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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 - II (Memory Based) MATHEMATICS Q.1 If the system of equations 2x + 3y – z = 0, x + ky – 2z = 0 and 2x – y + z = 0 has a non-trivial solution (x, y, z), then x z z y y x + k is equal to : (1) 2 1 (2) 4 3 (3) 4 1 (4) –4 Ans.  Sol. Given system of equations has non-trival solution = 0 1 1 2 2 K 1 1 3 2 = 0 K = 2 9 so equation are 2x + 3y – z = 0 .........(1) x + 2 9 y – 2z = 0 .........(2) 2x – y + z = 0 .........(3) (1) – (3) 4y – 2z = 0 2y = z .........(4) 2 1 z y From equation (1) & (4) 2x + 3y – 2y = 0 2x + y = 0 2 1 y x or 4 x z 2 1 K x z z y y x

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

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JEE Main Online Exam 2019

Questions & Solutions 9th April 2019 | Shift - II

(Memory Based)

MATHEMATICS

Q.1 If the system of equations 2x + 3y – z = 0, x + ky – 2z = 0 and 2x – y + z = 0 has a non-trivial solution (x, y, z),

then xz

zy

yx

+ k is equal to :

(1) 21 (2)

43 (3)

41

(4) –4

Ans.  Sol. Given system of equations has non-trival solution

= 0 1122K1132

= 0

K = 29

so equation are 2x + 3y – z = 0 .........(1)

x + 29 y – 2z = 0 .........(2)

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

21

zy

From equation (1) & (4) 2x + 3y – 2y = 0 2x + y = 0

21

yx

or 4xz

21K

xz

zy

yx CAREER POINT

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Q.2 The common tangent to the circles x2 + y2

= 4 and x2 + y2

+ 6x + 8y – 24 = 0 also passes through the point : (1) (–6, 4) (2) (4, –2) (3) (–4, 6) (4) (6, –2)

Ans.  Sol. circle x2 + y2 = 4 c1(0, 0) ; r1 = 2 and circle x2 + y2 + 6x + 8y – 24 = 0 c2 (–3, –4) ; r2 = 7 d = c1c2 = 5 also d = |r1 – r2| circles touch externally equation of common tangent s1 – s2 = 0 6x + 8y – 20 = 0 3x + 4y – 10 = 0 Point (6, –2) satisfy it Q.3 Some identical balls are arranged in rows to form an equilateral triangle. The first row consists of one ball,

the second row consists of two balls and so on. If 99 more identical balls are added to the total number of balls used in forming the equilateral triangle, then all these balls can be arranged in a square whose each side contains exactly 2 balls less than the number of balls each side of the triangle contains. Then the number of balls used to form the equilateral triangle is :

(1) 157 (2) 262 (3) 225 (4) 190 Ans. 

Sol. 2

)1n(n + 99 = (n – 2)2

n2 + n + 198 = 2n2 – 8n + 8 n2 – 9n – 190 = 0 (n – 19) (n + 10) = 0 n = 19

number of balls = 220.19 = 190

Q.4 If dx)xsecxtanx(sec)x(fxtanxsece 2xsec = esecx f(x) + C, then a possible choice of f(x) is :

(1) secx – tanx – 21 (2) secx + xtanx –

21

(3) secx + tanx + 21 (4) xsecx + tanx +

21

Ans. 

Sol. xsece (secx tanx f(x) + (secx tanx + sec2x)) dx

= esecx f(x) + C Diff both side w.r.t x esecx (secx tanx f(x) + (secx tanx + sec2x)) = esecx . secx tanx f(x) + esecx f '(x) f '(x) = sec2x + tanx secx f(x) = tanx + sec + C CAREER POINT

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Q.5 If the tangent to the parabola y2 = x at a point (, ), ( > 0) is also a tangent to the ellipse, x2 + 2y2 = 1, then is equal to :

(1) 2 + 1 (2) 2 2 + 1 (3) 2 – 1 (4) 2 2 – 1 Ans.  Sol. Equation of tangent to the parabola y2 = x at (, ) is T = 0

y = 2

x

y = 2

x 2 (2 = )

y = 2

x21

2c,

22m

this is also a tangent to ellipre x2 + 2y2 = 1

C = 222 bma

21

41

2 2

21

41

4 2

2

4 – 22 – 1 = 0 ( – 1)2 = 2 2 – 1 = 2 ( > 0) 2 = 2 + 1 = 2 = 2 + 1

Q.6 The total number of matrices A =

1yx21yx2

1y20, (x, yR, x y) for which ATA = 3I3 is :

(1) 2 (2) 4 (3) 3 (4) 6 Ans.  Sol. (AT) (A) = 3I3

300030003

1yx21yx2

1y20

111yyy2x2x20

300030003

3000y6000x8

2

2

8x2 = 3 x = 83

6y2 = 3 y = 21

4 materices are possible CAREER POINT

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Q.7 If f : R R is a differentiable function and f(2) = 6, then

)x(f

62x )2x(

dtt2lim is :

(1) 0 (2) 24 f ' (2) (3) 12 f ' (2) (4) 2 f ' (2) Ans. 

