JEE Main 2020 Maths paper September 2 Shift 2 solutions are available on this page. Students are recommended to go through these solutions so that they can improve their understanding of concepts and at the same time develop greater speed for solving problems. Using this paper and solutions to revise and practice will help students to find out their weaker areas and concentrate more on improving them. These solutions can be easily downloaded in PDF format for free.

JEE Main 2020 Maths Paper With Solutions Sep 2 Shift 2

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September 2 Shift 2 – Maths

Question 1.Let f:R -> R be a function which satisfies f(x+y) = f(x)+f(y) for all x, y ∈ R. If f(1) = 2 and g(n) =

\begin{array}{l} \sum_{k = 1}^{n - 1} f (k) \end{array}

, n ∈ N then the value of n for which g(n) = 20 is
a) 9
b) 5
c) 4
d) 20

f(1) = 2

f(x+y) = f(x)+f(y)

x = y = 1

f(2) = 2+2 = 4

x = 2, y = 1

f(3) = 4+2 = 6

g(n) = f(1)+f(2)+………+f(n-1)

= 2+4+6+ ….+ 2(n-1)

= 2Ʃ(n-1)

= 2(n-1)n/2

= n(n-1)

= n²-n

n²-n = 20

n²-n-20 = 0

(n-5)(n+4) = 0

n = 5

Answer:(b)

Question 2. If the sum of first 11 terms of an A.P, a₁,a₂,a₃… is 0 (a₁ ≠ 0) then the sum of the A.P, a₁, a₃, a₅…a₂₃ is ka₁, where k is equal to
a) -121/10
b) -72/5
c) 72/5
d) 121/10

Given

\begin{array}{l} \sum_{k = 1}^{11} a_{k} = 0 \end{array}

11a+55d = 0

a+5d = 0

d = -a/5

Now a₁+a₃+….+ a₂₃ = ka₁

12a+d (2+4+6+…..+22) = ka

12a+2d×66 = ka

12(a+11d) = ka

12(a+11(-a/5)) = ka

12(1-11/5) = k

k = -72/5

Answer: (b)

Question 3. Let E^C denote the complement of an event E. Let E₁, E₂ and E₃ be any pair wise independent events with P(E₁)>0 and P(E₁ ∩ E₂ ∩ E₃) = 0. Then P(E₂^C ∩ E₃^C/E₁) is equal to:
a) P(E₃^C)- P(E₂^C)
b) P(E₃)- P(E₂^C)
c) P(E₃^C)- P(E₂)
d) P(E₂^C)+ P(E₃)

JEE Main 2020 Paper With Solution Maths Sept 2 Shift 2

Answer: (c)

Question 4. If the equation cos⁴θ+sin⁴θ+λ = 0 has real solutions for θ, then λ lies in the interval:
a) (-1/2, -1/4]
b) [-1,-1/2]
c) [-3/2, -5/4]
d) (-5/4, -1)

cos⁴θ+sin⁴θ+λ = 0

λ = -{1-(1/2)sin²2θ}

2(λ+1) = sin²2θ

0 ≤2(λ+1) ≤1

0 ≤λ+1 ≤1/2

-1≤ λ ≤ -1/2

Answer: (b)

Question 5. The area (in sq. units) of an equilateral triangle inscribed in the parabola y² = 8x, with

one of its vertices on the vertex of this parabola, is:
a) 128√3
b) 192√3
c) 64√3
d) 256√3

A: (a cos 30⁰, a sin 30⁰) lies on parabola

(a²/4) = 8×a√3/2

a = 16√3

Area of equilateral triangle, Δ = (√3/4)a²

Δ= (√3/4)×16×16×3

Δ= 192√3

Answer: (b)

Question 6. The imaginary part of (3+2√-54)^1/2-(3-2√-54)^1/2 can be:
a) √6
b) -2√6
c) 6
d) -√6

(3+2i√54)^1/2-(3-i2√54)^1/2

= (9+6i²+2×3i√6)^1/2-(9+6i²-2×3i√6)^1/2

= ((3+√6i)²)^1/2 -((3-√6i)²)^1/2

= (3+√6i)( 3-√6i)

= -2√6i

Answer: (b)

Question 7.A plane passing through the point (3,1,1) contains two lines whose direction ratios are (1, -2, 2) and (2, 3, -1) respectively. If this plane also passes through the point (α,-3,5), then α is equal to:
a) -5
b) 10
c) 5
d) -10

