JEE Main 2021 July 20 Shift 1 Maths Question Paper

Question 9: If

Answer: 3

⇒ e^(2+1)2/(1+1)

= e^α

⇒ e³ = e^α

= α = 3

Question 10: Evaluate ∫_-1¹ log (√(1 – x) + √(1 + x))dx

Answer: 2 ln(2) – (1 – π/2)

f(-x) = f(x)

⇒ f is even

∴ I = 2∫₀¹ log ((1 – x) + √(1 + x))dx…(i)

We will solve it by integration by parts.

Let u = ln⁡(√(1-x)+√(1+x)

I = 2 ln(2) – (1 – π/2)

Question 11: The logical equivalence of the Boolean expression ~(p∧~q)∨(q∨~p) is

Answer: ≡ p → q

~(p∧~q) ∨ (q∨~p)

≡ (~p∨q) ∨ (q∨~p)

≡ ~p∨q

≡ p → q

Question 12: Consider the curve y² = 2x and point A(2, 2). If the normal at A intersects the curve again at point B and the tangent at A intersects the x-axis at C, then area of ∆ABC is

Answer: 25/2

Equation of tangent at A

yy₁ = x+x₁

⇒2y = x+2

Tangent at A intersects the x-axis at C

C(-2, 0)

Equation of normal at A

y+ 2x = 6

Solving normal equation with y² = 2x, we get B(9/2,-3)

So, area will be ½ × AC×AB

= ½ ×2√5×√[(25/4) + 25]

= 25/2

Question 13: If for vectors A and B, A.B = |A×B|, then |A-B| is

a. √(A² + B²√(2AB))
b. √(A² + B² – √(2AB))
c. √(A² + B² – √2 AB)
d. √(A² + B² + √2 AB)

Answer: (c)

A.B = |A×B|

⇒ cos θ = sin θ

⇒ tan θ = 1

⇒ θ = π/4

|A-B|² = A² + B² – 2A.B

= A² + B² – 2AB cos π/4

= A² + B² – √2AB

⇒ |A – B| = √(A²+ B² – √2 AB)

Hence option c is the answer.

Question 14: If the roots of the quadratic equation x² + 3^1/4x + 3^1/2 = 0 are α and β, then the value of α⁹⁶(α¹² – 1) + β⁹⁶(β¹² −1)

a. 50⋅3²⁴
b. 51⋅3²⁴
c. 52⋅3²⁴
d. 104⋅3²⁴

Answer: (c)

x² + 3^1/2 = – 3^1/4x

⇒ x⁴ + 3 = -3^1/2x²

⇒ x⁸ + 9 + 6x⁴ = 3x⁴

⇒ x⁸ + 9 + 3x⁴ = 0

α⁸ = -9 – 3α⁴

And α¹² = -9α⁴ – 3α⁸

= -9α⁴ -3(-9-3α⁴)

= 27

Similarly, β¹² = 27

α⁹⁶(α¹² – 1) + β⁹⁶(β¹² – 1) = 27⁸.26 + 27⁸.26

= 52.3²⁴

Hence option c is the answer.

Question 15: In ∆ABC, if AB = 5, ∠B=cos^-1 (3/5) and the radius of circumcircle of triangle is 5. Then the area of ∆ABC is

a. 6+8√3
b. 3+4√3
c. 3+8√3
d. 6+4√3

Answer: (a)

cos ⁡B = ⅗

⇒ sin⁡B = 4/5, R = 5

b = 2R sin⁡B = 8

AB = c = 5

By cosine rule,

cos ⁡B = (a²+c²-b²)/2ac = 3/5

⇒ (a²+25-64)/(2a(5))

= 3/5

⇒ a²-39 = 6a

⇒ a²-6a-39 = 0

⇒ a = ((6+8√3))/2

⇒ a = 3+4√3

∆ = abc/4R

= 6+8√3

Hence option a is the answer.

Question 16: If the shortest distance between the lines

a. 2
b. 4
c. 6
d. √6

Answer: (c)

⇒ |(α + 4) ×8 + 16 + 12| = 108

⇒ (α + 4) ×8 = 80 (since α >0)

⇒ α = 6

Hence option c is the answer.

Question 17: Let y = mx+c, m>0 be the focal chord of y² = −64x which is tangent to (x+10)²+ y² = 4. Then the value of 4√2(m+c) is equal to

Answer: 34

Focus of parabola is (−16,0)

So, −16m + c=0

⇒c = 16m …(i)

Now slope form of tangent to the circle is given by

y = m(x+10) ± 2√(1+m²)

So c = 10m ± 2√(1+m²) …(ii)

So, from (i) and (ii)

16m = 10m ± 2√(1+m²)

⇒9m² = 1 + m²

⇒ m = 1/2√2

⇒ c = 16m

= 8/2√2

4√2 (m+c) = 34

Question 18: A continuous differentiable function f(x) is increasing in (-∞, 3/2) and decreasing in (3/2, ∞). Then x = 3/2 is

a. point of local maxima
b. point of local minima
c. point of inflection
d. None of these

Answer: (a)

A rough graph of f(x) can be drawn as

Thus x = 3/2 is a point of local maxima.

