of 14 /14
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.  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
• Author

others
• Category

## Documents

• view

6

0

Embed Size (px)

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

CAREER POINT Ltd., CP Tower, IPIA, Road No.1, Kota (Raj.), Ph: 0744-6630500 www.careerpoint.ac.in 1

JEE Main Online Paper

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. 

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– CAREER POINT

CAREER POINT Ltd., CP Tower, IPIA, Road No.1, Kota (Raj.), Ph: 0744-6630500 www.careerpoint.ac.in 2

JEE Main Online Paper

Ans.  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.  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.  Sol. ~ (p p q) = ~ p (p ~ q) = (~ p p) (~ p ~ q) = c (~ p ~ q) = ~ p ~ q CAREER POINT

CAREER POINT Ltd., CP Tower, IPIA, Road No.1, Kota (Raj.), Ph: 0744-6630500 www.careerpoint.ac.in 3

JEE Main Online Paper

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. 

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.  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) CAREER POINT

CAREER POINT Ltd., CP Tower, IPIA, Road No.1, Kota (Raj.), Ph: 0744-6630500 www.careerpoint.ac.in 4

JEE Main Online Paper

so, 2

0cos

1

h

and 2sin

10k

h2

1cos and k21sin

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.  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 CAREER POINT

CAREER POINT Ltd., CP Tower, IPIA, Road No.1, Kota (Raj.), Ph: 0744-6630500 www.careerpoint.ac.in 5

JEE Main Online Paper

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.  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.  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. 

Sol. I = dxxeccosxsec 3/43/2

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

xcosxcosxsin

dxI2

3/4 CAREER POINT

CAREER POINT Ltd., CP Tower, IPIA, Road No.1, Kota (Raj.), Ph: 0744-6630500 www.careerpoint.ac.in 6

JEE Main Online Paper

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.  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.  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.  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 CAREER POINT

CAREER POINT Ltd., CP Tower, IPIA, Road No.1, Kota (Raj.), Ph: 0744-6630500 www.careerpoint.ac.in 7

JEE Main Online Paper

…………………. …………………. 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. 

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. 

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

22

mba

)ba(mmxy

2m1824

m4237 CAREER POINT

CAREER POINT Ltd., CP Tower, IPIA, Road No.1, Kota (Raj.), Ph: 0744-6630500 www.careerpoint.ac.in 8

JEE Main Online Paper

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. 

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. 

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. CAREER POINT

CAREER POINT Ltd., CP Tower, IPIA, Road No.1, Kota (Raj.), Ph: 0744-6630500 www.careerpoint.ac.in 9

JEE Main Online Paper

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.  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.  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.  Sol. Roots of eqn. x2 + x + 1 = 0 are and

, = 2

4–11– = 2

3i1–

= , = 2 (complex cube root of unity) CAREER POINT

CAREER POINT Ltd., CP Tower, IPIA, Road No.1, Kota (Raj.), Ph: 0744-6630500 www.careerpoint.ac.in 10

JEE Main Online Paper

=

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. 

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. 

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 CAREER POINT

CAREER POINT Ltd., CP Tower, IPIA, Road No.1, Kota (Raj.), Ph: 0744-6630500 www.careerpoint.ac.in 11

JEE Main Online Paper

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.  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. 

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] CAREER POINT

CAREER POINT Ltd., CP Tower, IPIA, Road No.1, Kota (Raj.), Ph: 0744-6630500 www.careerpoint.ac.in 12

JEE Main Online Paper

= 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. 

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.  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. CAREER POINT

CAREER POINT Ltd., CP Tower, IPIA, Road No.1, Kota (Raj.), Ph: 0744-6630500 www.careerpoint.ac.in 13

JEE Main Online Paper

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.  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. 

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. CAREER POINT

CAREER POINT Ltd., CP Tower, IPIA, Road No.1, Kota (Raj.), Ph: 0744-6630500 www.careerpoint.ac.in 14

JEE Main Online Paper

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. 

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)