Required plane is

\begin{array}{l} | \begin{array}{c} x - 3 & y - 1 & z - 1 \\ 1 & - 2 & 2 \\ 2 & 3 & - 1 \end{array} | = 0 \end{array}

-4(x-3)+5(y-1)+7(z-1) = 0

4x-5y-7z = 0

(α, -3,5) lies on 4x-5y-7z = 0

α = 5

Answer: (c)

Question 8.Let A = {X= (x,y,z)^T: PX = 0 and x²+y²+z² = 1}, where P =

\begin{array}{l} [\begin{array}{c} 1 & 2 & 1 \\ - 2 & 3 & - 4 \\ 1 & 9 & - 1 \end{array}] \end{array}

then the set A:
a) contains more than two elements.
b) is a singleton.
c) contains exactly two elements.
d) is an empty set.

Clearly |P|= 0

PX = 0 has infinite solutions

The line concurrence passes through (0,0,0) which is centre of sphere x²+y²+z² = 1.

Diameter will intersect at two points.

two solutions (exactly) exist.

Answer: (c)

Question 9. The equation of the normal to the curve y = (1+x)^2y +cos²(sin ^-1x) at x = 0 is:
a) y+4x = 2
b) 2y+x = 4
c) x+4y = 8
d) y = 4x+2

At x = 0, y = 1+cos²(0) = 2

P;(0,2)

y = (1+x)^2y+cos²(sin^-1x)

y = (1+x)^2y+cos(cos^-1√(1-x²))²

= (1+x)^2y+ (√(1-x²))²

y = (1+x)^2y+1-x²

Differentiating w.r.t x

Now y’ = (1+x)^2y{(2y/1+x)+ln(1+x)2y’}-2x

y’ at (0,2) = 4-0 = 4

N₀: y-2 = (-1/4)(x-0)

N₀: 4y-8 = -x

x+4y = 8

Answer: (c)

Question 10. Consider a region R = {(x,y)∈R² : x² ≤y ≤ 2x}. If a line y = α divides the area of region R into two equal parts, then which of the following is true.?
a) α³-6α²+16 = 0
b) 3α²-8α^3/2+8 = 0
c) α³-6α^3/2-16 = 0
d) 3α²-8α+8 = 0

JEE Main 2020 Maths Papers With Solutions Sept 2 Shift 2

Area =

\begin{array}{l} \int_{0}^{2} (2 x - x^{2}) d x \end{array}

= x²-(x³/3)

= 4/3 [Applying limits]

So

\begin{array}{l} \int_{0}^{α} (\sqrt{y} - \frac{y}{2}) d y = \frac{2}{3} \end{array}

\begin{array}{l} [\frac{2}{3} y^{\frac{3}{2}} - \frac{y^{2}}{4}]_{0}^{\infty} = \frac{2}{3} \end{array}

8.α^3/2-3α² = 8

3α²-8α^3/2+8 = 0

Answer: (b)

Question 11.Let f: (-1, ∞)→R be defined by f(0) = 1 and f(x) = (1/x)log_e(1+x), x ≠ 0. Then the function f:
a) increases in (-1, ∞)
b) decreases in (-1,0) and increases in (0, ∞)
c) increases in (-1,0) and decreases in (0, ∞)
d) decreases in (-1, ∞)

f(x) = (1/x)ln(1+x)

f’(x) = [x/(1+x) – ln(1 + x)]/x² = [x – (1+x)ln (1+x)]/x²(1+x)

f’<0 ∀ x ∈(-1,∞)

Answer: (d)

Question 12.Which of the following is a tautology?
a) (p→q) ˄( q→p)
b) (~p) ˄(p ˅q)→q
c) (q→p) ˅~(p→q)
d) (~q)˅( p˄q)→q

p	q	~p	(p ˅q)	~p ˄(p˅q)	~p ˄(p˅q)->q
T	T	F	T	F	T
T	F	F	T	F	T
F	T	T	T	T	T
F	F	T	F	F	T

Answer: (b)

Question 13. Let f(x) be a quadratic polynomial such that f(-1)+f(2) = 0. If one of the roots of f(x) = 0 is 3, then its other roots lies in:
a) (0,1)
b) (1,3)
c) (-1,0)
d) (-3,-1)

Let f(x) = a (x-3) (x-α)

f(-1)+f(2) = 0

a[(-1-3) (-1-α)+(2-3)(2-α)] = 0

a[4+4α-2+α]=0

5α+2 = 0

α = -2/5

Answer: (c)

Question 14. Let S be the sum of the first 9 terms of the series:

{x+ka}+{x²+(k+2)a}+{x³+(k+4)a}+{x⁴+(k+6)a}+… where a ≠ 0 and a ≠ 1.

