JEE Main 2019 Maths Paper with Solutions (April) - Download PDF

JEE Main 2019 solved Maths paper solutions are given on this page. The solutions are given in a step by step manner so that students can easily understand the problems. These solutions are prepared by subject experts. Students are recommended to revise and learn these solutions, so that they can easily improve their problem-solving skills. These solutions can be downloaded in PDF format for free.

April Paper – Maths

1. The sum of coefficients of even powers of x in (x+√(x³-1)⁶ + x-√(x³-1)⁶)
A) 23
B) 24
C) 18
D) 21

(x+√(x³-1)⁶ + x-√(x³-1)⁶) = 2[6C₀x⁶ +6C₂x⁴(x³-1)+6C₄x²(x³-1)² +6C₆(x³-1)³]

= 2[x⁶+15x⁴(x³-1)+15x²(x⁶-2x³+1)+(x³-1)³]

= 2[x⁶+15x⁷-15x⁴+15x⁸-30x⁵+15x²+x⁹-3x⁶+3(x³-1)]

Terms with even power of x =

\begin{array}{l} 2 [x^{6} - 15 x^{4} + 15 x^{8} + 15 x^{2} - 3 x^{6} - 1] \end{array}

Therefore Sum of coefficients = 2(1-15+15+15-3-1) = 24

Answer: B

2. Let sin(α-β) = 5/13 and cos (α+β) = 3/5 where α, β ∈ (0, π/4) then tan 2α
A) 63/16
B) 61/16
C) 65/16
D) 32/9

sin(α-β) = 5/13

cos (α-β) = 12/13

cos(α+β) = 3/5

sin(α+β) = 4/5

tan 2α = tan[(α+β)+(α-β)]

= (4/3)+(5/12)÷(1-(4/3)×(5/12))

= 63/16

Answer: A

3.The line x+y = n, n∈N, makes intercepts with x²+y² = 16. Then the sum of squares of all possible intercepts
A) 105/4
B) 105
C) 210
D) 105/2

x²+y² = 16

Centre = (0,0)

OA = P

Radius = 4

P = n/√2

JEE Main 2019 Paper With Solution Maths April

To make the intercepts n/√2 < 4

n< 4√2

Length of intercepts = √(r²-p²)

AB = √(16-n²/2)

Length of chord BC = 2√(16-n²/2) = √(64-2n²)

Square of intercepts = (64-2n²), n∈N

Possible value of n = 1,2,3,4,5

Sum of square of intercepts = (64-2)+(64-8)+(64-18)+(64-32)+(64-50)

= 62+56+46+32+50

= 210

Answer: C

4.

\begin{array}{l} \int \frac{\sin \frac{5 x}{2}}{\sin \frac{x}{2}} . d x = \end{array}

A) x+2sin x+sin 2x+c
B) x+2cos x+sin 2x+c
C) x-2sin x+sin 2x+c
D) x+2sin x-sin 2x+c

\begin{array}{l} \int \frac{\sin \frac{5 x}{2}}{\sin \frac{x}{2}} . d x \end{array}

=

\begin{array}{l} \int \frac{2 \sin \frac{5 x}{2} . \cos \frac{x}{2}}{2 \sin \frac{x}{2} . \cos \frac{x}{2}} . d x \end{array}

=

\begin{array}{l} \int \frac{\sin 3 x + \sin 2 x}{\sin 2 (\frac{x}{2})} . d x \end{array}

=

\begin{array}{l} \int \frac{\sin 3 x}{\sin x} . d x + \int \frac{\sin 2 x}{\sin x} . d x \end{array}

=

\begin{array}{l} \int \frac{3 \sin x - 4 \sin^{3} x}{\sin x} . d x + \int \frac{2 \sin x \cos x}{\sin x} d x \end{array}

=

\begin{array}{l} 3 \int d x - 4 \int \sin^{2} x . d x + 2 \int \cos x d x \end{array}

=

\begin{array}{l} 3 x - 4 \int \frac{1 - \cos 2 x}{2} . d x + 2 \sin x \end{array}