Sol.

)x(f

62x

dx)2x(

tdt2lim {given that f(2) = 6}

00 form. So we use L-Hospital Rule

= 1

)x(f2).x('flim2x

= f '(2) . 2f(2) = 12f ' (2) Q.8 If a unit vector a makes angles /3 with i , /4 with jand (0, ) with k , then a value of is :

(1) 125 (2)

65 (3)

4 (4)

32

Ans.  Sol. cos2 + cos2 + cos2 = 1

21

41 + cos2 = 1

cos2 = 1 – 41

43

cos = ±21

= 3

2or3

Q.9 The area (in sq. units) of the smaller of the two circles that touch the parabola, y2 = 4x at the point (1, 2) and

the x-axis is : (1) 4 (2 – 2 ) (2) 8 (3 – 2 2 ) (3) 4 (3 + 2 ) (4) 8 (2 – 2 )

Ans.  Sol.

C (3–r, r)

r

(3 – r, 0)

x

x+y–3=0

0

y y=x+1

P (1, 2)

equation of tangent to the perabola y2 = 4x at (1, 2) is

2y = 4

21x

y = x + 1 CAREER POINT

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equation of normal y = –x + 3 Let center be c(3 – r, r) Now PC2 = r2 (3 – r – 1)2 + (r – 2)2 = r2 2(2 – r)2 = r2

r2 – 8r + 8 = 0 r = 4± 2 2 for r = 4 + 2 2 (3 – r < 0) Not possible So r = 4 – 2 2 Area = r2 = (16 + 8 – 16 2 ) = 8 (3 – 2 2 ) Q.10 Two newspapers A and B are published in a city. It is known that 25 % of the city population reads A and

(1) 13.9 (2) 13.5 (3) 12.8 (4) 13 Ans.  Sol. Let population = 100 n(A) = 25 n(B) = 20 n(A B) = 8 n(A B ) = 17 n( A B) = 12

A B

17 8 12

Now % of the population who look advertiesement

= 81005012

1004017

10030

= 13.9 Q.11 The sum of the series 1 + 2 × 3 + 3 × 5 + 4 × 7 + .... upto 11th term is :

(1) 946 (2) 916 (3) 915 (4) 945 Ans.  Sol. S = 1 + 2 × 3 + 3 × 5 + 4 × 7 + ......+ upto 11th terms nth term of the sevies is Tn = n(2n – 1)

S =

11

1n

211

1n

)nn2(Tn

Sn = 2

)1n(n6

)1n2)(1n(n2

put n = 1

S11 = 2

)12(116

)23)(12)(11(2

S11 = 946 CAREER POINT

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Q.12 If the function f(x) =

5x,3xb5x,1xa

is continous at x = 5, then value of a – b

(1) 5

2

(2) 5

2 (3)

52

(4) 5

2

Ans. 

Sol. f(x) =

5x,3|x|b5x,1|x|a

cortinous at x = 5 L.H.L = R.H.L = f(5) b|– 5| + 3 = a| – 5| + 1 –b( – 5) +3 = –a(5 – ) + 1 (a – b) ( – 5) = –2

a – b =

52

52

Q.13 Let z C be such that |z| < 1. If = )z1(5

z35 then :

(1) 4 Im () > 5 (2) 5 Re () > 1 (3) 5 Im () < 1 (4) 5 Re () > 4 Ans. 

Sol. = )z1(5

325

5 – 5z = 5 + 3z

z = 53

55

given |z| < 1

53

55 < 1

|5 – 5 | < |3 + 5|

| – 1| < 43

53

Re() = 51

Re() > 51 CAREER POINT

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Q.14 If f(x) = [x] –

4x , x R, where [x] denotes the greatest integer function, then :

(1) 4x

lim f (x) exists but 4x

lim f (x) does not exist.

(2) Both 4x

lim f (x) and 4x

lim f (x) exist but are not equal.

(3) 4x

lim f (x) exists but 4x

lim f (x) does not exist.

(4) f is continuous at x = 4. Ans. 

Sol. f(x) = [x] –

4x

4x

lim f(x) = 4x

lim

4x]x[ = 4 – 1 = 3

4x

lim f(x) = 4x

lim

4x]x[ = 3 – 0 = 3

f(4) = 3 continous at x = 4 Q.15 A rectangle is inscribed in a circle with a diameter lying along the line 3y = x + 7. If the two adjacent vertices

of the rectangle are (–8, 5) and (6, 5), then the area of the rectangle (in sq. units) is : (1) 98 (2) 56 (3) 72 (4) 84

Ans. 

