## JEE Main 2020 Maths Paper With Solutions Shift 2 September 5

**1. **If x = 1 is a critical point of the function f(x) = (3x^{2}+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) = (3x^{2}+ax-2-a)e^{x}

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

e^{x} [3x^{2}+(a+6) x-2] =0

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

a = -7

f(x) = (3x^{2}-7x+5)e^{x}

f’(x) = (6x-7)e^{x} +(3x^{2}-7x+5)e^{x}

= e^{x}(3x^{2}-x-2) = 0

= 3x^{2}-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

p |
q |
q→ p |
p→( q→ p) |
p˅q |
p → (p˅q) |
(p→( q→ p)) → (p → (p ˅ q)) |

T |
T |
T |
T |
T |
T |
T |

T |
F |
T |
T |
T |
T |
T |

F |
T |
F |
T |
T |
T |
T |

F |
F |
T |
T |
F |
T |
T |

**Answer: (4)**

**4. **If L = sin^{2}(π/16) – sin^{2}(π/8) and M = cos^{2}(π/16) – sin^{2}(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) 7^{2}

3) e^{2}

4) 7^{46/21}

(2+3+4+…+21)log_{7}x = 460

⇒ (20×(21+2)/2)log_{7 }x = 460

230 log_{7}x = 460

log_{7}x = 2

x = 7^{2}

**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(^{5}C_{1}×^{5}C_{2}×^{5}C_{2})+3(^{5}C_{1} ×^{5}C_{1}×^{5}C_{3})

= 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) x^{2}-20x+18 = 0

2) x^{2}-10x+19 = 0

3) 2x^{2}-20x+19 = 0

4) x^{2}-10x+18 = 0

S.D =

2^{2} = (83+a^{2}+b^{2})/5 -(5)^{2}

4 = (83+a^{2}+b^{2})/5 -(25)

29×5-83 = a^{2}+b^{2}

⇒ a^{2}+b^{2} = 62

(a+b+15)/5 = 5

⇒ a+b = 10 ..(i)

2ab = 100-62 = 38

ab = 19 ..(ii)

from (i) and (ii)

x^{2}-10x+19 = 0

**Answer: (2)**

**8. **The derivative of tan^{-1}(√(1+x^{2})-1)/x with respect to tan^{-1}(2x√(1-x^{2}))/(1-2x^{2})) 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^{-1}x

du/dv = ½(1+x^{2})×√(1-x^{2})/2

= (√3/2×2)×(4/5×2)

= √3/10

**Answer: (4)**

9.** **If ∫ cos θ/(5+7sinθ-2+2sin^{2}θ)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+2sin^{2}θ)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, x^{2}+y^{2} = r^{2}(r>0) along the line, y-2x = 3 is r, then r^{2} is equal to:

1) 12

2) 24/5

3) 9/5

4) 12/5

AB = 2√(r^{2}-9/5) = r

r^{2}-9/5 = r^{2}/4

3r^{2}/4 = 9/5

r^{2} = 12/5

**Answer: (4)**

**11. **If α and β are the roots of the equation, 7x^{2}-3x-2 = 0, then the value of α/1-α^{2} and β/1-β^{2} is equal to:

1) 27/32

2) 1/24

3) 27/16

4) 3/8

α+β = 3/7

αβ = -2/7

[(α+β)-αβ(α+β)]/[1-(α^{2}+β

^{2})+(αβ)

^{2}]

= [(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)(3^{50}-1)

2) (1/26)(3^{49}-1)

3) (1/13)(3^{50}-1)

4) (1/26)(3^{50}-1)

(ar+ar^{2}+ar^{3})/(ar^{5}+ar^{6}+ar^{7}) = 3/243

(1+r+r^{2})/r^{4}(1+r+r^{2}) = 1/81

r = 3

a(3+9+27) = 3

a = 3/39 = 1/13

S_{50} = a(r^{50}-1)/(r-1)

= 1/13(3^{50}-1)/2

= (1/26)(3^{50}-1)

**Answer: (4)**

**13. **If the line y = mx+c is a common tangent to the hyperbola (x^{2}/100)-(y^{2}/64) = 1** **and the circle x^{2}+y^{2 }= 36, then which one of the following is true?