=

\begin{array}{l} 3 x - 2 x + \sin 2 x + 2 \sin x + c \end{array}

= x+2 sinx+sin 2x +c

Answer: A

5. The area bounded by the curve y ≤ x²+3x, 0 ≤ y ≤ 4, 0 ≤ x ≤3 is
A) 59/6
B) 57/4
C) 59/3
D) 57/6

JEE Main 2019 Papers With Solutions April Maths

y = x²+3x

y = 4

x²+3x= 4 = 0

x²+3x-4 = 0

x = 1 or x = -4

Area =

\begin{array}{l} = \int_{0}^{1} (x^{2} + 3 x) . d x \end{array}

+ area of rectangle

=

\begin{array}{l} {[\frac{x^{3}}{3} + \frac{3 x^{2}}{2}]}_{0}^{1} + 2 (4) \end{array}

= (1/3)+(3/2)+8

= 59/6

Answer: A

6. “If you are born in India then you are citizen of India” contrapositive of this statement is
A) If you are born in India then you are not citizen of India
B) If you are not citizen of India then you are not born in India
C) If you are citizen of India then you are not born in India
D) If you are citizen of India then you are born in India

Statement:

\begin{array}{l} p \Rightarrow q \end{array}

Contrapositive:

\begin{array}{l} \tilde{} p \Rightarrow \tilde{} q \end{array}

Therefore answer is if you are not citizen of India then you are not born in India.

Answer: B

7.

\begin{array}{l} A = [\begin{array}{c} \cos α & - \sin α \\ \sin α & \cos α \end{array}] \end{array}

and

\begin{array}{l} A^{32} = [\begin{array}{c} 0 & - 1 \\ 1 & 0 \end{array}] \end{array}

. then α may be
A) 0
B) π/32
C) π/64
D) π/16

A =

\begin{array}{l} [\begin{array}{c} \cos α & - \sin α \\ \sin α & \cos α \end{array}] \end{array}

A² =

\begin{array}{l} [\begin{array}{c} \cos 2 α & - \sin 2 α \\ \sin 2 α & \cos 2 α \end{array}] \end{array}

Aⁿ =

\begin{array}{l} [\begin{array}{c} \cos n α & - \sin n α \\ \sin n α & \cos n α \end{array}] \end{array}

cos 32α = 0 and sin 32α = 1

32α = π/2

α = π/64

Answer: C

8. Shortest distance between the curves y² = x-2 and y = x is
A) greater than 4
B) less than 2
C) greater than 3
D) greater than 2

y² = x-2

Differentiating we get,

2y (dy/dx) = 1

(dy/dx) = 1/2y

Slope of the line y = x is 1.

Slope of the tangent to the curve = 1/2y

1/2y = 1

→ y = 1/2

→ K = 1/2

∴ N = (9/4, 1/2)

Minimum distance = MN = perpendicular distance from point N to line y = x

\begin{array}{l} | \frac{a x_{1} + b y_{1} + c}{\sqrt{a^{2} + b^{2}}} | \end{array}

=

\begin{array}{l} | \frac{1 (\frac{9}{4}) - 1 (\frac{1}{2})}{\sqrt{1 + 1}} | \end{array}

= 7/4√2 which is less than 2.

Answer: B

9.

\begin{array}{l} lim_{x \to 0} \frac{\sin^{2} x}{\sqrt{2} - \sqrt{1 + \cos x}} = \end{array}

A) √2
B) 2
C) 4
D) 4√2

Rationalise the denominator

\begin{array}{l} lim_{x \to 0} \frac{\sin^{2} x [\sqrt{2} + \sqrt{1 + \cos x}]}{2 - (1 + \cos x)} \end{array}

=

\begin{array}{l} lim_{x \to 0} \frac{\sin^{2} x (\sqrt{2} + \sqrt{1 + \cos x})}{1 - \cos x} \end{array}

=

\begin{array}{l} \underset{x \to 0}{L t} \frac{\sin^{2} x (\sqrt{2} + \sqrt{1 + \cos x})}{2 \sin^{2} \frac{x}{2}} \end{array}

=

\begin{array}{l} \underset{x \to 0}{L t} \frac{\sin^{2} x}{x^{2}} . \frac{1}{2} \frac{1}{{(\frac{\sin \frac{x}{2}}{\frac{x}{2}})}^{2} \times \frac{1}{4}} (\sqrt{2} + \sqrt{1 + \cos x}) \end{array}

= (4/2) √2+√(1+1)

= 4√2

Answer: D

10. How many 9 digit numbers can be formed by using the digits 1, 1, 2, 2, 2, 2, 3, 4, 4 such that odd numbers occur at even places.
A) 160
B) 175
C) 180
D) 220

JEE Main Maths 2019 Solved Paper For April

Even places = 4

Odd numbers = 3

Even numbers = 6

Therefore number of numbers =

\begin{array}{l} = 4 c_{3} \times \frac{3!}{2!} \times \frac{6!}{4! 2!} = 180 \end{array}

Answer: C

11. Let g(x) = ln x and

\begin{array}{l} f (x) = (\frac{1 - x \cos x}{1 + x \cos x}) \end{array}

then

\begin{array}{l} \int_{- \frac{π}{4}}^{\frac{π}{4}} g [f (x)] . d x = \end{array}

A) ln 1
B) ln 2
C) ln e
D) ln 4

g(f(x)) =

\begin{array}{l} g [\frac{1 - x \cos x}{1 + x \cos x}] \end{array}

=

\begin{array}{l} ℓ n (\frac{1 - x \cos x}{1 + x \cos x}) \end{array}

I =

\begin{array}{l} \int_{- \frac{π}{4}}^{\frac{π}{4}} g [f (x)] \end{array}

=

\begin{array}{l} \int_{- \frac{π}{4}}^{\frac{π}{4}} ℓ n (\frac{1 - x \cos x}{1 + x \cos x}) d x \end{array}

..(1)

We have

\begin{array}{l} \int_{a}^{b} f (x) = \int_{a}^{b} f (a + b - x) \end{array}

I =

\begin{array}{l} \int_{- \frac{π}{4}}^{\frac{π}{4}} ℓ n (\frac{1 + x \cos x}{1 - x \cos x}) \end{array}

..(2)

Add (1) and (2)

2I =

\begin{array}{l} \int_{- \frac{π}{4}}^{\frac{π}{4}} [ℓ n (\frac{1 + x \cos x}{1 - x \cos x}) + (\frac{1 - x \cos x}{1 + x \cos x})] d x \end{array}

=

\begin{array}{l} \int_{- \frac{π}{4}}^{\frac{π}{4}} ℓ n (\frac{1 + x \cos x}{1 - x \cos x} . \frac{1 - x \cos x}{1 + x \cos x}) d x \end{array}

Because log ab = log a + log b

2I =

\begin{array}{l} \int_{- \frac{π}{4}}^{\frac{π}{4}} ℓ n 1 = 0 \end{array}

Answer: A

12. The sum of the series 2. 20C₀ + 5.20C₁+8.20C₂+62.20C₂₀ is equal to
A) 16.2²²
B) 8.2²²
C) 8.2²¹
D) 16.2²¹

2. 20C₀ + 5.20C₁+8.20C₂+62.20C₂₀

=

\begin{array}{l} \sum_{r = 0}^{20} (3 r + 2) 20 C_{r} \end{array}

=

\begin{array}{l} 3 \sum_{r = 0}^{20} r . 20 C_{r} + 2. \sum_{r = 0}^{20} 20 C_{r} . \end{array}

=

\begin{array}{l} 3.20 \sum_{r = 1}^{20} 19 C_{r - 1} + 2 (20 C_{0} + 20 C_{1} + \dots + 20 C_{20}) \end{array}

= 60.2¹⁹+2.2²⁰

= 2²¹(15+1)

= 2²¹.16

Answer: D

13. Sum of natural numbers between 100 and 200 whose HCF with 91 should be more than 1.
A) 1121
B) 3210
C) 3121
D) 1520

Given numbers: 101, 102, 103, … 198, 199

91=13 × 7

HCF of 91 and a number is more than 1 means the number should be either multiple of 7 or 13.

Therefore sum of the numbers = (Numbers divisible by 7) + (Numbers divisible by 13) – (Numbers divisible by 91)

= (105+112+…+196 +( 104+117+…195 )-( 182)

= (14/2)[105+196]+(8/2)[104+195]-182

= 7(301)+4(299)-182

= 2107+1196-182

= 3121

Answer: C

14. If mean and variance of 7 variates are 8 and 16 respectively and five of them are 2, 4, 10, 12, 14 then the product of remaining two variates is
A) 49
B) 48
C) 45
D) 40

Let the remaining two variates be x and y

Given mean = 8

(x+y+2+4+10+12+14)/7 = 8

x+y+42 = 56

x+y = 14 ..(1)

Given variance = 16

[(x²+y²+4+16+100+144+196)/7] – 8² = 16

(x²+y²+460)/7 = 80

(x²+y²+460) = 560

(x²+y²) = 100 ..(2)

