JEE Main 2021 August 26 Shift 1 Maths Question Paper with Solutions

Question 1. f(x) =

a. (1-x)² f'(x) – 2(f(x)) ² = 0
b. (1-x) ² f'(x) + 2(f(x)) ²= 0
c. (1+x) ² f'(x) – 2(f(x)) ²= 0
d. (1+x) ² f'(x)+2(f(x)) ²= 0

Answer: (b)

f(x) =

f(x) =

= cos((2 tan^-1(√x)

= cos((cos^-1 (((1 – x)/(1 + x)) ) = (1 – x)/(1 + x)

f'(x) = -2/(1+x)²

2(f(x))² = 2((1-x)/(1+x))²

(1 – x) ² f'(x) + 2(f(x)) ²= 0

Question 2. How many 3 digit numbers by using the digits 0,1,3,4,6,7 can be possible if repetition are allowed?

Answer: 180

digits 0,1,3,4,6

So, 5 x 6x 6 = 180

Question 3. The sum of the square of the distance of a point from (0, 0), (0, 1), (1, 1), (1, 0) is 18 and its locus is a circle. If d is the diameter of the circle, then find d².

Answer: 16

(OP) ² + (AP) ² + (PC) ² + (PB) ²=18

=> x² + y² + x² + (y – 1) ² + (x – 1) ² + y² + (x – 1) ² + (y – 1) ² =18

=> 4x² + 4y² – 4x – 4y=14

=> x² + y² – x – y=7/2

=> r=2

=> d = 4 => d² = 16

Question 4: If log_e (x+y) = 4xy. Find (d²y)/(dx²) at x = 0.

Answer: 40

log_e (x+y) = 4xy

or 1/((x + y)) [1 + (dy/dx)] = 4[x (dy/dx) + y]

or 1 + (dy/dx) = 4(x + y) [x (dy/dx) + y]⋯(i)

If x = 0, then y = 1

From (i)

1 + dy/dx = 4(0 + 1)[0 + 1] = 4

Or dy/dx = 3

From (i), again differentiate w.r.t. x

0 + (d² y)/(dx² ) = 4(x + y)[x (d²y)/(dx² ) + 2 (dy/dx)] + 4[x (dy/dx) + y](1 + (dy/dx))

At x = 0, y = 1, dy/dx = 3

(d² y)/(dx²) = 4(0 + 1)[0 + 2×3]+4[0 + 1](1 + 3) = 40

Therefore, (d² y)/(dx² ) = 40

Question 5: Find value of (5 – e² ) for ellipse x²/8 + y²/4 = 1

Answer: 9/2

x²/8 + y²/4 = 1

a² = 8, b² = 4

e = c/a = √((a² – b²)/a² ) = √(1-4/8) = √(1/2)

(5 – e² ) = 5 – 1/2 = 9/2

Question 6: In a hospital, 89% has disease A and 98% has disease B, K% is the number of patients who has both, then K can’t be the subset of ____.

Answer: Least k =87, Max k = 89

Let total =100

n(A)=89, n(B)=98

max (n(A),n(B)) ≤ n(AUB) ≤ 100

98 ≤ n(A)+n(B)-n(A n B) ≤ 100

=> 98 ≤ 98 + 89- k ≤ 100

Least k =87

Max k = 89

Question 7: P(A) = p, P(B) = 2p, P(Exactly one out of A and B occurs) = 5/9. What will be the maximum value of p?

Answer: 5/9

Given: P(A) = p and P(B) = 2p

P(A)-P(A∩B)+P(B)-P(A∩B) = 5/9

3p-2P(A∩B) = 5/9

0 ≤ P(A∩B) ≤ p

For maximum value: 3p – 2p = 5/9

So, p = 5/9

Question 8: Evaluate

Answer: 4 ln ½

Question 9: If cosx/(1+sinx) = |tanx|, then find the number of solutions in [0, 2π].

Answer: x = π/6

Case 1: tan x ≥ 0

cosx/(1+sinx ) = sinx/cosx

⇒cos²x = sin²x+sinx

⇒1-sin²x = sin²x + sinx

⇒2 sin²x + sin x – 1 = 0

⇒sin x = -1 ,1/2

sin x = 1/2 ⇒ x = π/6

sin x = -1 (not possible)

Case 2: tan x < 0

(Cos x)/(1+sinx) = -sin x/cos x

⇒cos²x = -sin²x – sinx

⇒1-sin²x = -sin²x – sinx

Sin x = -1 (Not possible)

Only one solution exists x = π/6

Question 10: Sum of infinite terms A, AR, AR², AR³,…. is 15. If the sum of the square of these terms is 150, then find the sum of AR², AR⁴, AR⁶, ⋯.

Answer: ½

A + AR + AR² + ⋯ = 15

⇒A/(1-R) = 15 ⋯(i)

Now, A² + A² R² + A² R⁴ + ⋯ = 150

⇒A²/(1-R² ) = 150

⇒(A⋅A)/((1-R)(1 + R)) = 150

⇒15A/(1 + R) = 150

⇒A = 10(1 + R)⋯(ii)

Solving (i) & (ii):

10(1 + R)/(1-R) = 15

⇒R = 1/5

So, A = 12

AR² + AR⁴ + AR⁶ + ⋯ = (AR^²)/(1-R² ) = (12⋅1/25)/(1-1/25) = 1/2

Question 11: If

Answer: √2

Question 12: A wire of length 36 units is cut into two parts which are bent respectively to form a square of side x units and a circle of radius of r units. If the sum of areas of the square and the circle so formed is minimum, then find the circumference of the circle.

Answer: 2π x 18/(4 + π) = 36π/(4 + π)

Sum of perimeter of square and circle = 36

4x + 2πr = 36⇒2x + πr = 18

x = (18 – πr)/2 = 9 – πr/2 …..(i)

Sum of areas: S = x² + πr² ….(ii)

S = πr² + (9 – πr/2) ² , diff. w.r.to r

dS/dr = 2πr + 2(9 – πr/2)( – π/2)

dS/dr = 0

⇒r – 9/2 + πr/4

⇒r(π + 4)/4 = 9/2

⇒r = 18/(4 + π)

Circumference of circle = 2π x 18/(4 + π) = 36π/(4 + π)

Question 13: Solve

Answer: [(1/2)] tan^-14 – 0

= [(1/2)] tan^-14 – 0

Question 14: If

Answer:

Given

This implies,

or

Question 15: Evaluate

Answer: 20 ⋅ 21 ⋅ 2¹⁸

Question 16: Solve (1 + y) tan²x + tan x (dy/dx) + y = 0

Answer: y tan x = -tan x + x + c

(1 + y) tan²x + tan x (dy/dx) + y = 0

tan x(dy/dx) + (tan²x + 1)y = -tan²x

dy/dx + (tan x + cot x )y = -tan x

I. F. =

= e ^{ln sec x + ln sin x}

= e ^{ln sec x . sin x}

= tan x

• y tan x = -∫tan² x dx
• y tan x = -∫(sec² x-1) dx
• y tan x = -tan x + x + c

Question 17: If

Answer: