numbers. Ans. [3] Sol. Four degree polynomial function f(x) have local extreme points at x = –1, 0, 1 f '(x) = (x + 1) (x – 0) (x – 1) = (x3 – x)

f(x) K2

x4

x 24

Now, f(x) = f(0)

KK2

x4

x 24

02

x4x 24

x = 0, 2 Two irrational and one rational number. Q.9 A committee of 11 members is to be formed from 8 males and 5 females. If m is the number of ways the

committee is formed with at least 6 males and n is the number of ways the committee is formed with at least 3 females, then :

(1) m + n = 68 (2) m = n = 78 (3) m = n = 68 (4) n = m – 8 Ans. [2] Sol. Given : (8 males, 5 females) Committee to be selected = 11 members m = no. of ways the committee is formed with at least 6 males. (6M, 5F) or (7M, 4F) or (8M, 3F) = 8C6 × 5C5 + 8C7 × 5C4 + 8C8 × 5C3 = 78 n = no. of ways the committee is formed with atleast 3 female (8M, 3F) or (7M, 4F) or (6M, 5F) = 8C8 × 5C3 + 8C7 × 5C4 + 8C6 × 5C5 = 10 + 40 + 28 = 78 m = n = 78 Q.10 The integral dxxeccosxsec 3/43/2 is equal to : (Here C is a constant of integration)

(1) – 3 tan–1/3x + C (2) – 43 tan–4/3x + C (3) 3 tan–1/3x + C (4) – 3 cot–1/3x + C

Ans. [1]

Sol. I = dxxeccosxsec 3/43/2

3/23/4 )xcos()x(sindxI dx

xcosxcosxsin

dxI2

3/4

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3/4

2

)x(tandxxsecI

put tan x = t sec2x dx = dt

C)3/1(

ttdtI

3/1

3/4

C)x(tan

3I 3/1

Q.11 Let p, q R. If 2 – 3 is a root of the quadratic equation, x2 + px + q = 0, then :

(1) p2 – 4q + 12 = 0 (2) q2 – 4p – 16 = 0 (3) q2 + 4p + 14 = 0 (4) p2 – 4q – 12 = 0 Ans. [4] Sol. If one root of equation x2 + px + q = 0 is 3–2 then other root will be 32 equation x2 – 4x + 1 = 0 p = –4 and q = 1 p2 – 4q – 12 = 0 Q.12 Let f(x) = 15 – |x – 10|; x R. Then the set of all values of x, at which the function, g(x) = f(f(x)) is not

differentiable, is : (1) {10, 15} (2) {5, 10, 15} (3) {10} (4) {5, 10, 15, 20} Ans. [2] Sol. f(x) = 15 – |x – 10| g(x) = f [f(x)] = 15 – | f(x) – 10 | = 15 – | 15 – | x – 10 | – 10 | = 15 – | 5 – | x – 10 ||

(5, 0) (15, 0)

(15, 15) (5, 15)

(10, 10)

y

x O

g(x) is not differentiable at x = 5, 10, 15

Q.13 Let

10

1k

)ka(f = 16(210 – 1), where the function f satisfies f(x + y) = f(x)f(y) for all natural numbers x, y

and f(1) = 2. Then the natural number ‘a’ is : (1) 2 (2) 3 (3) 16 (4) 4 Ans. [2] Sol. Given f(1) = 2 and f(x + y) = f(x) f(y) at x = 1, y = 1 f(2) = f(1) f(1) = 22 x = 2, y = 1 f(3) = f(2) f(1) = 23

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…………………. …………………. f(n) = 2n

Now )12(16)ka(f 1010

1k

f(a + 1) + f(a + 2) + ….. + f(a + 10) = 16(210 – 1) 2a+1 + 2a+2 + …… + 2a+10 = 16(210 – 1) 2a [21 + 22 + …… + 210] = 16(210 – 1)

2a )12(1612

)12(2 1010

2a+1 = 16 a = 3

Q.14 If the standard deviation of the numbers –1, 0, 1, k is 5 where k > 0, then k is equal to :

(1) 6 (2) 3

102 (3) 354 (4) 62

Ans. [4]

Sol. 22i )x(x

n1.D.S

Now mean 4k

4k101x

Given that S.D. = 5

16k)k101(

415

22

16k

4k25

22

80 = 8 + 4k2 – k2 3k2 = 72 k2 = 24 62k

62k ( k > 0)

Q.15 If the line y = mx + 37 is normal to the hyperbola 24x2

– 18y2

= 1, then a value of m is :

(1) 5

3 (2) 5

2 (3) 25 (4)

215

Ans. [2]

Sol. Equation of normal of hyperbola in slope form is 222

22

mba

)ba(mmxy

2m1824

m4237

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72 – 54 m2 = 36 m2 72 = 90 m2

54

9072m2

5

2m

5

2m

Q.16 If

1011

1021

1031

……..

