## JEE Main 2020 Maths Paper With Solutions Shift 2 September 3

**1.**If x

^{3}dy+xy dx = x

^{2}dy+2y dx; y(2) = e and x>1, then y(4) is equal to:

a) √e/2

b) (3/2)√e

c) (1/2)+√e

d) (3/2)+√e

(x^{3}-x^{2})dy = (2-x) ydx

p = 1, q = 2, r = -1

ln|y| = -ln x+(2/x)+ln (x-1)+c

x = 2, y = e

1 = 1-ln 2+c

⇒ c = ln 2

ln|y| = (2/x)-ln |x|+ln |x-1|ln 2

put x = 4

ln|y| = (1/2)-2ln 2+ ln 3+ ln 2

ln y = ln(3/2)+(1/2)

y = (3/2)×e^{1/2}

= (3/2)√e

**Answer: b**

**2.**Let A be a 3×3 matrix such that adj A =

^{-1})

^{T}| = μ, then the ordered pair, (|λ|,μ) is equal to:

a) (9, 1/81)

b) (9, 1/9)

c) (3, 1/81)

d) (3, 81)

adj A =

⇒ |adj A| = 9

⇒ |A^{2}| = 9

⇒ |A| = 3 = |λ| and |A| = -3 = |λ|

det (adj (adj A)) = ((|A|)^{(n-1)})^{2}

|B| = ((|A|)^{(3-1)})^{2} = |A|^{4} = 3^{4} = 81

|(B^{T})^{-1}| = 1/|(B^{T})|

= 1/|B|= 1/81=μ

Hence |λ|,μ = (3, 1/81)

**Answer: c**

**3.**Let a, b, cεR be such that a

^{2}+b

^{2}+c

^{2}= 1, if a cos θ = b cos(θ+2π/3) = c cos(θ+4π/3), where θ = π/9, then the angle between the vectors

a) π/2

b) 2π/3

c) π/9

d) 0

cos α =

= (ab+bc+ca)/a^{2}+b^{2}+c^{2}

**= **(ab+bc+ca)/1

cos α = ab+bc+ca

cos α = abc[(1/a)+(1/b)+(1/c)] ..(i)

a cos20^{0} = b cos(140^{0}) = c cos(260^{0}) = λ

⇒ (1/a) = cos 20^{0}/λ

(1/b) = cos 140^{0}/λ

(1/c) = cos 260^{0}/λ

Put in (i)

⇒** **cos α = (abc/λ)(cos 20^{0}+ cos 140^{0}+ cos 260^{0})

⇒** **cos α = (abc/λ)(cos 20^{0}+ 2cos 200^{0}+ cos 60^{0})

⇒** **cos α = (abc/λ)(cos 20^{0}– cos 20^{0})

⇒** **cos α = 0

⇒** **α = π/2

**Answer: a**

**4.**Suppose f(x) is a polynomial of degree four, having critical points at (-1,0,1). If T = {xεR f(x) = f(0) }, then the sum of squares of all the elements of T is:

a) 6

b) 2

c) 8

d) 4

f’(x) = k(x+1)x(x-1)

f’(x) = k[ x^{3}-x]

Integrating both sides

f(x) = k[(x^{4}/4)- (x^{2}/2)]+c

f(0) = c

f(x) = f(0) ⇒ k[(x^{4}/4)- (x^{2}/2)]+c = c

⇒ k(x^{2}/4)(x^{2}-2) = 0

⇒ x = 0, √2, -√2

Sum of squares of all elements = 0^{2}+(√2)^{2}+(-√2)^{2}

= 4

**Answer: d**

**5.**If the value of the integral

a) 2√3+π

b) 3√2+π

c) 3√2-π

d) 2√3-π

Put x = sinθ

=

⇒ (1/√3)-(π/6) = k/6

(2√3-π)/6 = k/6

k = 2√3-π

**Answer: d**

**6.**If the term independent of x in the expansion of ((3/2)x

^{2}-(1/3x))

^{9}is k, then 18 k is equal to:

a) 5

b) 9

c) 7

d) 11

T_{r+1} =

=

18-3r = 0

⇒ r = 6

=

⇒ 7/18 = k

⇒ 18k = 7

**Answer: c**

**7.**If a triangle ABC has vertices A(-1,7), B(-7,1) and C(5,-5), then its orthocentre has coordinates;

a) (-3,3)

b) (-3/5, 3/5)

c) (3/5, -3/5)

d) (3,-3)

m_{AH}. m_{BC} = -1

⇒ ((y-7)/(x+1))(1+5)/(-7-5) = -1

⇒ 2x-y+9 = 0 …..(i)

and m_{BH}. m_{AC} = -1

((y-1)/(x+7))(7–5)/(-1-5) = -1

x-2y+9 = 0 …..(ii)

