JEE Main 2021 August 27 Shift 1 Maths Question Paper with Solutions

Question 1:If

Answer: e² / 16

=

Question 2: Find the value of

Answer:2x²/(1-x) + x + ln(1-x)

Let S =

= 2[x² + x³ + ….∞] – [x²/2 + x³/3 + x⁴/4 + …∞]

= 2[x²/(1-x)] – [x + x²/2 + …. – x]

ln(1-x) = -x – x²/2 – x³/3 – x⁴/4 – …∞] = 2x²/(1-x) + x + ln(1-x)

Question 3: Solve

Answer: Solved below

Multiplying both the equations,

Comparing both the coefficients of x²⁰ on both the sides,

Question 4: If

Answer: 2

Equation of tangent is

(x/b)cosθ + (y/2a) sin θ = 1

Area(ΔOAb)=(1/2)(2ab)(secθ csc θ)

Area = 2ab/sin2θ

Minimum area = 2ab (Therefore, θ = 45 degrees)

So, K=2

Question 5:If 2x – y – z = 3,x + y – 2z = α,3x + 3y – βz = 3 has infinite solutions, then find the value of α + β – αβ.

Answer: 1 + 6 – 6 = 1

Δ = 0

= 2(-β + 6) + 1(-β + 6) – 1(3-3) = 0

β = 6

Δ₁ = 0

3(-β + 6) + 1(-αβ + 3) – 1(3α – 3) = 0+

Or α = 1

Therefore, α + β – αβ = 1 + 6 – 6 = 1

Question 6: The value of

a. 5
b. 8
c. 6
d. 10

Answer: (a)

Question 7: A wire of length 20 is cut into two parts one is made into a regular hexagon of side a and the other to square. Find ‘a’ if the combined area of a square and regular hexagon is minimum.

Answer: a = 10/(2√3 + 3)

4b + 6a=20

⇒ b=(20-6a)/4 = 5 – (3/2) a

A = b² + (3√3 a²)/2

A = (5 – (3/2) a) ² + (3√3 a²)/2

dA/db = 3√3a – 3/2×2(5 – 3a/2) = 0

⇒ 3√3 a + 9/2 a = 15

⇒ 2√3 a + 3a=10

⇒ a = 10/(2√3 + 3)

Question 8: If x² + y² + px + y(1 – p) = 0 is the equation of circle, r ∈ (0,5], q = p² then number of integral values of (p,q) satisfy is

a. 16
b. 14
c. 19
d. 21

Answer: (b)

Let S : x² + y² + px + y(1 – p) = 0

r = √((p/2) ² + ((1 – p)/2) ² )

0 < (p² + (1 – p) ²)/4 ≤ 25

⇒ 0 < 2p² + 1 – 2p ≤ 100

⇒ 0 < 2p² – 2p + 1 ≤ 100

⇒ ( – 1/4) < (p – 1/2)² ≤ 199/4

⇒ (p – 1/2) ² ≤ (√199/2) ²

Total number of integral pairs of (p,q) = 14

Question 9: In a certain biased die, the probability of getting a particular face is (1/6 + x) and the opposite is (1/6 – x) and 0 < x < 1/6. The probability for other faces is 1/6 each. The sum of opposite faces in a die is 7. If the probability of getting a sum 7 in a throw of 2 such die is 13/96, then x is.

a. 1/8
b. 1/12
c. 1/18
d. 1/20

Answer: (c)

For particular and opposite faces, if sum is 7.

Probability = 2(1/6 + x)(1/6 – x)

For remaining faces in two die, if sum is 7,

Probability = 2(1/6 x 1/6 + 1/6 x 1/6)

Given, 2[(1/6 + x)(1/6 – x) + 1/6 x 1/6 + 1/6 x 1/6] = 13/96

⇒ 1/36 – x² + 2/36 = 13/(96×2)

⇒ 1/12 – x² = 13/(96×2)

⇒ x² = 1/12 – 13/(96×2) = 1/12 (1 – 13/16) = 1/12 ((16 – 13)/16)

Therefore, x = 1/8

Question 10:Find y(π) if y(0) = 7 and dy/dx = 2(y – 2 sin x – 10)x + 2 cos x

Answer: y(π) = 3e^{π^2} + 10

dy/dx = 2(y – 2 sin x – 10)x + 2 cos x

dy/dx – 2 cos x = 2(y – 2 sin x – 10)x

(d/dx (y – 2 sin x – 10))/((y – 2 sin x – 10) ) = 2x

⇒ ∫d(y – 2 sin x – 10)/((y – 2 sin x – 10) ) = ∫2x dx

⇒ log |y – 2 sin x – 10| = x² + C

At x = 0, y = 7

⇒ log |7 – 0 – 10| = 0 + C⇒ C = log 3

At x = π

⇒ log |y – 2 sin π – 10| = π² + log 3

⇒ log ((y – 10)/3) = π²

⇒ y(π) = 3e^{π^2} + 10

Question 11:If α, β are distinct roots of x^2 + bx + c = 0 then find

Answer: 2(β – α)² = 2(b² – 4c)

=

Let (x – α)(x – β) = y

then

Using L hospital rule,

=

=

This implies, 2(β – α)² = 2(b² – 4c)

Question 12:If (z + i)/(z + 2i) is purely real, then the locus of z is

a. x – axis
b. y – axis
c. y = x
d. y = x/2

Answer: (b)

(z + i)/(z + 2i) is purely real, so (z + i)/(z + 2i) =

⇒

⇒

⇒

z is purely imaginary

Therefore, z lies on y – axis.