10

1–n1=

10781

, then the inverse of

10n1

is :

(1)

11201

(2)

11321

(3)

1012–1

(4)

1013–1

Ans. [4]

Sol.

1011

1021

1031

…….

10

1–n1 =

10781

10

1–n....3211 =

10781

1 + 2 + 3 + …… + (n – 1) = 78

2

)2–n(n = 78

n = 13

Now inverse of

10n1

i.e.

10

131 is

1013–1

Ans.

Q.17 If the function f defined on

3,

6 by f(x) =

4x,k

4x

1–xcot1–xcos2

is continuous, then k is equal to :

(1) 2

1 (2) 1 (3) 21 (4) 2

Ans. [3]

Sol. 4

xlim

1–xcot1–xcos2 = f

4

4

xlim

1–xcot

1–xcos2 = k

using L-Hospital Rule

4

xlim

xeccos–

)xsin(–22 = k

23

21

= k

k = 21 Ans.

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Q.18 Four persons can hit a target correctly with probabilities 21 ,

31 ,

41 and

81 respectively. If all hit at the target

independently, then the probability that the target would be hit, is :

(1) 19225 (2)

3225 (3)

327 (4)

1921

Ans. [2] Sol. Let four persons are A, B, C, D. Probability of Hitting target = 1 – (None of four person Hit the target) = 1 – P( A )P( B )P( C )P( D )

= 1 – 21

32

43

87

= 3225 Ans.

Q.19 If one end of a focal chord of the parabola, y2 = 16x is at (1, 4), then the length of this focal chord is : (1) 24 (2) 20 (3) 25 (4) 22 Ans. [3] Sol. Parabola y2 = 16x {4a = 16 a = 4}

O S(4, 0)

y

x

P(1, 4) = (at2, 2at)

One end (at2, 2at) = (1, 4) 2at = 4 2(4) t = 4 t = 1/2

Length of focal chord = a2

t1t

= 4

2

221

= 25 Ans.

Q.20 Let and be the roots of the equation x2 + x + 1 = 0. Then for y 0 in R,

y11y

1y is equal to :

(1) y(y2 – 3) (2) y3 (3) y(y2 – 1) (4) y3 – 1 Ans. [2] Sol. Roots of eqn. x2 + x + 1 = 0 are and

, = 2

4–11– = 2

3i1–

= , = 2 (complex cube root of unity)

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=

y11y

1y

2

2

2

R1 R1 + R2 + R3

=

y1

1yyyy

2

2 ( 1 + + 2 = 0)

= y

y1

1y111

2

2

= y(y2) = y3 Ans.

Q.21 All the points in the set S =

R:

i–i (i = 1– ) lie on a :

(1) circle whose radius is 1. (2) straight line whose slope is 1.

(3) straight line whose slope is –1. (4) circle whose radius is 2 . Ans. [1]

Sol. Let i–i

= z

i–i

= |z|

|z| = 1 Circle of radius = 1

Q.22 The value of dxxcosxsin

xsin2/

0

3

is :

(1) 8

2– (2) 2

1– (3) 4

1– (4) 4

2–

Ans. [3]

Sol. I = dxxcosxsin

xsin2/

0

3

…(1)

I = dxx–

2cosx–

2sin

x–2

sin2/

0

3

I = dxxsinxcos

xcos2/

0

3

…(2)

Adding (1) & (2) we get

2I = dxxcosxsinxcosxsin

2/

0

33

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2I = dx)xcosx(sin

)xcosxsin–xcosx)(sinxcosx(sin2/

0

22

2I = dx)xcosxsin–1(2/

0

2I = dxx2sin21–1

2/

0

2I = 2

04x2cosx

2I =

41–

2 –

41 =

2 –

21

I =

41–

Q.23 The value of cos210º – cos10º cos50º + cos250º is :

(1) 23 (1 + cos20º) (2) 3/2 (3) 3/4 (4)

43 + cos 20º

Ans. [3] Sol. cos210º – cos10º cos50º + cos250º

= 21 [2cos210º – 2cos10º cos50º + 2cos250º]

= 21 [(1 + cos20º) – (cos60º + cos40º) + (1 + cos100º)]

= 21 [2 – cos60º + cos20º + (cos100º – cos40º)]

= 21 [2 –

21 + cos20º + 2sin70º sin(–30º)]

= 21 [

23 + cos20º – sin70º]

= 21 [

23 + cos20º – sin(90º – 20º]

= 43 Ans.

