JEE Main 2021 July 20 Shift 2 Maths Question Paper

Question 1:∫_-π/2^π/2([x] – [sin⁡x ])dx = (Where [.] represents G.I.F)

a. -2
b. 1
c. 0
d. -1

Answer: (c)

I = ∫_-π/2^π/2([x] – [sin⁡x ])dx

Using the property ∫_-a^a f(x)dx = ∫₀^a f(x)dx + ∫₀^a f(-x)dx

I = ∫₀^π/2 ([x] + [-x] )dx – ∫₀^π/2([sin ⁡x] + [-sin ⁡x])dx

= 0

Hence option c is the answer.

Question 2: If

Answer: (3)

⇒ α – β = 0

⇒ α = β

⇒ α + β/2 + γ = 0

⇒ γ = -3β/2

⇒ α/2 – β/3 – γ = 10

⇒ β = 6, α = 6, γ = -9

So, the value of α + β + γ = 3

Question 3: The value of x satisfying the equation log_(x+1)⁡(2x²+7x+5) + log_(2x+5)⁡(x+1)² = 4 is:

a. -2
b. 2
c. -4
d. 4

Answer: (b)

log_(x+1) ⁡(2x+5)(x+1) + log_(2x+5) (x+1)² = 4

1+ log_(x+1) (2x+5)⁡ + 2log_(2x+5) (x+1) = 4

Put log_(x+1) (2x+5) = t

∴1+ t + 2/t = 4

⇒ t² + t + 2 = 4t

⇒ t² – 3t + 2 = 0

t = 1, t = 2

For t = 1

2x+5 = x + 1

⇒ x = -4 (Rejected)

For t = 2

2x+5 = (x + 1)²

x = 2, x = -2 (Rejected)

Hence option b is the answer.

Question 4: Mean of 6 observations is 10 and their variance is 20/3 . If observations are 15, 11, 10, 7, a, b then |a-b| is equal to:

a. 2
b. 1
c. 3
d. 4

Answer: (b)

Mean = 10

(7+10+11+15+a+b)/6 = 10

a + b = 17

Variance = 20/3

(49+100+121+225+a²+b²)/6 – 100 = 20/3

a²+b² = 145

(a+b)² = 289

ab = 72

(a-b)² = (a+b)² – 4ab

(a-b)² = 289 – 288 = 1

|a-b| = 1

Hence option b is the answer.

Question 5: If f(x) = x+1, then find

a. 7/2
b. 3/2
c. 5/2
d. 1/2

Answer: (a)

= ∫₀¹ f(5x)

= ∫₀¹(5x+1) dx

= ((5x²/2) + x)₀¹

= (5/2) + 1

= 7/2

Hence option a is the answer.

Question 6: Sum of 21 terms of series

a. 7
b. 243
c. 9
d. 81

Answer: (c)

2 log₉ x + 3 log₉ x + 4 log₉ x………….21 terms

= (2 + 3 + 4 + 5…+ 22) log₉x

= (21/2)(2 + 22) log₉x

= 21×12 log₉x

= 252 log₉x

Given sum = 252

⇒ log₉ x = 1

⇒ x = 9

Hence option c is the answer.

Question 7: If Re [(1+cos θ + 2i sin⁡ θ)] = 4, then the value of θ is

a. π/2
b. π/3
c. -π/2
d. π

Answer: (d)

= (1 + cos θ – 2i sin θ)/((1 + cos θ)²+ 4 sin²θ)

⇒ (1 + cos θ)/(1 + cos²θ + 2 cos θ + 4 sin²θ) = 4

⇒ (1 + cos θ)/(1 + cos²θ + 2cosθ + 4 – 4 cos²θ) = 4

⇒ (1 + cos θ)/(5 + 2 cos θ – 3 cos² θ) = 4

⇒ 1 + cos θ = 20 + 8 cos θ – 12 cos² θ

12cos²θ – 7cos θ -19 = 0

⇒ 12cos²θ -19cos θ + 12cos θ – 19 = 0

⇒ cos θ(12cos θ -19) + 1(12cos θ -19) = 0

⇒ cos θ = -1 or cosθ = 19/12 (rejected)

⇒ θ = π

Hence option d is the answer.

Question 8: If x = ay-1 = z-2, and x = 3y-2 = bz-2 lie in same plane, then the value of a, b is

a. a = 2,b = 3
b. a = 1, b = 1
c. b = 1, a ∈ R-{0}
d. a = 3,b = 2

Answer:(c)

(x/1) = (y-1/a)/(1/a) = (z-2)/1

x = (y-⅔)/(⅓) = (z-2/b)/(1/b)

⇒ (1/ab) – (1/a) = 0

⇒ b = 1, a∈ R-{0}

Hence option c is the answer.

Question 9: If

P(A∩B∩C) = k², k ∈(0, 1). Then the value of P (at least one of A, B, C) is

a. >1/2
b. [⅛, ¼]
c. < 1/4
d. 1/4

Answer: (a)

P(A) + P(B) – 2P(A∩B) = 1 – k

P(A) + P(C) – 2P(A∩C) = 1 – 2k

P(B) + P(C) – 2P(B∩C) = 1 – k

P(A⋃B⋃C) = P(A) + P(B) + P(C) – P(A∩B) – P(B∩C) – P(A∩C) + P(A∩B∩C)

= (3 – 4k)/2) + k²

= (2k² – 4k + 3)/2

Since the value of 2k² – 4k + 3 is greater than 1

P(A⋃B⋃C) = 1/2

Hence option a is the answer.

