JEE Main 2020 Solved Maths Paper Shift 2 (Sept 3rd) - Download PDF

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JEE Main 2020 Maths Paper With Solutions Shift 2 September 3

September 3 Shift 2 – Maths

1. If x³dy+xy dx = x²dy+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

(x³-x²)dy = (2-x) ydx

\(\begin{array}{l}\int \frac{dy}{y}=\int \frac{2-x}{x^{2}(x-1)}dx\end{array} \)

\(\begin{array}{l}\int \frac{dy}{y}=-\int \frac{x-2}{x^{2}(x-1)}dx\end{array} \)

\(\begin{array}{l}\int \frac{dy}{y}=-\int (\frac{p}{x}+\frac{q}{x^{2}}+\frac{r}{x-1})dx\end{array} \)

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

\(\begin{array}{l}\int \frac{dy}{y}=\int \frac{-1}{x}dx-\int\frac{2}{x^{2}}dx-\int\frac{-1}{x-1}dx\end{array} \)

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 =

\(\begin{array}{l}\begin{bmatrix} 2 & -1 &1 \\ -1 &0 &2 \\ 1 & -2 & -1 \end{bmatrix}\end{array} \)

and B = adj(adj A) . If |A| = λ and |(B^-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 =

\(\begin{array}{l}\begin{bmatrix} 2 & -1 &1 \\ -1 &0 &2 \\ 1 & -2 & -1 \end{bmatrix}\end{array} \)

⇒ |adj A| = 9

⇒ |A²| = 9

⇒ |A| = 3 = |λ| and |A| = -3 = |λ|

det (adj (adj A)) = ((|A|)^(n-1))²

|B| = ((|A|)^(3-1))² = |A|⁴ = 3⁴ = 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 a²+b²+c² = 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

\(\begin{array}{l}a\hat{i}+b\hat{j}+c\hat{k}\end{array} \)

and

\(\begin{array}{l}b\hat{i}+c\hat{j}+a\hat{k}\end{array} \)

is:
a) π/2
b) 2π/3
c) π/9
d) 0

cos α =

\(\begin{array}{l}\frac{\vec{p}.\vec{q}}{\left | p \right |\left | q \right |}\end{array} \)

= (ab+bc+ca)/a²+b²+c²

= (ab+bc+ca)/1

cos α = ab+bc+ca

cos α = abc[(1/a)+(1/b)+(1/c)] ..(i)

a cos20⁰ = b cos(140⁰) = c cos(260⁰) = λ

⇒ (1/a) = cos 20⁰/λ

(1/b) = cos 140⁰/λ

(1/c) = cos 260⁰/λ

Put in (i)

⇒ cos α = (abc/λ)(cos 20⁰+ cos 140⁰+ cos 260⁰)

⇒ cos α = (abc/λ)(cos 20⁰+ 2cos 200⁰+ cos 60⁰)

⇒ cos α = (abc/λ)(cos 20⁰– cos 20⁰)

⇒ 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[ x³-x]

Integrating both sides

f(x) = k[(x⁴/4)- (x²/2)]+c

f(0) = c

f(x) = f(0) ⇒ k[(x⁴/4)- (x²/2)]+c = c

⇒ k(x²/4)(x²-2) = 0

⇒ x = 0, √2, -√2

Sum of squares of all elements = 0²+(√2)²+(-√2)²

= 4

Answer: d

5. If the value of the integral

\(\begin{array}{l}\int_{0}^{\frac{1}{2}}\frac{x^{2}}{(1-x^{2})^{\frac{3}{2}}}dx\end{array} \)

is k/6, then k is equal to:
a) 2√3+π
b) 3√2+π
c) 3√2-π
d) 2√3-π

\(\begin{array}{l}\int_{0}^{\frac{1}{2}}\frac{x^{2}}{(1-x^{2})^{\frac{3}{2}}}dx\end{array} \)

Put x = sinθ

\(\begin{array}{l}\int_{0}^{\frac{\pi }{6}}\frac{\sin ^{2}\theta }{\cos ^{3}\theta }\cos \theta .d\theta\end{array} \)

=

\(\begin{array}{l}\int_{0}^{\frac{\pi }{6}}\tan ^{2} \theta .d\theta = [\tan \theta -\theta]_{0}^{\frac{\pi }{6}}\end{array} \)

⇒ (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)x² -(1/3x))⁹ is k, then 18 k is equal to:
a) 5
b) 9
c) 7
d) 11

