**1.**If 3x+4y =12√2 is a tangent to the ellipse (x

^{2}/a

^{2}) + (y

^{2}/9) = 1 for some a∈R then the distance between the foci of the ellipse is :

a. 2√5

b. 2√7

c. 2√2

d. 4

3x+4y =12√2 is a tangent to the ellipse (x^{2}/a^{2}) + (y^{2}/9) = 1

Equation of tangent to ellipse (x^{2}/a^{2}) + (y^{2}/9) = 1 is y = mx + √(a^{2} m^{2}+ 9)

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

m = -3/4

And √(a^{2} m^{2}+ 9) = 3√2

(a^{2} (-3/4)^{2}+ 9) = 18

a^{2} (9/16) = 9

a^{2} = 16

a = 4

e = √(1-b^{2}/a^{2})

e = √(1-9/16)

= √7/4

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

= 2√7

**Answer: **(b)

**2.** Let A, B, C and D be four non-empty sets. The Contrapositive statement of “If A ⊆ B and B ⊆ D

then A ⊆ C ” is :

a. If A ⊆ C, then B ⊂ A or D ⊂ B

b. If A ⊈ C, then A ⊆ B and B ⊆ D

c. If A ⊈ C, then A ⊈ B and B ⊆ D

d. If A ⊈ C, then A ⊈ B or B ⊈ D

Given statements: A ⊆ B and B ⊆ D

Let A ⊆ B be p

B ⊆ D be q

A ⊆ C be r

Modified statement: (p

Contrapositive: ~r ⇒ ~ (p

~r ⇒ (~p

A ⊈ C, then A ⊈ B or B ⊈ D

**Answer: **(d)

**3.**The coefficient of x

^{7}in the expression (1 + x)

^{10}+ x(1 + x)

^{9}+ x

^{2}(1 + x)

^{8}+ ⋯ +x

^{10}is :

a. 420

b. 330

c. 210

d. 120

Coefficient of x^{7} in (1+x)^{10}+ x(1+x)^{9}+ x^{2}(1+x)^{8}+⋯ +x^{10}

Applying sum of terms of G.P =

= (1 + x)^{11} − x^{11}

Coefficient of x^{7} ⟹ ^{11}C_{7} = 330

**Answer:** (b)

**4.**In a workshop, there are five machines and the probability of any one of them to be out of service on a day is ¼ . If the probability that at most two machines will be out of service on the same day is (3/4)

^{3}k, then k is equal to:

a. 17/2

b. 4

c. 17/4

d. 17/8

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)

= ^{5}C_{0} p^{0}q^{5} + ^{5}C_{1} p^{1}q^{4} + ^{5}C_{2} p^{2}q^{3}

= q^{5} + 5pq^{4} + 10p^{2}q^{3}

= (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)

**5.**The locus of mid points of the perpendiculars drawn from points on the line x = 2y to the line x = y is:

a. 2x − 3y = 0

b. 3x − 2y = 0

c. 5x − 7y = 0

d. 7x − 5y = 0

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)

**6.**The value of α for which

a. log

_{e}2

b. log

_{e}√2

c. log

_{e}(4/3)

d. log

_{e}(3/2)

⇒ 4[1 − e^{−2α} − e^{−α} + 1] = 5

Let e^{−α} = t

⇒ 4t^{2} + 4t − 3 = 0

⇒ t = 1/2 = e^{−α}

⇒ α = log_{𝑒} 2

**Answer.** (a)

**7.**If the sum of the first 40 terms of the series, 3 + 4 + 8 + 9 + 13 + 14 + 18 + 19 + ….. is (102)m, then m is equal to :

a. 10

b. 25

c. 5

d. 20

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)

**8.**If

a. π- tan

^{-1}(4/3)

b. – tan

^{-1}(3/4)

c. π- tan

^{-1}(4/3)

d. tan

^{-1}(4/3)

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)

**9.**Let 𝐴 = [𝑎

_{𝑖𝑗}] and B = [𝑏

_{𝑖𝑗}] be two 3 × 3 real matrices such that 𝑏

_{𝑖𝑗}= (3

^{)(𝑖+𝑗−2)}𝑎

_{𝑗𝑖}, where 𝑖, 𝑗 = 1, 2, 3. If the determinant of 𝐵 is 81, then the determinant of 𝐴 is :

a. 1/9

b. 1/81

c. 1/3

d. 3

𝑏_{𝑖} = (3)^{(𝑖+𝑗−2)}𝑎_{𝑗𝑖}

B =

Taking 3^{2} common each from C_{3} and R_{3}

Taking 3 common each from C_{2} and R_{2}

Given ǀBǀ = 81

81 = 81×9 ǀAǀ

ǀAǀ = 1/9

**Answer.** (a)

**10.**Let f(x) be a polynomial of degree 5 such that x = ± 1 are its critical points. If

a. f(1) –4f(–1) = 4

b. x = 1 is a point of maxima and x = −1 is a point of minimum of f.

c. f is an odd function.

d. x = 1 is a point of minima and x = −1 is a point of maxima of f.