1) 4c^{2} = 369

2) c^{2} = 369

3) 8m+5 = 0

4) 5m = 4

c = √(a^{2}m^{2}-b^{2})

c = √(100m^{2}-64)

y = mx√(100m^{2}-64)

d_{(0,0) }= 6

|√(100m^{2}-64)/√(m^{2}+1)| = 6

100m^{2 }-64 = 36m^{2}+36

64m^{2} = 100

m = 10/8

c^{2} = 100×(100/64)-64

⇒ 164×36/64

4c^{2} = 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]

=

y^{2 }≤ 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,

=

_{3}→ C

_{3}-C

_{1})

=

_{2}→ C

_{2}-C

_{3}) =

R_{2} → R_{2} – R_{1} and R_{3} → R_{3} – R_{1}

=

= 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 L_{1} : (x+1)/2 = (y-2)/-1 = (z-1)/1 and L_{2} : (x+2)/α = (y+1)/(5-α) = (z+1)/1 are coplanar, then the line L_{2} 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

L_{2}: (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) 2^{15}i

2) -2^{15}

3) -2^{15}i

4) 6^{5}

^{30}⇒ [((-1+i√3)/2)(1+i)]

^{30}

ω^{30}(1+i)^{30} = 2^{15}(-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) }= sec^{2}x

y(sec^{2}x) = 2∫sinx/cos^{2}x dx

= 2∫secx tan x dx

= 2 secx+C

y(π/3) = 0

0 = 2×2+c

⇒ c = -4

y(sec^{2}x) = 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+k^{2}z = 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-k^{2})-(3-3k^{2}) + 3(-8)=0

9-k^{2}-3+3k^{2}-24 = 0

2k^{2}-18 = 0

k^{2} = 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 4x^{3}e^{y}+x^{4}e^{y}+2√(y+1) = 3 at the point (1,0)?

1) (2,6)

2) (2,2)

3) (-2,6)

4) (-2,4)

4x^{3}e^{y}+x^{4}e^{y}y’ +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

= ^{3}C_{2} ^{3}C_{1} = 9

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

= ^{3}C_{1} ^{3}C_{1} = 9

Total = 19

**Answer: 19**

**22.** The coefficient of x^{4} in the expansion of (1+x+x^{2}+x^{3})^{6} in powers of x, is:

(1+x+x^{2}+x^{3})^{n} = (1+x)^{n}(1+x^{2})^{n}

(1+x+x^{2}+x^{3})^{6}= (1+x)^{6}(1+x^{2})^{6}

Coefficient of x^{4 }⇒ ^{n}C_{0}^{ n}C_{2}+^{ n}C_{2}.^{ n}C_{1}+^{ n}C_{4}.^{ n}C_{6}

= ^{6}C_{0}^{ 6}C_{2}+^{ 6}C_{2}.^{ 6}C_{1}+^{ 6}C_{4}.^{ 6}C_{6}

= 15+(15×6)+15

= 120

**Answer: 120**

**23.** Let the vectors

⇒

⇒

⇒

⇒

^{2}+b

^{2}+c

^{2 }+2a.b-2b.c-2a.c)

= √(4+16+16)

= 6

**Answer: 6**

**24. **If the lines x+y = a and x-y = b touch the curve y = x^{2}-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-[^{n}C_{0}(p)^{0}q^{n}+^{n}C_{1}(p)^{1}(q)^{n-1}] ≥ 0.99

1-[q^{n}+pnq^{n-1}] ≥0.99

1-[(1/2^{n})+(n/2)×(1/2^{n-1})] ≥0.99

1-(1/2^{n})(n+1)≥ 0.99

0.01 ≥ (1/2^{n})(n+1)

2^{n} ≥100+100n

n ≥ 11

**Answer: 11**

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