Solving equation (i) and (ii) we get

(x, y) = (-3,3)

**Answer: a**

**8.**Let e

_{1}and e

_{2}be the eccentricities of the ellipse, (x

^{2}/25)+(y

^{2}/b

^{2}) = 1 (b<5) and the hyperbola, (x

^{2}/16)-(y

^{2}/b

^{2}) = 1 respectively satisfying e

_{1}e

_{2 }= 1. If α and β are the distances between the foci of the ellipse and the foci of the hyperbola respectively, then the ordered pair (α,β) is equal to:

a) (8,12)

b) (24/5,10)

c) (20/3,12)

d) (8,10)

α = 10e_{1}

b^{2} = 25(1-e_{1}^{2})

β = 8e_{2}

b^{2} = 16(e_{2}^{2}-1)

(e_{1}e_{2})^{2} = 1

(1-b^{2}/25)( 1+b^{2}/16) = 1

⇒ 1+(b^{2}/16)-(b^{2}/25)-(b^{4}/400) = 1

⇒ (9b^{2}/16×25)-(b^{4}/400) = 0

⇒ 9b^{2}-b^{4} = 0

⇒ b^{2}(9-b^{2}) = 0, b ≠ 0

Therefore, 9-b^{2} = 0

b^{2} = 9

e_{1} = 4/5

α = 2ae_{1}

= 10×4/5

= 8

e_{2} = 5/4

β = 2ae_{2}

= 8×5/4

= 10

(α, β) = (8, 10)

**Answer: d**

**9.**If z

_{1}, z

_{2}are complex numbers such that Re(z

_{1}) = |z

_{1}-1|, Re(z

_{2}) = |z

_{2}-1| and

**arg (z**

_{1}-z

_{2}) = π/6, then Im(z

_{1}+z

_{2}) is equal to:

a) 2√3

b) 2/√3

c) 1/√3

d) √3/2

z_{1} = x_{1}+iy_{1}

z_{2} = x_{2}+iy_{2}

x_{1}^{2} = (x_{1}-1)^{2}+y_{1}^{2}

⇒ y_{1}^{2}-2x_{1}+1 = 0 ..(i)

x_{2}^{2} = (x_{2}-1)^{2}+y_{2}^{2}

⇒ y_{2}^{2}-2x_{2}+1 = 0 ..(ii)

From equation (ii)-(i)

(y_{1}^{2}-y_{2}^{2})+2(x_{2}-x_{1}) = 0

(y_{1}+y_{2})(y_{1}-y_{2}) = 2(x_{1}-x_{2})

(y_{1}+y_{2}) = 2(x_{1}-x_{2})/ (y_{1}-y_{2})

arg (z_{1}-z_{2}) = π/6

tan^{-1}(y_{1}-y_{2})/(x_{1}-x_{2}) = π/6

⇒ (y_{1}-y_{2})/(x_{1}-x_{2}) = 1/√3

So y_{1}+y_{2} = 2√3

**Answer: a**

**10.**The set of all real values of λ for which the quadratic equations, (λ

^{2}+1)x

^{2}-4λx+2 = 0 always have exactly one root in the interval (0,1) is:

a) (-3,-1)

b) (2,4)

c) (1,3)

d) (0,2)

f(0) f(1) ≤ 0

⇒ (2) [λ^{2}-4λ +3] ≤0

(λ-1)(λ- 3) ≤0

⇒ λ ε [1, 3]

at λ = 1

2x^{2} -4x+2 = 0

⇒ (x-1)^{2} = 0

x = 1, 1

⇒λ ε (1, 3]

**Answer: c**

**11.**Let the latus rectum of the parabola y

^{2}= 4x be the common chord to the circles C

_{1 }and C

_{2}each of them having radius 2√5. Then, the distance between the centres of the circles C

_{1}and C

_{2 }is:

a) 8

b) 8√5

c) 4√5

d) 12

C_{1}C_{2} = 2C_{1}A

(C_{1}A)^{2} + 4 = ( 2√5)^{2}

C_{1}A = 4

C_{1}C_{2} = 8

**Answer: a**

**12.**The plane which bisects the line joining the points (4,-2,3) and (2,4,-1) at right angles also passes through the point:

a) (0,-1,1)

b) (4,0,1)

c) (4,0,-1)

d) (0,1,-1)

a = 2, b = -6

c = 4

Equation of plane is

2(x-3)+(-6)(y -1)+ 4(z-1) = 0

⇒ 2x-6y+4z = 4 passes through (4, 0, -1).