Given

f(x) = ax^{5} + bx^{4} + cx^{3}

c = 2

also f’(x) = 5ax^{4} + 4bx^{3} + 6x^{2}

f’(1) = 5a + 4b + 6 = 0

f’(-1) = 5a – 4b + 6 = 0

b = 0 and a = -6/5

f(x) = -(6/5)x^{5} + 2x^{3}

f(x) is odd

f’(x) = -6x^{4}+ 6x^{2}

f’’(x) = -24x^{3} +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)

**11.**The number of ordered pairs (r, k) for which 6 ⋅

^{35}C

_{r}= (k

^{2}− 3) ⋅

^{36}C

_{r+1}, where k is an integer, is :

a. 4

b. 6

c. 2

d. 3

Using ^{36}C_{r+1} = (36/r+1)×^{35}C_{r}

(36/r+1)×^{35}C_{r}×(k^{2}-3) = ^{35}C_{r}×6

k^{2} – 3 = (r+1)/6

k^{2} = ((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)

**12.**Let a

_{1}, a

_{2}, a

_{3}, … be a G.P. such that a

_{1}< 0, a

_{1}+ a

_{2}= 4 and a

_{3}+ a

_{4}= 16. If

a. 171

b. 511/3

c. -171

d. -513

a_{1} + a_{2 }= 4 ⇒ a + ar = 4 ⇒ a(1 + r) = 4

a_{3} + a_{4} = 16 ⇒ ar^{2} + ar^{3} = 16 ⇒ ar^{2}(1 + r) = 16 ⇒ 4r^{2} = 16

⇒ r = ±2

If r = 2, a = 4/3

whch is not possible as a_{1} < 0

If r = −2, a = −4

^{9}-1)/(r-1)

= (-4)[(-2)^{9} -1]/-3

= 4/3)(-512-1)

= 4(-171)

λ = -171

**Answer:** (c)

**13.**Let a, b,c be three unit vectors such that

a. (3/2, 3a×c)

b. (-3/2, 3c×b)

c. (-3/2, 3a×b)

d. (3/2, 3b×c)

Given a, b,c be three unit vectors.

a + b + c =0

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

ǀa + b+cǀ^{2} = ǀ10ǀ^{2}

ǀaǀ^{2} + ǀbǀ^{2} +ǀcǀ^{2} + 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)

**14.**Let y = y(x) be the solution curve of the differential equation (y

^{2}-x)(dy/dx) = 1 satisfying y(0) = 1

This curve intersects the x − axis at a point whose abscissa is :

a. 2+e

b. 2

c. 2-e

d. -e

(y^{2}– x)dy/dx = 1

(dx/dy) + x = y^{2}

xe^{y} = ∫ y^{2}e^{y}dy

x = y^{2} – 2y + 2 + ce^{-y}

Given y(0) = 1

c = -e

Solution is x = y^{2} -2y + 2 – e^{-y+1}

the value of x where the curve cuts the x axis will be at x = 2-e

**Answer: **(c)

**15.**If θ

_{1}and θ

_{2 }be respectively the smallest and the largest values of θ in (0,2π) − {π} which satisfy the equation

a. 2π/3

b. π/3

c. (π/3) + (1/6)

d. π/9

⇒ 2cosec^{2}θ – 2 – 5cosec θ + 4 = 0

⇒2 cosec^{2}θ – 4cosec θ – cosec θ + 2 = 0

⇒ cosec θ= 2 or ½ (not possible)

As θ ε[0,2π),

⇒ θ_{1} =π /6, θ_{2} = 5π/6

⇒

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

= π/3

**Answer:** (b)

**16.**Let α and β are the roots of the equation x

^{2}− x − 1 = 0. If p

_{k}= (α)

^{k}+ (β)

^{k}, k ≥ 1 then which one of the following statements is not true?