(x+y)²-2xy = 100

Substitute (1) in above equation

14²-2xy = 100

196-2xy = 100

2xy = 96

xy = 48

Answer: B

15. If α and β are the roots of x²-2x+2 = 0 then the minimum value of n such that (α/β)ⁿ = 1
A) 4
B) 3
C) 2
D) 5

x²-2x+2 = 0

x²-2x+1+1 = 0

(x-1)² = -1

(x-1) = ±i

x = 1+i or 1-i

Let α = 1+i

β = 1-i

(α/β) = (1+i)/(1-i)

= (1+i)²/2

= (1+i²+2i)/2

= (1-1+2i)/2

= i

(α/β)ⁿ = iⁿ = 1

Least value of n is 4.

Answer: A

16. Let y = y(x) be the solution of the differential equation, (x²+1)²dy/dx + 2x(x²+1)y = 1 such that y(0) = 0. If √a y(1) = π/32, then the value of a is:
A) 1/2
B) 1/4
C) 1/16
D) 1

(x²+1)²dy/dx + 2x(x²+1)y = 1

\begin{array}{l} \frac{d y}{d x} + \frac{2 x}{x^{2} + 1} . y = \frac{1}{{(x^{2} + 1)}^{2}} \end{array}

dy/dx +Py = Q

P = 2x/x²+1

Q = 1/(x²+1)²

I.F =

\begin{array}{l} e^{\int \frac{2 x}{1 + x^{2}}} d x \end{array}

= 1+x²

The general solution of linear differential equation is

y.(I.F) =

\begin{array}{l} \int Q . (I . F) d x + C \end{array}

y(1+x²) =

\begin{array}{l} \int \frac{1}{1 + x^{2}} d x + C \end{array}

= tan^-1 x+C

At x = 0, y = 0

0(1) = tan^-10+C

C = 0

y.(x²+1) = tan^-1x

put x = 1

y.(2) = tan^-11

y.(2) = π/4

y = π/8

(1/4)y = π/32

i.e √a = ¼

a = 1/16

Answer: C

17. If f(x) = log((1-x)/(1+x)) then f(2x/1+x²) is equal to
A) f(x)
B) 2f(x)
C) -2f(x)
D) [f(x)]²

f(x) =

\begin{array}{l} \log (\frac{1 - x}{1 + x}) \end{array}

f(2x/1+x²) =

\begin{array}{l} \log [\frac{1 - \frac{2 x}{1 + x^{2}}}{1 + \frac{2 x}{1 + x^{2}}}] \end{array}

=

\begin{array}{l} \log (\frac{1 + x^{2} - 2 x}{1 + x^{2} + 2 x}) \end{array}

=

\begin{array}{l} \log {(\frac{1 - x}{1 + x})}^{2} \end{array}

=

\begin{array}{l} 2 l o g (\frac{1 - x}{1 + x}) \end{array}

= 2 f(x)

Answer: B

18. Given that

\begin{array}{l} A \subset B \end{array}

then identify the correct statement.
A)

\begin{array}{l} P (\frac{A}{B}) = P (A) \end{array}

B)

\begin{array}{l} P (\frac{A}{B}) \leq P (A) \end{array}

C)

\begin{array}{l} P (\frac{A}{B}) \geq P (A) \end{array}

D)

\begin{array}{l} P (\frac{A}{B}) = P (A) - P (B) \end{array}

\begin{array}{l} P (\frac{A}{B}) = \frac{P (A \cap B)}{P (B)} \end{array}

=

\begin{array}{l} \frac{P (A)}{P (B)} \geq P (A) \end{array}

19. Find the value of ‘c’ for which the following equations have non-trivial solutions.

cx-y-z = 0

-cx+y-cz = 0

x+y-cz = 0

A) 1/2
B) -1
C) 2
D) 0

For non trivial solutions,

\begin{array}{l} | \begin{array}{c} c & - 1 & - 1 \\ - c & 1 & - c \\ 1 & 1 & - c \end{array} | = 0 \end{array}

c(-c+c) +1(c²+c)-1(-c-1) = 0

c²+c+c+1 = 0

(c+1)² = 0

c+1 = 0

c = -1

Answer: B

20. Let

\begin{array}{l} 2 y = {[\cot^{- 1} (\frac{\sqrt{3} \cos x + \sin x}{\cos x - \sqrt{3} \sin x})]}^{2} \end{array}

then

\begin{array}{l} \frac{d y}{d x} = \end{array}

A) x-(π/6)
B) x+(π/6)
C) 2x-(π/6)
D) 2x-(π/3)