Q.24 Let the sum of the first n terms of a non-constant A.P., a1, a2, a3, …….. be 50n + 2

)7–n(n A, where A is a

constant. If d is the common difference of this A.P., then the ordered pair (d, a50) is equal to : (1) (50, 50 + 46A) (2) (50, 50 + 45A) (3) (A, 50 + 45A) (4) (A, 50 + 46A) Ans. [4]

Sol. Sn = 50n + 2

)7–n(n A

Tn = Sn – Sn–1

Tn = 50n +

2)7–n(n A – 50(n – 1) –

2)8–n)(1–n( A

= 50 + 2A [n2 – 7n – n2 + 9n – 8]

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= 50 + A(n – 4) Now, d = Tn – Tn–1 = 50 + A(n – 4) – 50 – A(n – 5) = A and T50 = 50 + 46A (d, A50) = (A, 50 + 46A) Ans.

Q.25 If the line, 2

1–x = 3

1y = 4

2–z meets the plane, x + 2y + 3z = 15 at a point P, then the distance of P from

the origin is :

(1) 2/5 (2) 7/2 (3) 52 (4) 9/2

Ans. [4]

Sol. Line 2

1–x = 3

1y = 4

2–z = k (say)

any point on this line P(2k + 1, 3k – 1, 4k + 2) This point P lies on plane x + 2y + 3z = 15 (2k + 1) + 2(3k – 1) + 3(4k + 2) = 15 20k + 5 = 15 20k = 10

k = 1/2 P

4,21,2

Distance of P from origin is

= 16414 =

29 Ans.

Q.26 A plane passing through the points (0, –1, 0) and (0, 0, 1) and making an angle 4 with the plane y – z + 5 = 0,

also passes through the point :

(1) ( 2 , –1, 4) (2) (– 2 , 1, –4) (3) (– 2 , –1, –4) (4) ( 2 , 1, 4) Ans. [4] Sol. Let ax + by + cz = 1 be the eqn. of plane it passed through (0, –1, 0) and (0, 0, 1) b = – 1 and c = 1 other plane is y – z + 5 = 0

Given that angle b/w them is 4

cos = |b||a|

ba

2

1 = 11011a

|1–1–0|2

a2 + 2 = 4 a = 2 eqn. of plane 2 x – y + z = 1 Now for –ve sign – 2 ( 2 ) – 1 + 4 = 1 ( 2 , 1, 4) satisfy the eqn. of plane.

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Q.27 If the tangent to the curve, y = x3 + ax – b at the point (1, – 5) is perpendicular to the line, –x + y + 4 = 0, then which one of the following points lies on the curve?

(1) (2, – 2) (2) (– 2, 2) (3) (– 2, 1) (4) (2, – 1) Ans. [1] Sol. y = x3 + ax – b (1, – 5) lies on curve – 5 = 1 + a – b a – b = – 6 …(1)

dxdy = 3x2 + a

Slope of tangent at (1, – 5)

dxdy = 3 + a

This tangent is perpendicular to – x + y + 4 = 0 (3 + a) (1) = – 1 a = – 4 …(2) By (1) & (2) a = – 4, b = 2 So, eqn. of curve y = x3 – 4x – 2 (2, – 2) lies on this curve

Q.28 If the fourth term in the Binomial expansion of 6

xlog8xx2

(x > 0) is 20 × 87, then a value of x is :

(1) 82 (2) 83 (3) 8 (4) 8–2

Ans. [1]

Sol. 6

xlog8xx2

(x > 0)

T4 = 20 × 87

6C3

3

x2

3xlog8x = 20 × 87

xlog33

8xx

160 = 20 × 87

3–xlog3 8x = 86 3–xlog2x = 86 = 218 log2 3–xlog2x = log2218 (log2x – 3) (log2x) = 18 Let log2x = t t2 – 3t – 18 = 0 t = 6, – 3 log2x = 6 x = 26 = 82 log2x = –3 x = 2–3 = 1/8 Q.29 The area (in sq. units) of the region A = {(x, y) : x2 y x + 2} is :

(1) 29 (2)

631 (3)

310 (4)

613

Ans. [1]

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Sol.

y

x 2 O –1

x2 y x + 2 x2 = y; y = x + 2 x2 = x + 2 x = 2, – 1

So, area = 2

1–

2 dx}x–)2x{( = 29

Q.30 If the function f : R – {1, – 1} A defined by f(x) = 2

2

x–1x , is surjective, then A is equal to :

(1) R – {– 1} (2) R – [– 1, 0) (3) R – (– 1, 0) (4) [0, ) Ans. [2]

Sol. y = 2

2

x–1x

y – x2y = x2

x2 = y1

y

x = y1

y

y1

y

0

+ – 1 O +

– +

Range of y is R – [–1, 0) For surjective function codomain = Range A is R – [–1, 0)