JEE Main 2020 Maths Paper With Solutions Shift 2 Sept 5

1. If x = 1 is a critical point of the function f(x) = (3x²+ax-2-a)e^x, then:

1) x = 1 is a local minima and x = -2/3 is a local maxima of f.
2) x = 1 is a local maxima and x = -2/3 is a local minima of f.
3) x = 1 and x = -2/3 are local minima of f.
4) x = 1 and x = -2/3 are local maxima of f.

f(x) = (3x²+ax-2-a)e^x

f’(x) = (3x²+ax-2-a)e^x +(6x+a)e^x = 0

e^x [3x²+(a+6) x-2] =0

at x = 1, 3+a+6-2 = 0

a = -7

f(x) = (3x²-7x+5)e^x

f’(x) = (6x-7)e^x +(3x²-7x+5)e^x

= e^x(3x²-x-2) = 0

= 3x²-3x+2x-2 = 0

= (3x+2)(x-1) = 0

x = 1, -2/3

x = 1 is point of local minima.

x = –2/3 is point of local maxima.

Answer: (1)

2.

1) is equal to √e
2) is equal to 1
3) is equal to 0
4) does not exist

Answer: (2)

3. The statement (p → (q → p)) → (p → (p ˅ q)) is:

1) equivalent to (p˅q)˄ (~ p)
2) equivalent to (p˄q)˅(~ p)
3) a contradiction
4) a tautology

Answer: (4)

4. If L = sin²(π/16) – sin²(π/8) and M = cos²(π/16) – sin²(π/8), then

1) M = (1/2√2)+(1/2)cos (π/8)
2) M = (1/4√2)+(1/4)cos (π/8)
3) L = -(1/2√2)+(1/2)cos (π/8)
4) L = (1/4√2)-(1/4)cos (π/8)

L = sin(3π/16) sin(-π/16)

L = (-1/2)[cos(π/8)-cos(π/4)]

L = (1/2√2)-(1/2)cos (π/8)

M = cos(3π/16)cos(π/16)

M = (1/2)[cos(π/4)+cos(π/8)]

M = (1/2√2)+(1/2)cos(π/8)

Answer: (1)

5. If the sum of the first 20 terms of the series is 460, then x is equal to:

1) 7^1/2
2) 7²
3) e²
4) 7^46/21

(2+3+4+…+21)log₇x = 460

⇒ (20×(21+2)/2)log₇ x = 460

230 log₇x = 460

log₇x = 2

x = 7²

Answer: (2)

6. There are 3 sections in a question paper and each section contains 5 questions. A candidate has to answer a total of 5 questions, choosing at least one question from each section. Then the number of ways, in which the candidate can choose the questions, is:

1) 2250
2) 2255
3) 1500
4) 3000

3(⁵C₁×⁵C₂×⁵C₂)+3(⁵C₁ ×⁵C₁×⁵C₃)

= 3(5×5×2×5×2)+3(5×5×10)

= 750 + 1500

= 2250

Answer: (1)

7. If the mean and the standard deviation of the data 3,5,7,a,b are 5 and 2 respectively, then a and b are the roots of the equation:

1) x²-20x+18 = 0
2) x²-10x+19 = 0
3) 2x²-20x+19 = 0
4) x²-10x+18 = 0

S.D =

2² = (83+a²+b²)/5 -(5)²

4 = (83+a²+b²)/5 -(25)

29×5-83 = a²+b²

⇒ a²+b² = 62

(a+b+15)/5 = 5

⇒ a+b = 10 ..(i)

2ab = 100-62 = 38

ab = 19 ..(ii)

from (i) and (ii)

x²-10x+19 = 0

Answer: (2)

8. The derivative of tan^-1(√(1+x²)-1)/x with respect to tan^-1(2x√(1-x²))/(1-2x²)) at x = 1/2 is:

a) 2√3/3
b) 2√3/5
c) √3/12
d) √3/10

x = tan θ

u = tan^-1(sec θ-1)/tan θ

= tan^-1 (tan θ/2)

= θ/2

= (tan^-1 x)/2

x = sin θ

v = tan^-1(2 sinθ cosθ/cos 2θ) = 2θ

= 2 sin^-1x

du/dv = ½(1+x²)×√(1-x²)/2

= (√3/2×2)×(4/5×2)

= √3/10

Answer: (4)

9. If ∫ cos θ/(5+7sinθ-2+2sin²θ)dθ = A log_e |B(θ)| + C where C is a constant of integration, then B(θ)/A can be:

1) 5(2sinθ+1)/(sinθ+3)
2) 5(sinθ+3)/(2sinθ+1)
3) (2sinθ+1)/(sinθ+3)
4) (2sinθ+1)/5(sinθ+3)