T_r+1 =

\(\begin{array}{l}^{^{9}}C_{r}\left ( \frac{3}{2}x^{2} \right )^{9-r}\left ( \frac{-1}{3x} \right )^{r}\end{array} \)

=

\(\begin{array}{l}^{^{9}}C_{r}\frac{3^{9-r}}{2^{9-r}}(\frac{-1}{3})^{r}x^{18-3r}\end{array} \)

18-3r = 0

⇒ r = 6

=

\(\begin{array}{l}^{^{9}}C_{r}(\frac{3^{-3}}{2^{3}})=k\end{array} \)

⇒ 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)

JEE Main 2020 Paper With Solution Maths Sept 3 Shift 2

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 e₁ and e₂ be the eccentricities of the ellipse, (x²/25)+(y²/b²) = 1 (b<5) and the hyperbola, (x²/16)-(y²/b²) = 1 respectively satisfying e₁e₂= 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 = 10e₁

b² = 25(1-e₁²)

β = 8e₂

b² = 16(e₂²-1)

(e₁e₂)² = 1

(1-b²/25)( 1+b²/16) = 1

⇒ 1+(b²/16)-(b²/25)-(b⁴/400) = 1

⇒ (9b²/16×25)-(b⁴/400) = 0

⇒ 9b²-b⁴ = 0

⇒ b²(9-b²) = 0, b ≠ 0

Therefore, 9-b² = 0

b² = 9

e₁ = 4/5

α = 2ae₁

= 10×4/5

= 8

e₂ = 5/4

β = 2ae₂

= 8×5/4

= 10

(α, β) = (8, 10)

Answer: d

9. If z₁, z₂ are complex numbers such that Re(z₁) = |z₁-1|, Re(z₂) = |z₂-1| and arg (z₁-z₂) = π/6, then Im(z₁+z₂) is equal to:
a) 2√3
b) 2/√3
c) 1/√3
d) √3/2

z₁ = x₁+iy₁

z₂ = x₂+iy₂

x₁² = (x₁-1)²+y₁²

⇒ y₁²-2x₁+1 = 0 ..(i)

x₂² = (x₂-1)²+y₂²

⇒ y₂²-2x₂+1 = 0 ..(ii)

From equation (ii)-(i)

(y₁²-y₂²)+2(x₂-x₁) = 0

(y₁+y₂)(y₁-y₂) = 2(x₁-x₂)

(y₁+y₂) = 2(x₁-x₂)/ (y₁-y₂)

arg (z₁-z₂) = π/6

tan^-1(y₁-y₂)/(x₁-x₂) = π/6

⇒ (y₁-y₂)/(x₁-x₂) = 1/√3

So y₁+y₂ = 2√3

Answer: a

10. The set of all real values of λ for which the quadratic equations, (λ²+1)x²-4λx+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) [λ²-4λ +3] ≤0

(λ-1)(λ- 3) ≤0

⇒ λ ε [1, 3]

at λ = 1

2x² -4x+2 = 0

⇒ (x-1)² = 0

x = 1, 1

⇒λ ε (1, 3]

Answer: c

11. Let the latus rectum of the parabola y² = 4x be the common chord to the circles C₁and C₂ each of them having radius 2√5. Then, the distance between the centres of the circles C₁ and C₂is:
a) 8
b) 8√5
c) 4√5
d) 12

C₁C₂ = 2C₁A

(C₁A)² + 4 = ( 2√5)²

C₁A = 4

C₁C₂ = 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.

\(\begin{array}{l}\lim_{x\to a}\frac{(a+2x)^{\frac{1}{3}}-(3x)^{\frac{1}{3}}}{(3a+x)^{\frac{1}{3}}-(4x)^{\frac{1}{3}}}\end{array} \)

is equal to:
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)4^1/3a^-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

\(\begin{array}{l}\sum_{i=1}^{10}(x_{i}-p)=3\end{array} \)

and

\(\begin{array}{l}\sum_{i=1}^{10}(x_{i}-p)^{2}=9\end{array} \)

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) 134/10⁴
b) 121/10⁴
c) 135/10⁴
d) 50/10⁴

Total case = 9(10⁴)

First case : Choose two non-zero digits = ⁹C₂

Now, number of 5-digit numbers containing both digits = 2⁵-2

Second case : Choose one non-zero and one zero as digit = ⁹C₁

Number of 5-digit numbers containing one non-zero and one zero both = 2⁴-1.

fav. case = ⁹C₂(2⁵-2)+⁹C₁(2⁴-1)