Sol.

C D

A B (–8, 5) (6, 5)

(6, K) 3y = x + 7

Let vertex c is (6, K) then center of circle

2K5,1 it lies on diameter 3y = x + 7

3

2K5 = –1 + 7

K = –1 So AB = 14 and BC = 6 Area = 14 × 6 = 84

Q.16 The domain of the definition of the function f(x) = 2x41

+ log10 (x3 – x) is :

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

Ans. CAREER POINT

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Sol. f(x) = 2x41

+ log10 (x3 – x)

Let f1 = 2x41

and f2 = log10 (x3 – x)

4 – x2 0 x3 – x > 0 x ±2 x(x – 1) (x + 1) > 0

– + –

–1 0 +

1

x(–1, 0 ) (1, ) – {2} x(–1, 0 ) (1, 2) (2, ) Q.17 Two poles standing on a horizontal ground are of heights 5 m and 10 m respectively. The line joining their

tops makes and angle of 15° with the ground. Then the distance (in m) between the poles, is :

(1) 10 ( 3 –1) (2) 5(2 + 3 ) (3) 5( 3 + 1) (4) 25 (2+ 3 )

Ans. 

Sol.

D E

A

B

x

5 cm 10 cm 15°

5 C

in ABC tan15° = x5

2 – 3 = x5

x = 5(2 + 3 ) Q.18 If m is chosen in the quadratic equation (m2

+ 1) x2 – 3x – (m2 + 1)2 = 0 such that the sum of its roots is greatest, then the absolute difference of the cubes of its roots is :

(1) 38 (2) 34 (3) 510 (4) 58

Ans.  Sol. (m2 + 1)x2 – 3x + (m + 1)2 = 0

+ = 1m

32

= 1m)1m(

2

2

+ is maximum m2 + 1 is minimum m = 0 + = 3 and = 1 |3 – 3| = |( – ) (2 + + 2)|

= 22 )(4)(

= )19(49

= 58 CAREER POINT

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Q.19 The value of the integral 1

0

1cotx (1–x2+x4) dx is :

(1) 21

4

loge 2 (2) 2 – loge 2 (3)

4 – loge 2 (4)

21

2

loge 2

Ans. 

Sol. I = 1

0

1cotx (1 – x2 + x4) dx

I =

42

1

0

1

xx11tanx dx

I =

)1x(x1)1x(xtanx 22

221

0

1 dx

I = 1

0

x {tan–1 x2 – tan–1 (x2 – 1)}dx

let x2 = t 2x dx = dt

I = 1

0

11 )1t(tanttan21 dt

= 1

0

11

0

1 )1t(tan21dtttan

21 dt

= 1

0

11

0

1 )t(tan21tdttan

21 dt

= 1

0

11

0

1 )t(tan21dtttan

21 dt

= 1

0

1 )t(tan dt

=

1

02

10

1

t1t)tan.(tan dt

= 102t1log(21

4

= 2log21

4 e

Q.20 Let P be the plane, which contains the line of intersection of the planes, x + y + z – 6 = 0 and 2x + 3y + z + 5

= 0 and it is perpendicular to the xy-plane. Then the distance of the point (0, 0, 256) from P is equal to : (1) 205 5 (2) 11/ 5 (3) 63 5 (4) 17/ 5

Ans.  Sol. Equation of plane P1 + P2 = 0 (x + y + z – 6) + (2x + 3y + z + 5) = 0 x(1 + 2) + y(1 + 3) +z (1 + ) – 6 + 5 = 0 This plane is to xy - plane 1 + = 0 = –1 CAREER POINT

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So, equation of plane –x – 2y – 11 = 0 x + 2y + 11 = 0 distance of the point (0, 0, 256) from this plane

= 5

1141100

Q.21 If some three coefficients in the binomial expansion of (x + 1)n in powers of x are in the ratio 2 : 15 : 70,

then the average of these three coefficients is : (1) 232 (2) 964 (3) 625 (4) 227

Ans. 

Sol. given : 152

1rnr

152

CC

rn

1rn

15r = 2n – 2r + 2 17r = 2n + 2 ......(1)

also given 143

rn1r

7015

CC

1rn

rn

3n – 3r = 14r + 14 17r = 3n – 14 Solving (1) & (2) n = 16, r = 2

average of coefficient = 3

CCC 316

216

116

= 3

56012016

= 232

Q.22 A water tank has the shape of an inverted right cirular cone, whose semi-vertical angle is tan–1

21 . Water is

poured into it at a constant rate of 5 cubic meter per minute. Then the rate (in m/min.), at which the level of water is rising at the instant when the depth of water in the tank is 10 m; is :

(1) 1/5 (2) 1/10 (3) 1/15 (4) 2/ Ans. 