∫ cos θ/(5+7sinθ-2+2sin²θ)dθ

Put sinθ = t, cosθ dθ = dt

= (1/5)ln |(sinθ+1/2)/(sinθ+3)|+C

B(θ)/A = 5(2sinθ+1)/(sinθ+3)

Answer: (1)

10. If the length of the chord of the circle, x²+y² = r²(r>0) along the line, y-2x = 3 is r, then r² is equal to:

1) 12
2) 24/5
3) 9/5
4) 12/5

AB = 2√(r²-9/5) = r

r²-9/5 = r²/4

3r²/4 = 9/5

r² = 12/5

Answer: (4)

11. If α and β are the roots of the equation, 7x²-3x-2 = 0, then the value of α/1-α² and β/1-β² is equal to:

1) 27/32
2) 1/24
3) 27/16
4) 3/8

α+β = 3/7

αβ = -2/7

= [(3/7)+(2/7)×(3/7)](1-{(9/49)+(4/7)}+(4/49))

= (21+6)/49(16/49)

= 27/16

Answer: (3)

12. If the sum of the second, third and fourth terms of a positive term G.P. is 3 and the sum of its sixth, seventh and eighth terms is 243, then the sum of the first 50 terms of this G.P. is:

1) (2/13)(3⁵⁰-1)
2) (1/26)(3⁴⁹-1)
3) (1/13)(3⁵⁰-1)
4) (1/26)(3⁵⁰-1)

(ar+ar²+ar³)/(ar⁵+ar⁶+ar⁷) = 3/243

(1+r+r²)/r⁴(1+r+r²) = 1/81

r = 3

a(3+9+27) = 3

a = 3/39 = 1/13

S₅₀ = a(r⁵⁰-1)/(r-1)

= 1/13(3⁵⁰-1)/2

= (1/26)(3⁵⁰-1)

Answer: (4)

13. If the line y = mx+c is a common tangent to the hyperbola (x²/100)-(y²/64) = 1 and the circle x²+y² = 36, then which one of the following is true?

1) 4c² = 369
2) c² = 369
3) 8m+5 = 0
4) 5m = 4

c = √(a²m²-b²)

c = √(100m²-64)

y = mx√(100m²-64)

d_(0,0) = 6

|√(100m²-64)/√(m²+1)| = 6

100m² -64 = 36m²+36

64m² = 100

m = 10/8

c² = 100×(100/64)-64

⇒ 164×36/64

4c² = 369

Answer: (1)

14. The area (in sq. units) of the region A = {(x,y): (x-1)[x] ≤y ≤2√x, 0 ≤x ≤2} where [t] denotes the greatest integer function, is:

1) (4/3)√2-(1/2)
2) (8/3)√2-(1/2)
3) (8/3)√2-1
4) (4/3)√2+1

y = f(x) = (x-1) [x]

=

y² ≤ 4x

= (4/3)+{(4/3)×2√2-2+2)-((4/3)+(1/2))}

= (4/3)+(8√2/3)-(4/3)-(1/2)

= (8√2/3)-(1/2)

Solution: (2)

15. If a+x = b+y = c+z+1, where a,b,c,x,y,z are non-zero distinct real numbers, then

1) y(a-b)
2) 0
3) y(b-a)
4) y(a-c)

Given a+x = b+y = c+z+1

Now,

=

=

R₂ → R₂ – R₁ and R₃ → R₃ – R₁

=

= y[x×0-1{(y-x)(c-a)-(b-a)(z-x)}+a×0]

= y[bz-bx-az+ax-(cy-ay-cx+ax)]

= y[bz-bx-az-cy+ay+cx]

= y[b(z-x)+a(y-z)+c(x-y)]

= y[b{a-c-1}+a(c-b+1)+c(b-a)]

= y[ab-bc-b+ac-ab+a+bc-ac]

= y(a-b)

Answer: (1)

16. If for some α∈ R , the lines L₁ : (x+1)/2 = (y-2)/-1 = (z-1)/1 and L₂ : (x+2)/α = (y+1)/(5-α) = (z+1)/1 are coplanar, then the line L₂ passes through the point:

1) (2, -10, -2)
2) (10, -2, -2)
3) (10, 2, 2)
4) (-2, 10, 2)

A(-1,2,1), B(-2,-1,-1)

-1(-1+α-5)+3(2-α)-2(10-2α+α) = 0

6-α+6-3α+2α-20 = 0

-8-2α = 0

α = -4

L₂: (x+2)/-4 = (y+1)/9 = (z+1)/1

Check options. (2, -10, -2) satisfies above equation.