If S = (x¹⁰-x+45a(x-1))/(x-1), then k is equal to:

a) 3
b) -3
c) 1
d) 5

S = {x+ka} + {x²+(k+2)a} +{x³+(k+4)a} up to 9 terms

S = (x+x²+….+x⁹)+a{k+(k+2)+(k+4)+……up to 9 term)}

S = x(1-x⁹)/(1-x) + a(9k+2×36)

= (x¹⁰-x)/(x-1) +9ak+72a

S = (x¹⁰-x+45a(x-1))/(x-1)

= (x¹⁰-x+(9k+72)a(x-1))/(x-1)

45 = 9k+72

9k = -27

k = -3

Answer: (b)

Question 15. The set of all possible values of θ in the interval (0, π) for which the points (1, 2) and (sin θ, cos θ ) lie on the same side of the line x+y = 1 is:
a) (0, π/4)
b) (0, π/2)
c) (0, 3π/4)
d) (π/4, 3π/4)

L₁₁×L₂₂ >0

sinθ +cosθ > 1

θ ∈(0, π/2)

Answer: (b)

Question 16. Let n>2 be an integer. Suppose that there are n metro stations in a city located along a circular path. Each pair of stations is connected by a straight track only. Further, each pair of nearest stations is connected by blue line, whereas all remaining pairs of stations are connected by red line. If the number of red lines is 99 times the number of blue lines, then the value of n is:
a) 201
b) 199
c) 101
d) 200

Red line = 99 blue line

ⁿC₂-n = 99n

n(n-1)/2 = 100n

n-1 = 200

n = 201

Answer: (a)

Question 17. If a curve y = f(x), passing through the point (1,2) is the solution of the differential equation, 2x²dy = (2xy+y²)dx, then f(1/2) is equal to:

a) -1/(1+log_e2)
b) 1+log_e 2
c) 1/(1+log_e2)
d) 1/(1-log_e2)

2(dy/dx) = 2(y/x)+(y/x)² HDE

Since y = vx

2(v+x(dv/dx)) = 2v+v²

2(dv/v²) = dx/x

-(2/v) = ln x+c

-2x/y = ln x+c

c = -1

c: ln x+2x/y = 1

for f(1/2) ln(1/2) +(2/2y) = 1

y = 1/(1+ln 2)

Answer (c)

Question 18. For some θ∈(0, π/2) , if the eccentricity of the hyperbola, x²-y²sec² θ = 10 is √5 times the eccentricity of the ellipse, x²sec²θ+y² = 5, then the length of the latus rectum of the ellipse, is:
a) 4√5/3
b) 2√5/3
c) 2√6
d) √30

H: x²-y²sec²θ = 10

E: x²sec²θ+y² = 5

√(1+10cos²θ/10)= √5√(1-5cos²θ/5)

1+cos²θ = 5-5cos²θ

6 cos²θ = 4

cosθ = √(2/3)

LR of ellipse = 2×5×cos²θ/√5

= 2√5×2/3

= 4√5/3

Answer (c)

Question 19.

\begin{array}{l} lim_{x \to 0} {(\tan (\frac{π}{4} + x))}^{\frac{1}{x}} \end{array}

is equal to:
a) e
b) e²
c) 2
d) 1

\begin{array}{l} lim_{x \to 0} {(\tan (\frac{π}{4} + x))}^{\frac{1}{x}} \end{array}

JEE Main Solution Sept 2 Shift 2 2020 Maths Papers

Answer: (b)

Question 20. Let a, b, c ∈ R be all non-zero and satisfy a³+b³+c³ = 2. If the matrix A =

\begin{array}{l} [\begin{array}{c} a & b & c \\ b & c & a \\ c & a & b \end{array}] \end{array}

satisfies A^TA = I, then a value of abc can be:
a) 2/3
b) 3
c) -1/3
d) 1/3

a³+b³+c³ = 2

A^TA = I

\begin{array}{l} [\begin{array}{c} a & b & c \\ b & c & a \\ c & a & b \end{array}] [\begin{array}{c} a & b & c \\ b & c & a \\ c & a & b \end{array}] = [\begin{array}{c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}] \end{array}

a²+b²+c² = 1

ab+bc+ca = 0

Now (a+b+c)² = Ʃa²+2Ʃab

(Ʃa)² = 1+0 = 1

Ʃa = 1

Now Ʃa³-3abc = (Ʃa)(Ʃa²– Ʃab)