April JEE Main 2019 Paper With Solutions Maths

April JEE Main 2019 Paper With Solutions Maths

2y = [(π/2) -(π/3)+x]²

2y = [(π/6)-x]²

Differentiating we get,

2 dy/dx = 2[(π/6)-x](-1)

dy/dx = x-(π/6)

Answer: A

21. Let S₁ is set of minima and S₂is set of maxima for the curve y = 9x⁴+12x³-36x²-25 then
A) S₁= {-2,-1} S₂= {0}
B) S₁= {-2,1} S₂= {0}
C) S₁= {-2,1} S₂= {-1}
D) S₁= {-2,2} S₂= {0}

y = 9x⁴+12x³-36x²-25

dy/dx = 36x³ + 36x²-72x

= 36x(x²+x-2)

= 36x(x²+2x-x-2)

= 36x(x+2)(x-1)

Critical points are 0,-2,1

At {-2,1} -> points of minima

{0} -> points of maxima.

Answer: B

22. Let f:[0,2] -> R be a twice differentiable functions such that f’’(x) >0 for all x ∈ (0,2). If f(x) = f(x)+f(2-x). then f is:
A) Increasing in (0, 1) and decreasing in (1, 2)
B) Decreasing in (0, 1) and increasing in (1, 2)
C) Increasing in (0, 2)
D) Decreasing in (0, 2)

Given f’’(x) >0, x ∈(0,2)

i.e f’(x) is an increasing function

f(x) = f(x)+f(2-x)

f’(x) = f’(x)-f’(2-x)

f(x) is increasing (x) > 0

f’(x)-f’(2-x) >0

f’(x) > f’(2-x)

x > 2-x

2x > 2

x>1

i.e x ∈ (1,2)

f(x) is decreasing (x) < 0

f’(x)-f’(2-x) <0

f’(x) < f’(2-x)

x < 2-x

2x < 2

x <1

i.e x ∈ (0,1)

Answer: B

23. Let vertices of the triangle ABC is A(0, 0), B(0, 1) and C(x, y) and perimeter is 4 then locus of ‘C’ is
A) 9x² + 8y²+8y = 16
B) 8x² + 9y²+9y = 16
C) 9x² + 8y²-8y = 16
D) 8x² + 9y²-9x = 16

Maths April JEE Main 2019 Paper With Solutions

Given AB+AC+BC = 4

1+

\begin{array}{l} \sqrt{x^{2} + {(y - 1)}^{2}} + \sqrt{x^{2} + y^{2}} \end{array}

= 4

√(x²+y²)² = (3-√(x²+(y-1)²)²

x²+y² = 9+x²+y²-2y+1-6√(x²+(y-1)²)

3√(x²+(y-1)²) = 5-y

9(x²+(y-1)² = (5-y)²

9x²+8y²-8y = 16

Answer: C

24. A point on the straight line, 3x + 5y = 15 which is equidistant from the coordinate axes will lie only in
A) 1^st, 2^nd and 4^th quadrants
B) 1^st and 2^nd quadrants
C) 4^th quadrants
D) 1^st quadrants

3x+5y = 15 is the equation of the straight line.

If x = y

3x+5x = 15

8x = 15

x = 15/8

y = 15/8

P(15/8, 15/8) lies in 1^st quadrant.

If x = -y

3x-5x = 15

-2x = 15

x = -15/2

y = 15/2

Q(-15/2, 15/2) lies in 2^nd quadrant.

Answer: B

25. The perpendicular distance of point ( 2,-1,4) from the line

\begin{array}{l} \frac{x + 3}{10} = \frac{y - 2}{- 7} = \frac{z}{1} \end{array}

lies between
A) (2, 3)
B) (3, 4)
C) (4, 5)
D) (1, 2)

Let A be (2,-1,4).