**Answer: c**

**13.**

a) (2/9)

^{4/3}

b) (2/3)

^{4/3}

c) (2/3)(2/9)

^{1/3}

d) (2/9)(2/3)

^{1/3}

Apply L hospital rule

⇒ [(2/3)(3a)^{-2/3}-1/3^{2/3}(a^{-2/3})]/[(1/3)(4a)^{-2/3}-(1/3)4^{1/3}a^{-2/3}]

= (2/3)×(2/9)^{1/3}

**Answer: c**

**14.**Let x

_{i}(1≤ i ≤10)be ten observations of a random variable X. If

where 0 ≠ p εR , then the standard deviation of these observations is :

a) 7/10

b) 9/10

c) √(3/5)

d) 4/5

Standard deviation is free from shifting of origin

S.D = √variance

= √[(9/10)-(3/10)^{2}]

= √[(9/10)-(9/100)]

= √(81/100)

= 9/10

**Answer: b**

**15.**The probability that a randomly chosen 5 digit number is made from exactly two digits

is :

a) 134/10^{4}

b) 121/10^{4}

c) 135/10^{4}

d) 50/10^{4}

Total case = 9(10^{4})

First case : Choose two non-zero digits = ^{9}C_{2}

Now, number of 5-digit numbers containing both digits = 2^{5}-2

Second case : Choose one non-zero and one zero as digit = ^{9}C_{1}

Number of 5-digit numbers containing one non-zero and one zero both = 2^{4}-1.

fav. case = ^{9}C_{2}(2^{5}-2)+^{ 9}C_{1}(2^{4}-1)

= 1080+135

= 1215

Probability = 1215/9×10^{4}

= 135/10^{4}

**Answer: c**

**16.**If

a) (x+1, -√x)

b) (x-1, -√x)

c) (x+1, √x)

d) (x-1, √x)

Given

∫tan^{-1}√x. 1 dx

Here tan^{-1}√x is the first function and 1 dx is the second function.

Integrating we get

(tan^{-1}√x)x-∫(x/1+x)×(1/2√x)dx

Put x = t^{2}

dx = 2t dt

(tan^{-1}√x)x-∫(x/1+x)×(1/2√x)dx = x(tan^{-1}√x)- ∫(t^{2})(2tdt)/(1+t^{2})2t

= x(tan^{-1}√x)-t+tan^{-1}t+c

= x(tan^{-1}√x)-√x+tan^{-1}√x+c

= tan^{-1}√x(x+1)-√x+c

A(x) = x+1

B(x) = -√x

**Answer: a**

**17.**If the sum of the series

^{th }term is 488 and the n

^{th}term is negative, then:

a) n = 60

b) n = 41

c) n

^{th}term is -4

d) n

^{th}term is

S_{n} = 488

⇒ (n/2)(2×20+(n-1)(-2/5)) = 488

⇒ 20n-(n^{2}/5)+(n/5) = 488

⇒ 100n-n^{2}+n = 2440

⇒ n^{2}-101n+2440 = 0

⇒ n = 61 or 40

For n = 40, T_{n} = 20+39(-2/5) = +ve

n = 61, T_{n} = 20+60(-2/5) =20-24 = -4

**Answer: c**

**18.**Let p, q, r be three statements such that the truth value of (p˄q) -> (∼q˅r)

**is F. Then the truth values of p, q, r are respectively:**

a) F, T, F

b) T, F, T

c) T, T, F

d) T, T, T

(p˄q) -> (∼q˅r) possible when p˄q -> T

∼q˅r -> F

p -> T ; p˄q⇒ T

q -> T ; ∼q˅r -> F˅F ⇒ F

r -> F ; T F ⇒ F

**Answer: c**

**19.**If the surface area of a cube is increasing at a rate of 3.6 cm

^{2}/sec, retaining its shape; then the rate of change of its volume (in cm

^{3}/sec), when the length of a side of the cube is 10cm, is :

a) 9

b) 10

c) 18

d) 20

A = 6a^{2}

A is the side of cube.

dA/dt = 6×2a da/dt

⇒ 3.6 = 12×10 (da/dt)