2-3abc = ± 1(1-0)

2-3abc = ± 1

abc = 1/3 or abc = 1

Answer: (d)

Question 21. Let the position vectors of points ‘A’ and ‘B’ be

\begin{array}{l} \hat{i} + \hat{j} + \hat{k} \end{array}

and

\begin{array}{l} 2 \hat{i} + \hat{j} + 3 \hat{k} \end{array}

respectively. A point ‘P’ divides the line segment AB internally in the ratio λ:1 (λ>0) . If O is the origin and

\begin{array}{l} \vec{O B} . \vec{O P} - 3 {| \vec{O A} \times \vec{O P} |}^{2} = 6 \end{array}

, then λ is equal to:

JEE Main Sept 2 Shift 2 2020 Solved Maths Papers

Shift 2 JEE Main Sept 2 2020 Solved Maths Papers

Now (14λ+6)/(λ+1) -3(6λ²)/(λ+1)²= 6

(14λ+6) (λ+1)—18λ² = 6(λ+1)²

-4λ² + 20λ + 6 = 6λ² + 12λ + 6

10λ²-8λ = 0

λ(10λ-8) = 0

Since λ >0

λ = 0.8

Answer: (0.8)

Question 22.Let [t] denote the greatest integer less than or equal to t. Then the value of

\begin{array}{l} \int_{1}^{2} | 2 x - [3 x] | d x \end{array}

is:

Solution:

To find

\begin{array}{l} \int_{1}^{2} | 2 x - [3 x] | d x \end{array}

Put 3x = t

=

\begin{array}{l} \frac{1}{3} \int_{3}^{6} | \frac{2 t}{3} - [t] | d t \end{array}

=

\begin{array}{l} \frac{1}{3} \int_{3}^{6} | \frac{2 t}{3} - [t] | d t \end{array}

Shift 2 2020 JEE Main Sept 2 Solved Maths Papers

Answer: (1)

Question 23. If y =

\begin{array}{l} \sum_{k = 1}^{6} k \cos^{- 1} {\frac{3}{5} \cos k x - \frac{4}{5} \sin k x} \end{array}

, then (dy/dx) at x = 0 is:

y =

\begin{array}{l} \sum_{k = 1}^{6} k \cos^{- 1} {\cos k x + θ} \end{array}

where tanθ = 4/3

cos^-1[cos(kx+a)] = kx+a as kx+a∈[0, π]

y = cos^-1(cos (x+θ)+2 cos^-1(2x+θ))….+6cos^-1(cos (6x+θ))

y = x+θ+2(2x+θ)+…6(6x+θ)

dy/dx = 1×1+2×2+3×3+….+6×6

dy/dx at x = 0

= 1×1+2×2+3×3+….+6×6

= Ʃ6²

= 6×7×13/6

= 91

Answer: (91)

Question 24.If the variance of the terms in an increasing A.P., b₁, b₂, b₃, …., b₁₁ is 90, then the common difference of this A.P. is

Var(x) = (Ʃbi²/11)- (Ʃbi/11)²

90 = [(11a²+385d²+110ad)/11]-[(11a+55d)/11]²

90 = a²+35d²+10ad- (a+5d)²

90 = a²+35d²+10ad- a²-25d²-10ad

90 = 10d²

9 = d²

d = 3

Answer: (3)

Question 25. For a positive integer n, (1+1/x)ⁿ is expanded in increasing powers of x. If three consecutive coefficients in this expansion are in the ratio, 2:5:12, then n is equal to

Let 3 consecutive coefficients are ⁿC_r-1 :ⁿC_r : ⁿC_r+1::2:5:12

ⁿC_r-1/ⁿC_r = 2/5

and ⁿC_r/ⁿC_r+1 = 5/12

r/(n-r+1) = 2/5

and (r+1)/(n-r) = 5/12

7r = 2n+2 and 17r = 5n-12

(2n+2)/7 = (5n-12)/17

34n+34 = 35n-84

n = 118

Answer (118)

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