JEE Main 2020 Maths Paper With Solutions Jan 7 Shift 2

3x+4y =12√2 is a tangent to the ellipse (x²/a²) + (y²/9) = 1

Equation of tangent to ellipse (x²/a²) + (y²/9) = 1 is y = mx + √(a² m²+ 9)

Now, 3x + 4y = 12√2 ⇒ y = -(3/4)x + 3√2

m = -3/4

And √(a² m²+ 9) = 3√2

(a² (-3/4)²+ 9) = 18

a² (9/16) = 9

a² = 16

a = 4

e = √(1-b²/a²)

e = √(1-9/16)

= √7/4

Distance between foci is 2ae = 2 × 4 × √7/4

= 2√7

Answer: (b)

Coefficient of x⁷ in (1+x)¹⁰+ x(1+x)⁹+ x²(1+x)⁸+⋯ +x¹⁰

Applying sum of terms of G.P =

= (1 + x)¹¹ − x¹¹

Coefficient of x⁷ ⟹ ¹¹C₇ = 330

Answer: (b)

P(Machine being faulty) = p = (1/4)

q = ¾

P(at most two machines being faulty)=P(zero machine being faulty)+P(one machine being

faulty)+P(two machines being faulty)

= ⁵C₀ p⁰q⁵ + ⁵C₁ p¹q⁴ + ⁵C₂ p²q³

= q⁵ + 5pq⁴ + 10p²q³

= (3/4)⁵ + 5×(1/4)×(3/4)⁴ + 10(1/4)²×(3/4)³

= (3/4)³[(9/16) + (15/16) + (10/16)]

= (3/4)³× (34/16)

= (3/4)³× (17/8)

k = 17/8

Answer: (d)

Let R be the midpoint of PQ

PQ is perpendicular on line y = x

Equation of the line PQ can be written as y = −x + c

y = −x+c intersects y = x at Q: (c/2, c/2)

y = −x+c intersects x = 2y at P: (2c/3, c/3)

Midpoint R = (7c/12 , 5c/12)

Locus of R: x = 7c/12, y = 5c/12

5x – 7y = 0

Answer. (c)

⇒ 4[1 − e^−2α − e^−α + 1] = 5

Let e^−α = t

⇒ 4t² + 4t − 3 = 0

⇒ t = 1/2 = e^−α

⇒ α = log_𝑒 2

Answer. (a)

S = 3 + 4 + 8 + 9 + 13 + 14 +…..40 terms

S = (3+4)+(8+9)+(13+14)+…..40 terms

S = 7 + 17 + 27 + 37 +….20 terms

This is an AP with a = 7, d = 10

S = (n/2)(2a+(n-1)d)

= (20/2)(14 + (19)10)

= (20/2)204

= 20 × 102

m = 20

Answer.(d)

Let

𝑧 is real.

4 sin 𝜃 + 3 cos 𝜃 = 0

⇒ tan 𝜃 = -3/4 [𝜃 lies in 2^𝑛𝑑 quadrant]

arg (sin θ+ i cosθ ) =π + tan^-1 (cos 𝜃 /sin 𝜃) =π – tan^-1(4/3)

Answer: (a)

𝑏_𝑖 = (3)^{(𝑖+𝑗−2)}𝑎_𝑗𝑖

B =

Taking 3² common each from C₃ and R₃

Taking 3 common each from C₂ and R₂

Given ǀBǀ = 81

81 = 81×9 ǀAǀ

ǀAǀ = 1/9

Answer. (a)

Given

f(x) = ax⁵ + bx⁴ + cx³

c = 2

also f’(x) = 5ax⁴ + 4bx³ + 6x²

f’(1) = 5a + 4b + 6 = 0

f’(-1) = 5a – 4b + 6 = 0

b = 0 and a = -6/5

f(x) = -(6/5)x⁵ + 2x³

f(x) is odd

f’(x) = -6x⁴+ 6x²

f’’(x) = -24x³ +12x

(f’’(1) < 0)

(f’’(-1) > 0)

At x = -1 there is local minima and at x = 1 there is local maxima.

And f(1) -4f(-1) = 4

Answer: (d)

Using ³⁶C_r+1 = (36/r+1)׳⁵C_r

(36/r+1)׳⁵C_r×(k²-3) = ³⁵C_r×6

k² – 3 = (r+1)/6

k² = ((r+1)/6) +3

k∈I

r → Non-negative integer 0 ≤ r ≤ 35

r = 5 ⇒ k = ±2

r = 35 ⇒ k = ±3

No. of ordered pairs (r, k) = 4

Answer: (a)

a₁ + a₂ = 4 ⇒ a + ar = 4 ⇒ a(1 + r) = 4

a₃ + a₄ = 16 ⇒ ar² + ar³ = 16 ⇒ ar²(1 + r) = 16 ⇒ 4r² = 16

⇒ r = ±2

If r = 2, a = 4/3

whch is not possible as a₁ < 0

If r = −2, a = −4

= (-4)[(-2)⁹ -1]/-3

= 4/3)(-512-1)