Answer:(1)

17. The value of [((-1+i√3)/(1-i))]³⁰ is:

1) 2¹⁵i
2) -2¹⁵
3) -2¹⁵i
4) 6⁵

ω³⁰(1+i)³⁰ = 2¹⁵(-i)

Answer: (3)

18. Let y = y(x) be the solution of the differential equation cos x (dy/dx)+2y sinx = sin2x, x∈(0, π/2). If y(π/3) = 0, then y(π/4) is equal to:

1) 2+√2
2) √2-2
3) (1/√2)-1
4) 2-√2

(dy/dx) +(2tanx) y = 2sinx

I.F = e^{2ln(sec x)} = sec²x

y(sec²x) = 2∫sinx/cos²x dx

= 2∫secx tan x dx

= 2 secx+C

y(π/3) = 0

0 = 2×2+c

⇒ c = -4

y(sec²x) = 2 secx-4

x = π/4

2y = 2√2-4

y = √2-2

Answer: (2)

19. If the system of linear equations

x+y+3z = 0

x+3y+k²z = 0

3x+y+3z = 0

has a non-zero solution (x,y,z) for some k∈R, then x+(y/z) is equal to:

1) -9
2) 9
3) -3
4) 3

(9-k²)-(3-3k²) + 3(-8)=0

9-k²-3+3k²-24 = 0

2k²-18 = 0

k² = 9

k = 3, -3

x+y+3z = 0 ..(i)

x+3y+9z = 0 ..(ii)

Equation (ii)-(i)

2y +6z= 0

y = -3z

y/z = -3

2x = 0

x = 0

x+(y/z) = -3

Answer: (3)

20. Which of the following points lies on the tangent to the curve 4x³e^y+x⁴e^y+2√(y+1) = 3 at the point (1,0)?

1) (2,6)
2) (2,2)
3) (-2,6)
4) (-2,4)

4x³e^y+x⁴e^yy’ +2y’/2√(y+1) = 0

At (1,0)

4+y’+(2y’/2) = 0

2y’ = -4

⇒ y’ = -2

E.O.T.:

y = -2(x-1)

2x+y = 2

Check options given.

(-2,6) satisfies the equation 2x+y = 2.

Answer: (3)

21. Let A = {a,b,c} and B = {1,2,3,4}. Then the number of elements in the set C = {f : A → B 2∈ f(A)and f is not one-one} is:

C = {f : A → B 2 ∈ f(A) and f is not one-one}

Case-I : If f(x) = 2 ∀ x∈ A, then number of function = 1

Case-II : If f(x) = 2 for exactly two elements then total number of many-one function

= ³C₂ ³C₁ = 9

Case -III : If f(x) = 2 for exactly one element then total number of many-one function

= ³C₁ ³C₁ = 9

Total = 19

Answer: 19

22. The coefficient of x⁴ in the expansion of (1+x+x²+x³)⁶ in powers of x, is:

(1+x+x²+x³)ⁿ = (1+x)ⁿ(1+x²)ⁿ

(1+x+x²+x³)⁶= (1+x)⁶(1+x²)⁶

Coefficient of x⁴ ⇒ ⁿC₀ ⁿC₂+ ⁿC₂. ⁿC₁+ ⁿC₄. ⁿC₆

= ⁶C₀ ⁶C₂+ ⁶C₂. ⁶C₁+ ⁶C₄. ⁶C₆

= 15+(15×6)+15

= 120

Answer: 120

23. Let the vectors

⇒

⇒

⇒

⇒

= √(4+16+16)

= 6

Answer: 6

24. If the lines x+y = a and x-y = b touch the curve y = x²-3x+2 at the points where the curve intersects the x-axis, then a/b is equal to:

y -0 = -1(x-1)

x+y = 1

⇒ a = 1

y-0 = x-2

x-y = 2 = b = 2

a/b = 1/2

Answer: 1/2

25. In a bombing attack, there is 50% chance that a bomb will hit the target. At least two independent hits are required to destroy the target completely. Then the minimum number of bombs, that must be dropped to ensure that there is at least 99% chance of completely destroying the target, is

Let n is total no. of bombs being dropped

at least 2 bombs should hit

⇒ prob ≥ 0.99

p(x≥2) ≥0.99

1-p(x<2) ≥ 0.99

1-p(x=0)+p(x=1) ≥ 0.99

1-[ⁿC₀(p)⁰qⁿ+ⁿC₁(p)¹(q)^n-1] ≥ 0.99

1-[qⁿ+pnq^n-1] ≥0.99

1-[(1/2ⁿ)+(n/2)×(1/2^n-1)] ≥0.99

1-(1/2ⁿ)(n+1)≥ 0.99

0.01 ≥ (1/2ⁿ)(n+1)

2ⁿ ≥100+100n

n ≥ 11

Answer: 11