= 1080+135

= 1215

Probability = 1215/9×10⁴

= 135/10⁴

Answer: c

16. If

\(\begin{array}{l}\int \sin ^{-1}\left ( \sqrt{\frac{x}{1+x}} \right )dx= A(x)\tan ^{-1}(\sqrt{x})+B(x)+C\end{array} \)

, where C is a constant of integration, then the ordered pair (A(x), B(x)) can be:
a) (x+1, -√x)
b) (x-1, -√x)
c) (x+1, √x)
d) (x-1, √x)

JEE Main Solution 2020 Maths Papers Sept 3 Shift 2

Given

\(\begin{array}{l}\int \sin ^{-1}\left ( \sqrt{\frac{x}{1+x}} \right )dx= A(x)\tan ^{-1}(\sqrt{x})+B(x)+C\end{array} \)

∫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 = t²

dx = 2t dt

(tan^-1√x)x-∫(x/1+x)×(1/2√x)dx = x(tan^-1√x)- ∫(t²)(2tdt)/(1+t²)2t

= x(tan^-1√x)-t+tan^-1t+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

\(\begin{array}{l}20+19\frac{3}{5}+19\frac{1}{5}+18\frac{4}{5}+…\end{array} \)

upto n^thterm 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

\(\begin{array}{l}-4\frac{2}{5}\end{array} \)

\(\begin{array}{l}20+19\frac{3}{5}+19\frac{1}{5}+18\frac{4}{5}+…\end{array} \)

S_n = 488

⇒ (n/2)(2×20+(n-1)(-2/5)) = 488

⇒ 20n-(n²/5)+(n/5) = 488

⇒ 100n-n²+n = 2440

⇒ n²-101n+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 cm²/sec, retaining its shape; then the rate of change of its volume (in cm³/sec), when the length of a side of the cube is 10cm, is :
a) 9
b) 10
c) 18
d) 20

A = 6a²

A is the side of cube.

dA/dt = 6×2a da/dt

⇒ 3.6 = 12×10 (da/dt)

⇒ (da/dt) = 3/100

V = a³

dV/dt = 3a²da/dt

= 3×100×3/100

= 9 cm³/sec

Answer: a

20. Let R₁ and R₂ be two relations defined as follows:

R₁ = {(a,b) ∈ R²: a²+b² ∈ Q} and

R₂ = {(a,b) ∈ R²: a²+b² ∉Q}, where Q is the set of all rational numbers. Then :

a) R₁ is transitive but R₂is not transitive
b) R₁ and R₂ are both transitive
c) R₂ is transitive but R₁ is not transitive
d) Neither R₁ nor R₂ is transitive

For R₁

Let a = 1+√2, b = 1-√2, c = 8^1/4

aR₁b ⇒ a²+b² = (1+√2)²+(1-√2)² = 6 ∈ Q

bR₁c ⇒ b²+c² = (1-√2)²+(8^1/4)² = 3 ∈ Q

aR₁c ⇒ a²+c² = (1+√2)²+(8^1/4)² = 3+4√2 ∉Q

R₁ is not transitive

For R₂

Let a = 1+√2, b = √2, c = 1-√2

aR₂b ⇒ a²+b² = 5+2√2 ∉Q

bR₂c ⇒ b²+c² = 5-2√2 ∉Q

aR₂c ⇒ a²+c² = 6 ∈ Q

R₂ 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^thA.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 = 3r⁴

r = 3

2^ndG.M = ar² = 27

Given that 4^thA.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

\(\begin{array}{l}\vec{r} = \hat{i}+\lambda (\hat{i}+\hat{j})\end{array} \)

, λ ∈ R and

\(\begin{array}{l}\vec{r} = -\hat{j}+\mu (\hat{j}-\hat{k})\end{array} \)

, μ∈R. If Q(α,β,γ) is the foot of the perpendicular drawn from the point M(1,0,1) to P, then 3(α+β+γ) equals:

\(\begin{array}{l}\vec{r} = \hat{i}+\lambda (\hat{i}+\hat{j})\end{array} \)

\(\begin{array}{l}\vec{r} = -\hat{j}+\mu (\hat{j}-\hat{k})\end{array} \)

\(\begin{array}{l}\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} &\hat{k} \\ 1 & 1 & 0\\ 0& 1 & -1 \end{vmatrix}\end{array} \)

= (-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≤x²+y²+z²≤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 ≤4y²+y² ≤150

⇒ 3≤y²≤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+3-1C_3-1 = ¹¹C₂ = 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, y² = 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

Video Lessons – September 3 Shift 2 Maths

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