= 4(-171)

λ = -171

Answer: (c)

Given a, b,c be three unit vectors.

a + b + c =0

= a.b + b.c + c.a

ǀa + b+cǀ² = ǀ10ǀ²

ǀaǀ² + ǀbǀ² +ǀcǀ² + 2(a.b + b.c + c.a) = 0

λ= -3/2

Also vector d = a×b + b×c + c×a

vector d = a×b + b×(-a-b) + (-a-b)×a

vector d = a×b + a×b + a×b

vector d = 3(a×b)

Answer: (c)

⇒ 2cosec²θ – 2 – 5cosec θ + 4 = 0

⇒2 cosec²θ – 4cosec θ – cosec θ + 2 = 0

⇒ cosec θ= 2 or ½ (not possible)

As θ ε[0,2π),

⇒ θ₁ =π /6, θ₂ = 5π/6

⇒

= (1/2) [(5π/6)-(π/6)] + (sin 6θ/12) ǀ^5π/6 _π/6

= π/3

Answer: (b)

Given α, β are the roots of x² − x − 1 = 0

⇒ α + β = 1 & αβ = −1

⇒ α² = α + 1 & β² = β + 1

p_k = α^k−2α² + β^k-2β²

pk = α^k−2(α + 1) + β^k−2(β + 1)

p_k = α ^k−1 + β ^k−1 + α ^k−2 + β ^k−2

⇒ p_k = p _k−1 + p _k−2

⇒ p₃ = p₂ + p₁ = 4

p₄ = p₃ + p₂ = 7

p₅ = p₄ + p₃ = 11

∴ p₅ ≠ p₂ ⋅ p₃ & p₁ + p₂ + p₃ + p₄ + p₅ = 26

& p₃ = p₅ − p₄

Answer: (c)

For point of intersection,

4x² = 8x + 12

⇒ x² − 2x − 3 = 0

⇒ x = 3, −1

Area bounded is given by

A = (36 + 36 -36) – (4-12 + 4/3)

= 44 – (4/3)

= 128/3

Answer: (b)

LMVT is applicable on f(x) in [0,1], therefore it is continuous and differentiable in [0,1]

Now, f(0) = 11, f(1) = 16

f ^′(x) = 3x² − 8x + 8

f’(c) = (f(1) – f(0))/(1-0)

= (16 – 11)/1

⇒ 3c² − 8c + 8 = 5

⇒ 3𝑐² − 8𝑐 + 3 = 0

c = (8±2√7)/6 = (4±√7)/3

As c ∈ (0,1)

We get c = (4 – √7)/3

Answer: (a)

y√(1 − x²)= k − x√(1 − y²)

Answer: (a)

x² + y² − 8x − 4y + 16 = 0

(x − 4)² + (y − 2)² = 4 ⇒ Centre (4, 2) , radius = 2

OA = 4 = OB

In ∆OBC

tan θ= 4/2 = 2 ⇒ sin θ = 2/√5

In ∆BDC

sin θ= BD/2 ⇒ BD = 4/√5

Length of chord of contact (𝐴𝐵) = 8/√5

Alternative

(𝑙) length of tangent = 4 and (𝑟) radius = 2

⇒Length of chord of contact = 2lr /(√(l²+ r²)

Square of length of chord of contact = 64/5

Answer: (b)

Given points (1, 0, 3) and Q (5/3, 7/3, 17/3)

Direction ratios of line L : (α-5/3, 7-7/3, 1-17/3)

= ((3α-5)/3, 14/3, -14/3)

Direction ratios of line PQ : (-2/3, -7/3, -8/3)

As line L is perpendicular to PQ

((3α-5)/3)(-2/3) + (14/3)( -7/3) + (-14/3)(-8/3) = 0

-6α + 10 – 98 + 112 = 0

6α = 24

α= 4

Answer: (4)

Mean = 10

(64 + x + y)/8 = 10

x + y = 80 – 64 = 16

Variance =

25 = ((3² + 7² + 9² + 12² + 13² + 20² + x² + y² )/8 )- 100

1000 = 852 + x² + y²

(x + y)² – 2xy = 148

256 – 2xy = 148

2xy = 108

So xy = 54

Answer: (54)

A = {x: x is multiple of 2} = {2,4,6,8, … }

B = {x: x is multiple of 7} = {7,14,21, … }

X = {x ∶ 1 ≤ x ≤ 50, x ∈N}

Smallest subset of X which contains elements of both A and B is a set with multiples of 2 or 7 less than 50.

P = {x: x is a multiple of 2 less than or equal to 50}

Q = {x: x is a multiple of 7 less than or equal to 50}

n(P ∪ Q) = n(P) + n(Q) − n(P ∩ Q)

= 25 + 7 − 3

= 29

Answer: (29)