Question 10: If f(x) = (5x+3)/(6x+a) and f(f(x) = x then the value of a is:

a. -5
b. 5
c. 6
d. -6

Answer: (a)

f(f(x) = (5f(x)+3)/(6f(x)+a) = x

⇒ 5f(x)+3 = 6xf(x)+ax

⇒ 25x+15+18x+3a = 30x²+18x+6ax²+a²x

⇒(30+6a)x² + (a²-25)x – (3a+15)=0

⇒6(a+5)x² + (a²-25)x – (3a+15) = 0,

a + 5 = 0

a = -5

Hence option a is the answer.

Question 11: If

a. 1
b. 3
c. 2
d. 4

Answer: (b)

y = t³– 6t² + 9t – 3

y’ = 3t²-12t + 9

= 3(t²– 4t + 3)

So total 3 points.

Hence option b is the answer.

Question 12: Given below are three conditions:

A: If 2+4 = 7, then 3+4 = 8

B: if 3+5 = 8, then earth is flat

C: if A and B are true, then 5+4 = 11

Choose the correct option.

a. A is true, B and C are false

b. B is true, A and C are false

c. C is true, A and B are false

d. B is false, A and C are true

Answer: (d)

Truth table p → q

A is true, B is false, C is true

Hence option d is the answer.

Question 13: If A =

a. 2021
b. 2021²
c. -2021
d. 0

Answer: (b)

A = I, B= I+I+I+….,2021 times

|B| = 2021²

Hence option b is the answer.

Question 14: If element of matrix A = [a_ij]_(3×3) where

a. 72
b. 36
c. 108
d. 48

Answer: (c)

A =

|A| = 2(4-1)+1(-2+1)+1(1-2)

= 2(3)+1(-1)+1(-1)

= 4

|3 adj(2A^-1)| = 3³|adj(2A^-1)|

= 3³×|2A^-1|²

= 3³×2⁶×|A⁻¹|²

= 3³×2⁶×1/|A|²

= 108

Hence option c is the answer.

Question 15: In a ∆ABC, if

Answer: 5/2

Projection of

5|(3²+5²-7²)/(2.3.5)| = 5|-15/30|

= 5/2

Question 16: The value of tan (2 tan^-1 (3/5) + sin^-1(5/13)) is:

a. 220/21
b. 110/21
c. 55/21
d. 20/11

Answer: (a)

tan (tan^-1 (6/5)/(1-(9/5)) + tan^-1 (5/12))

= tan (tan^-1 (15/8) + tan^-1 (5/12))

= ((15/8) + (5/12))/(1 – (15/8)(5/12)

= 220/21

Hence option a is the answer.

Question 17: If (α, β) is the point on y² = 6x, that is closest to (3, 3/2) then find 2(α+β)

a. 6
b. 9
c. 7
d. 5

Answer: (b)

Equation of normal at (α, β) is

y-β = (-β/2a)(x-α)

⇒ 3y-3β = -β(x-α)

For shortest distance normal will pass through (3, 3/2)

So (9/2) – 3β = -3β-βα

αβ = 9/2⋯(1)

Now from y² = 6x

β² = 6α⋯(2)

From (1) and (2) we get α = 3/2, β = 3

2(α+β) = 2(3/2 + 3)

= 2(9/2)

= 9

Hence option b is the answer.

Question 18: Two circles pass through (-1, 4) and their centres lie on x²+y²+2x+4y = 4. If r₁ and r₂ are maximum and minimum radii and r₁/r₂ = a+b√2, then the value of a+b is

Answer: 3

Given circle

(x+1)²+(y+2)² = 3²

Any point on the circle is (3 cos θ – 1,3 sin θ – 2)

Radius of the circles which passes through (-1, 4)

r = √((3 cos θ – 1 + 1)² + (3 sin θ – 2 – 4)²)

=√(9 cos² θ + 9 sin² θ + 36 – 36 sin θ)

= √(45 – 36 sin θ)

r_max = 9 = r₁ and r_min = 3 = r₂

⇒ r₁/r₂ = 9/3 = 3

⇒ 3 + 0√2

= a+b

= 3

Question 19: If ∆ABC is a right-angled triangle with sides a,b and c and smallest angle θ. If 1/a, 1/b and 1/c are also the sides of the right-angle triangle then find sin θ.

a. √((3-√5)/2)

b. (3-√5)/2

c. √((3+√5)/2)

d. (3+√5)/2

Answer: a

Let a>b>c

sin θ = c/a

1/a < 1/b < 1/c

1/c² = 1/a² + 1/b²

1 = (c²/a²) + (c²/b²)

1 = (c²/a²) + (c²/(a² – c²)) (As b² + c² = a²)

1 = sin²θ + 1/((a²/c²)-1)

= (sin² θ) + (1/(cosec²θ-1))

(1-sin² θ + 1)/(cosec² θ -1) = 1

⇒ sin² θ + cosec² θ = 3

Put sin² θ = t

⇒ t⁴ – 3t² + 1 = 0

⇒ t² = (3±√5)/2

So sin θ = √((3-√5)/2)

Hence option a is the answer.