Sol. given = tan–1

21

r

h = 10m

tan =

hr

21

r = 2h CAREER POINT

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v = 31r2h

v = 31

4h3

dtdh)h3(

12dtdv 2

S =

51

dtdh

dtdh)100(

4

Q.23 The value of sin10° sin30° sin50° sin70° is :

(1) 161 (2)

361 (3)

181 (4)

321

Ans.  Sol. = sin10° sin30° sin50° sin70° = sin30°{sin10° sin(60°–10°) sin(60° + 10°)}

= sin30°

)10(3sin

41

=

21

41

21

= 161

Q.24 If p (q r) is false, then the truth values of p, q, r are respectively : (1) F, F, F (2) T, F, F (3) F, T, T (4) T, T, F

Ans.  Sol. p (q r) is false. ( T F = F) So, P = T, q = F and r = F

Q.25 The vertices B and C of ABC lie on the line, 4z

01y

32x

such that BC = 5 units. Then the area (in sq.

units) of this triangle, given that the point A(1, –1, 2), is :

(1) 5 17 (2) 34 (3) 2 34 (4) 6

Ans. 

Sol.

A(1, –1, 2)

D 5

B

4z

01y

32x

C

Let any point on given line is D(3 –2, 1, 4) Now AD BC D.R. of BC a1 = 3, b1 = 0, c1 = 4 D.R. of AD a2 = 3 – 3, b2 = 2, c2 = 4 – 2 CAREER POINT

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a1a2 + b1b2 + c1c2 = 0 3(3 – 3) + 0 + 4(4 – 2) = 0 25 = 17

= 2517

co-oridnation of point D

2568,1,

251

253244

625576

Area of ABC = 21 × BC × AD

= 21 × 5 × 34

52

= 34

Q.26 If cosx dxdy – ysinx = 6x, (0<x<

2 ) and y

3= 0, then y

6is equal to :

(1) – 34

2 (2) – 2

2 (3) 32

2 (4) – 32

2

Ans. 

Sol. cosx dxdy – ysinx = 6x

dxdy – y tanx = 6x sec x

I.F = xsec

1ee xseclogdxxtane

solution of equation

y. dxxsec

1.xsecx6xsec

1

xsec

y = 3x2 + c ........(1)

given y

3= 0

So, 0 = 9

3 2 + C

C = – 3

2

Now from (1)

3

x3xsec

y 22

at x = 6

336

32

y3 22

y = – 32

2 CAREER POINT

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Q.27 If the sum and product of the first three terms in an A.P. are 33 and 1155, respectively, then a value of its 11th term is :

(1) –25 (2) –35 (3) –36 (4) 25 Ans.  Sol. Let the three numbers in A.P. are a – d, a, a + d given that : a – d + a + d = 33 a = 11 and (a – d)(a)(a + d) = 1155 a(a2 – d2) = 1155 11(121 – d2) = 1155 d2 = 16 d = ±4 If d = 4 then first term a – d = 7 If d = –4 then first term a – d = 15 T11 = 7 + 10(4) = 47 or T11 = 15 + 10(–4) = –25 Q.28 The mean and the median of the following ten numbers in increasing order 10, 22, 26, 29, 34 x, 42, 67, 70, y

are 42 and 35 respectively, then xy is equal to :

(1) 7/2 (2) 9/4 (3) 8/3 (4) 7/3 Ans.  Sol. mean = 42

10

y706742x3429262210 = 45

420 = 300 + x+ y x + y = 120 ......(1) and medium = 35

2

x34 = 35

x = 36 From (1) y = 84

37

3684

xy

Q.29 If the two lines x + (a – 1) y = 1 and 2x + a2y = 1 (a R –{0, 1}) are perpendicular, then the distance of their

point of intersection from the origin is :

(1) 52 (2)

52 (3)

52 (4)

52

Ans.  Sol. Two lines are perpendicular m1m2 = –1

2a2

1a1 = –1

a3 – a2 + 2 = 0 (a + 1) (a2 – 2a + 2) = 0 CAREER POINT

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a = –1 so lines are L1 : x – 2y + 1 = 0 L2 : 2x + y – 1 = 0

Solving there equation we get point of intersetion P

53,

51

Now distance of P from origin

OP = 52

259

251

Q.30 The area (in sq. units) of the region A = {(x, y) : 2y2

x y + 4} is :

(1) 18 (2) 30 (3) 3

53 (4) 16

Ans.  Sol.

(8, 4)

0

y

(2, –2)

x

y2 = 2x .......(1) and x – y – 4 = 0 .......(2) solving (1) & (2) (x – 4)2 = 2x x2 – 10x + 16 = 0 x = 81 + 2 and y = 41 –2

A =

4

2

2

2y4y dy

A = 4

2

32

3yy4

4y

A =

6881

664164 = 18