Direction ratio of AB = (10λ-5, -7λ+3, λ-4)

Drs of given line = (10, -7, 1)

10(10λ-5)-7(-7λ+3)+(λ -4) = 0

100λ -50+49-21+λ-4 = 0

150λ = 75

λ = 1/2

Point B (2, -3/2, 1/2)

Length AB = 5/√2

= 3.53

Answer: B

26. If a plane passes through intersection of planes 2x-y-4 = 0 and y+2z-4 = 0 and also passes through the point (1, 1, 0). Then the equation of the plane is
A) x-y-z = 0
B) 2x-z = 0
C) x+2z-1 = 0
D) x-z-1 = 0

Required equation of the plane is (2x-y-4)+λ(y+2z-4) = 0

Given it pass through (1,1,0)

(2-1-4)+λ(1+0-4) = 0

λ= -1

(2x-y-4)-(y+2z-4) = 0

2x-2y-2z = 0

x-y-z = 0

Answer: A

27. The sum of the solutions of the equation |√x-2|+√x(√x-4)+2 = 0, (x>0) is equal to
A) 4
B) 10
C) 9
D) 12

Given equation is |√x-2|+√x(√x-4)+2 = 0, (x>0)

Put √x = t

|t-2|+t(t-4)+2 = 0

If t ≥2

t-2+t²-4t+2 = 0

t²-3t = 0

t(t-3) = 0

t = 3 or t = 0

t = 0 not possible

if t = 3, x = 9

If t < 2

2-t+t²-4t+2 = 0

t²-5t+4 = 0

t = 1 or t = 4

t = 4 not possible

so t = 1

x = 1

Sum of solution = 9+1 = 10

Answer: B

28. If the tangents on the ellipse 4x²+y² = 8 at the points (1,2) and (a,b) areperpendicular to each other, then a² is equal to
A) 2/17
B) 64/17
C) 128/17
D) 4/17

The equation of the ellipse 4x²+y² = 8

dy/dx = -4x/y

The tangent at (1,2) and (a,b) are perpendicular

(-y/2)(-4a/b) = -1

b = -8a ..(i)

(a,b) is on the ellipse.

4a²+b² = 8 [from eq(i)]

4a²+64a² = 8

a² = 8/68 = 2/17

Answer: A

29. Find the magnitude of projection of vector

\begin{array}{l} 2 \hat{i} + 3 \hat{j} + \hat{k} \end{array}

on a vector which is perpendicular to the plane containing vectors

\begin{array}{l} \hat{i} + \hat{j} + \hat{k} \end{array}

and

\begin{array}{l} \hat{i} + 2 \hat{j} + 3 \hat{k} \end{array}

A) √3/2
B) √(3/2)
C) 3√6
D) √6

The vector perpendicular to given vectors is

\begin{array}{l} \vec{a} = | \begin{array}{c} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 1 & 2 & 3 \end{array} | \end{array}

So

\begin{array}{l} \vec{a} = \hat{i} - 2 \hat{j} + \hat{k} \end{array}

Projection of vector b in direction of vector a =

\begin{array}{l} \frac{| \vec{a} . \vec{b} |}{| \vec{a} |} \end{array}

=

\begin{array}{l} \frac{| (\hat{i} - 2 \hat{j} + \hat{k}) . (2 \hat{i} + 3 \hat{j} + \hat{k}) |}{| \sqrt{6} |} \end{array}

= 2-6+1/√6

= 3/√6

= √(3/2)

Answer: B

30. If α = cos^-1(3/5), β = tan^-1 (1/3), where 0<α, β < π/2, then α-β is equal to:
A) tan^-1(9/5√10)
B) sin^-1(9/5√10)
C) tan^-1(9/15)
D) cos^-1(9/5√10)

α = cos^-1(3/5) = tan^-1(4/3)

β = tan^-1 (1/3)

α-β = tan^-1(4/3) – tan^-1 (1/3)

=

\begin{array}{l} \tan^{- 1} (\frac{\frac{4}{3} - \frac{1}{3}}{1 + \frac{4}{3} \times \frac{1}{3}}) \end{array}

=

\begin{array}{l} \tan^{- 1} (\frac{1}{1 + \frac{4}{9}}) \end{array}

= tan^-1(9/13)

=

\begin{array}{l} \sin^{- 1} \frac{9}{\sqrt{13^{2} + 9^{2}}} \end{array}

= sin^-1(9/√250)

= sin^-1(9/5√10)

Answer: B

Video Lessons – Maths Paper April

13,029

42,860

3,00,921

1,22,329

JEE Main 2019 April Maths Paper with Solutions

carouselExampleControls112

Comments

Leave a Comment Cancel reply