Question 13:Tangent and normal at P(2, -4) to parabola y² = 8x meet directrix at A and B. Q(a,b) is such that APBQ is a square. Find 2a + b.

a.-12
b.-16
c.-20
d.-18

Answer: (b)

y² = 8x

⇒ a = 2

Tangent equation

-4y = 4(x+2)⇒ x+y+2 = 0

Normal: x-y+6 = 0

Directrix x+2 = 0

A = (-2, 0), B = (-2, -8)

Because, mid-point AB is (-2, -4)

So, Q = (-6, -4)

2a+b = -16

Question 14:If y = log x + log x^(1/3) + log x^(1/9) + .. and (2 + 4 + 6 + ⋯ + 2y)/(3 + 6 + 9 + ⋯ + 3y) = 1/9 log x then the values of x and y are

a. x = 10⁵, y = 8
b. x = 10⁶, y = 9
c. x = 10⁶, y = 8
d. x = 10⁵, y = 9

Answer: (b)

y = log x + log x^(1/3) + log x^(1/9) + ⋯

y = (1 + 1/3 + 1/9 + ⋯) log x

y = [1/(1 – 1/3)] log x = 3/2 log x ⋯(i)

(2 + 4 + 6 + ⋯ + 2y)/(3 + 6 + 9 + ⋯ + 3y) = 2(1 + 2 + 3 + ⋯ + y)/3(1 + 2 + 3 + ⋯ + y)

⇒2/3 = (1/9) log x

log x = 6

x = 10⁶

y = 3/2 log x ⋯ From (i)

y = 3/2×6 = 9

Question 15:In a ∆ABC, if sinA/sinB = sin(A-b)/sin(A-c) and a,b,c are the sides of the ∆ABC, then the correct relationship between a,b,c is ___.

a. c² (a² + b² ) = a² (c² – b²)
b. a² (b² – c² ) = b² (a² – c²)
c. ac(a² – c² ) = b² (a² – b²)
d. bc(a² – c² ) = a² (b² – c²)

Answer: (c)

Question 16:A plane has equation x – y + z = 5 and a line has direction ratios as (2,3,-6), then the distance of the point P(1,3,5) along the line from the given plane is:

a. 2 m
b. 2√3 m
c. 3m
d. 3√2 m

Answer: (a)

Equation of line: (x – 1)/2 = (y – 3)/3 = (z – 5)/( – 6) = k

⇒ x = 2k + 1 , y = 3k + 3 , z = – 6k + 5

Equation of plane: x – y + z = 5

⇒ 2k + 1 – 3k – 3 – 6k + 5 = 5

⇒ – 7k = 2 ⇒ k = – 2/7

(x, y, z)≡ (3/7, 15/7, 47/7)

d = 1/7 √(16 + 36 + 144)

d = 14/7 = 2

Question 17:In (sin^-1 x )² – (cos^-1 x )² = a for 0 < x < 1. Then find 2x² – 1.

a. sin(2a/π)
b. cos(4a/π)
c. cos(2a/π)
d. sin(4a/π)

Answer: (a)

(sin^-1 x )² – (cos^-1 x )² = a

⇒ (sin^-1 x )² – (π/2 – sin^-1 x )² = a

⇒(sin^-1 x )² – (π²/4) – (sin^-1x )² + π sin^-1 x = a

⇒ π sin^-1 x = a + π²/4

⇒x = sin (a/π + π/4)

Now, 2x² – 1 = 2 sin²(a/π + π/4) – 1

= – cos(2a/π + π/2)

= sin(2a/π)

Question 18:If (-2, 2) satisfies y + x (dy/dx) = x², then the correct equation is

a. x² + 2xy + 12 = 0
b. x² + 2xy – 12 = 0
c. x² + 2xy + 4 = 0
d. x³ – 3xy – 4 = 0

Solution:

Answer: (d)

x=-2 and y=2

y + x( dy/dx) = x²

⇒ y dx+x dy = x² dx

⇒∫d(xy) =∫ x² dx

⇒ x y = x³/3 + c

At x=-2 & y=2

⇒ -4 = (-8)/3 + c

⇒c = -4 + 8/3

Therefore, xy = x³/3 – 4/3

⇒3xy = x³ – 4 ⇒x³ – 4 – 3xy = 0

Question 19: Solve