Hence, option a is the answer.

Question 19: If z and ω are complex numbers such that |zω| = 1, arg⁡(z) – arg⁡(ω) = (3π )/2 then find arg

a. π/4
b. -π/4
c. 3π/4
d. -3π/4

Answer:

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

= -½ – ½ i

arg( -½ – ½ i) = -3π/4

Hence option d is the answer.

Question 20: If an invertible function f(x) is defined as f(x) = 3x – 2, g(x) is also an invertible function such that f^-1 (g^-1 (x) )= x – 2, then g(x) is

a. (x-8)/3
b. (x+8)/3
c. (x-3)/8
d. (x+3)/8

Answer: (b)

f^-1 (g^-1(x) ) = x-2

⇒ f(x-2) = g^-1 (x)

3(x – 2) – 2 = g^-1(x)

3x – 8 = g^-1 (x)

g^-1 (x) = 3x – 8

or x = 3g(x) – 8

So g(x)=(x + 8)/3

Hence option b is the answer.

Question 21: The probability of selecting integers a ∈ [-5,30], such that x² + 2(a+4)x – 5a + 64 >0 for all x∈ R is:

Answer: 1/2

x²+ 2(a+4)x – (5a – 64)>0

D<0

Therefore 4(a+4)² + 4(5a – 64) < 0

⇒ a² + 13a – 48<0

⇒ (a + 16)(a – 3)<0

So, a ∈ (-16,3)

Total integers are 18

Probability = 18/36 = 1/2

Question 22: If ∫₀^a e^x-[x] dx = 10e – 9, then the value of a is (where [.] is greatest integer function)

a. 9 + ln⁡ 2
b. 10 + ln⁡ 2
c. 10
d. 9

Answer: (b)

Let a = K + λ, 0 ≤ λ < 1

∫₀^a e^{x} dx = ∫₀^K e^{x} dx + ∫_K^K+λ e^{x} dx

= K ∫₀¹ e^x dx + ∫₀^λ e^x dx

= K(e – 1)+(e^λ – 1)

=(Ke + e^λ) – (K+1)

Given, ∫₀^a e^x-[x] dx = 10e – 9

⇒ (Ke+e^λ )- (K+1) = 10e – 9

⇒ K = 10 and e^λ = 11 – 9 = 2

⇒ λ = ln ⁡2

a = 10 + ln ⁡2

Hence option b is the answer.

Question 23:

a. 20/27
b. −20/27
c. 88/27
d. -88/27

Answer: (d)

= 4x³– 4x² – 4x

f’(x) = 4(3x²– 2x – 1)

f’(x) = 4(x-1)(x + ⅓)

f(1) = 4 – 4 – 4 = -4

f(-1/3) = (-4/27) – (4/9) + (4/3)

= (-4-12+36)/27

= 20/27

(20/27) – 4 = (20-108)/27 = -88/27

Hence option d is the answer.

Question 24:Find the coefficient of a³b⁴c⁵ in (ab+bc+ca)⁶.

a. 60
b. 45
c. 40
d. 90

Answer: (a)

(ab + bc + ca)⁶ =

=

= For a³b⁴c⁵, we need

p + r = 3

p + q = 4

q + r = 5

Solving we get, p = 1,q = 3,r = 2

Coefficient of a³b⁴c⁵ in (ab+bc+ca)⁶ is 6!/1!3!2! = 60.

Hence option a is the answer.

Question 25: x (dy/dx)⋅ tan⁡ (y/x) = y tan (y/x) + x, y(1/2) = π/6. Area bounded by x = 0,x = 1/√2, y = y(x).

Answer: (⅛)(π-1)

x (dy/dx). tan ⁡(y/x) = y tan (⁡y/x) + x

(dy/dx) tan⁡(y/x) = (y/x) tan⁡(y/x) + 1

Let y = vx

dy/dx = v+x (dv/dx)

(v+x dv/dx) tan⁡v = v tan⁡v + 1

⇒ v+x dv/dx = v + cot⁡ v

⇒ dv/cot⁡v = dx/x

tan⁡v dv = dx/x

-log⁡|cos⁡v| = log⁡|cx|

-cos ⁡v = cx

So y = x cos^-1 ⁡x

∫₀^1/√2x cos^-1 ⁡x = (⅛)(π-1)