a. (p

_{1}+ p

_{2 }+ p

_{3}+ p

_{4}+ p

_{5}) = 26

b. p

_{5}= 11

c. p

_{5}= p

_{2}⋅ p

_{3}

d. p

_{3}= p

_{5}− p

_{4}

Given α, β are the roots of x^{2} − x − 1 = 0

⇒ α + β = 1 & αβ = −1

⇒ α^{2} = α + 1 & β^{2} = β + 1

p_{k} = α^{k−2}α^{2 }+ β^{k-2}β^{2}

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}

⇒ p_{3} = p_{2} + p_{1} = 4

p_{4 }= p_{3} + p_{2} = 7

p_{5 }= p_{4 }+ p_{3} = 11

∴ p_{5} ≠ p_{2} ⋅ p_{3} & p_{1} + p_{2} + p_{3} + p_{4} + p_{5} = 26

& p_{3} = p_{5} − p_{4}

**Answer:** (c)

**17.**The area (in sq. units) of the region {(x, y)ϵR|4x

^{2}≤ y ≤ 8x + 12} is :

a. 125/3

b. 128/3

c. 124/3

d. 127/3

For point of intersection,

4x^{2} = 8x + 12

⇒ x^{2} − 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)

**18.**The value of c in Lagrange’s mean value theorem for the function f(x) = x

^{3}− 4x

^{2}+ 8x + 11, where x ∈ [0,1] is

a. (4 – √7)/3

b. 2/3

c. (√7 – 2)/3

d. (4 – √5)/3

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) = 3x^{2} − 8x + 8

f’(c) = (f(1) – f(0))/(1-0)

= (16 – 11)/1

⇒ 3c^{2} − 8c + 8 = 5

⇒ 3𝑐^{2} − 8𝑐 + 3 = 0

c = (8±2√7)/6 = (4±√7)/3

As c ∈ (0,1)

We get c = (4 – √7)/3

**Answer:** (a)

**19.**Let y = y(x) be a function of x satisfying y√(1 − x

^{2})= k − x√(1 − y

^{2}) where k is a constant and y(1/2) = -1/4. Then dy/dx at x = ½ is equal to:

a. -√5/2

b. √5/2

c. -√5/4

d. 2/√5

y√(1 − x^{2})= k − x√(1 − y^{2})

**Answer:** (a)

**20.**Let the tangents drawn from the origin to the circle, x

^{2}+ y

^{2}− 8x − 4y + 16 = 0 touch it at the points 𝐴 and 𝐵. The (𝐴𝐵)

^{2}is equal to

a. 32/5

b. 64/5

c. 52/5

d. 56/5

x^{2} + y^{2} − 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 /(√(l^{2}+ r^{2})

Square of length of chord of contact = 64/5

**Answer:** (b)

**21.** If system of linear equations

x + y + z = 6

x + 2y + 3z = 10

3x + 2y + 𝜆z = 𝜇

has more than two solutions, then μ – λ^{2} is equal to

The system of equations has more than 2 solutions

∴ D = D_{3} = 0

2 λ− 6 −λ + 9 + 2 − 6 = 0

λ= 1

⇒ μ = 14

So, μ − λ^{2} = 13

**Answer: **(13)

**22.**If the foot of perpendicular drawn from the point (1,0,3) on a line passing through (α, 7,1) is (5/3, 7/3, 17/3) then α is equal to:

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)

**23.**If the function 𝑓 defined on (-1/3, 1/3) by

is continuous, then k is equal to

As f(x) is continous

3+2 = k

k = 5

**Answer:** (5)

**24.**If the mean and variance of eight numbers 3, 7, 9, 12, 13, 20, x and y be 10 and 25 respectively then xy is equal to

Mean = 10

(64 + x + y)/8 = 10

x + y = 80 – 64 = 16

Variance =

25 = ((3^{2} + 7^{2} + 9^{2} + 12^{2} + 13^{2} + 20^{2} + x^{2} + y^{2} )/8 )- 100

1000 = 852 + x^{2} + y^{2}

(x + y)^{2} – 2xy = 148

256 – 2xy = 148

2xy = 108

So xy = 54

**Answer:** (54)

**25.**Let 𝑋 = {𝑛 ∈ 𝑵: 1 ≤ 𝑛≤ 50}. If A = {𝑛 ∈ 𝑋: 𝑛 is a multiple of 2} and B = {𝑛 ∈ 𝑋: 𝑛 is a multiple of 7}, then the number of elements in the smallest subset of 𝑋 containing both A and B is .

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)

## Video Lessons – January 7 Shift 2 Maths

## JEE Main 2020 Maths Paper With Solutions Jan 7 Shift 2

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