⇒ (da/dt) = 3/100

V = a^{3}

dV/dt = 3a^{2}da/dt

= 3×100×3/100

= 9 cm^{3}/sec

**Answer: a**

**20. **Let R_{1} and R_{2} be two relations defined as follows:

R_{1} = {(a,b) ∈ R^{2}: a^{2}+b^{2} ∈ Q} and

R_{2} = {(a,b) ∈ R^{2}: a^{2}+b^{2} ∉Q}, where Q is the set of all rational numbers. Then :

a) R_{1} is transitive but R_{2 }is not transitive

b) R_{1} and R_{2} are both transitive

c) R_{2} is transitive but R_{1} is not transitive

d) Neither R_{1} nor R_{2} is transitive

For R_{1}

Let a = 1+√2, b = 1-√2, c = 8^{1/4}

aR_{1}b ⇒ a^{2}+b^{2} = (1+√2)^{2}+(1-√2)^{2} = 6 ∈ Q

bR_{1}c ⇒ b^{2}+c^{2} = (1-√2)^{2}+(8^{1/4})^{2} = 3 ∈ Q

aR_{1}c ⇒ a^{2}+c^{2} = (1+√2)^{2}+(8^{1/4})^{2} = 3+4√2 ∉Q

R_{1} is not transitive

For R_{2}

Let a = 1+√2, b = √2, c = 1-√2

aR_{2}b ⇒ a^{2}+b^{2} = 5+2√2 ∉Q

bR_{2}c ⇒ b^{2}+c^{2} = 5-2√2 ∉Q

aR_{2}c ⇒ a^{2}+c^{2} = 6 ∈ Q

R_{2} is not transitive

**Answer: d**

**21.**If m arithmetic means (A.Ms) and three geometric means (G.Ms) are inserted between 3 and 243 such that 4

^{th }A.M. is equal to 2

^{nd}G.M., then m is equal to

3, …., 243 (m A.M)

d = (b-a)/(n+1)

= (243-3)/(m+1)

= 240/(m+1)

4^{th} AM = 3+4d

= 3+4(240/m+1)

3, …, 243 (3G.M)

243 = 3r^{4}

r = 3

2^{nd }G.M = ar^{2} = 27

Given that 4^{th }A.M. is equal to 2^{nd} G.M.

3+(960/m+1) = 27

**= **960/m+1 = 24

⇒ m = 39

**Answer: 39**

**22.**Let a plane P contain two lines

= (-1, 1 ,1)

Equation of plane

-1(x -1)+1(y-0) +1(z-0) = 0

⇒x-y-z-1 = 0

Foot of the perpendicular from m (1,0,1)

(x-1)/1 = (y-0)/-1 = (z-1)/-1 = -(1-0-1-1)/3

(x-1) = 1/3

y/-1 = 1/3

(z-1)/-1 = 1/3

x = 4/3

y = -1/3

z = 2/3

⇒α = 4/3

β = -1/3

γ= 2/3

α+β+ γ= (4/3)-(1/3)+(2/3) = 5/3

3(α+β+γ) = 5

**Answer: 5**

**23. **Let S be the set of all integer solutions, (x, y, z), of the system of equations

x-2y+5z = 0

-2x+4y+z = 0

-7x+14y+9z = 0

such that 15≤x^{2}+y^{2}+z^{2}≤150. Then, the number of elements in the set S is equal to:

x-2y+5z = 0…..(i)

-2x+4y+z = 0….(ii)

-7x+14y+9z = 0…..(iii)

2.(i) + (ii) we get z = 0, x = 2y

15 ≤4y^{2}+y^{2} ≤150

⇒ 3≤y^{2}≤30

y ∈ [-√30, -√3] U [√3, √30]

y = 2, 3, 4, 5, -2, -3, -4, -5

Number of elements in S is 8.

**Answer: 8**

**24.**The total number of 3 digit numbers, whose sum of digits is 10, is:

Let xyz be 3 digit number

x+y+z = 10 where x ≥1, y ≥0, z ≥0

x-1≥0

Put x-1 = t

t≥0

⇒ t+ y+z = 9

^{9+3-1}C_{3-1} = ^{11}C_{2} = 55

But for t = 9, x = 10 not possible

Total numbers = 55 – 1 = 54

**Answer: 54**

**25.**If the tangent to the curve, y = e

^{x}at a point (c, e

^{c}) and the normal to the parabola,

**y**

^{2}= 4x at the point (1,2) intersect at the same point on the x-axis, then the value of c

**is:**

Tangent at (c, e^{c})

y-e^{c} = e^{c}(x-c) ….(i)

normal to parabola y-2 = -1(x-1)

x+y = 3 …(ii)

At x-axis y = 0

in (i), x = c-1

in (ii), x = 3

c-1 = 3

⇒ c = 4

**Answer: 4**

## Comments