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JEE Main 2021 August 27th Shift 1 Maths Question Paper

Question 1:If

\(U(n) =(1+\frac{1^2}{n^2})(1+\frac{2^2}{n^2})^2(1+\frac{3^2}{n^2})^3.....(1+\frac{n^2}{n^2})^n\)

, then find
\(\lim_{n \rightarrow \infty}[U(n)]^{\frac{-4}{n^2}}\)

Solution:

Answer: e² / 16

\(log(U(n)) = \sum_{r=1}^n log(1+\frac{r^2}{n^2})^r\)

=
\(\sum_{r=1}^n r log(1+\frac{r^2}{n^2})\)

Question 2: Find the value of

\(\frac{3}{2} x^2 + \frac{5}{3} x^3 + \frac{7}{4} x^4 +......\infty\)

if 0 < x < 1.

Solution:

Answer:2x²/(1-x) + x + ln(1-x)

Let S =
\(\sum_{r=1}^{\infty}\frac{2r+1}{r+1}x^{r+1}=2 \sum_{r=1}^{\infty}x^{r+1} - \sum_{r=1}^{\infty}\frac{x^{r+1}}{r+1}\)

= 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

\(\sum_{n=0}^{20} (^{20}C_n)^2\)

Solution:

Answer: Solved below

\(\sum_{n=0}^{20} (^{20}C_n)^2\)

\((^{20}C_r) \rightarrow\)
coefficient of x^r in (1+x)²⁰ expansion.

\((1+x)^{20}=^{20}C_0+^{20}C_1x+^{20}C_2x^2+ ..+^{20}C_{20}x^{20}\)

\((x+1)^{20}=^{20}C_0 x^{20} + ^{20}C_1x^{19} +^{20}C_2x^{18}+ ..+^{20}C_{20}\)

Multiplying both the equations,

\((x+1)^{40}= [^{20}C_0+^{20}C_1x+^{20}C_2x^2+ ..+^{20}C_{20}x^{20}] \times [^{20}C_0 x^{20} + ^{20}C_1x^{19} +^{20}C_2x^{18}+ ..+^{20}C_{20}] \)

Comparing both the coefficients of x²⁰ on both the sides,

\(^{40}C_{20} = (^{20}C_{0})^2 + (^{20}C_{1})^2 +..+(^{20}C_{20})^2\)

\(^{40}C_{20}=\sum_{n=0}^{20} (^{20}C_n)^2\)

Question 4: If

\(\frac{x^2}{b^2} +\frac{y^2 }{4a^2 }=1\)

and minimum area with coordinate axes of tangent is Kab, then the value of K is

Solution:

Answer: 2

\(\frac{x^2}{b^2} +\frac{y^2 }{4a^2 }=1\)

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.

Solution:

Answer: 1 + 6 – 6 = 1

Δ = 0

\(\begin{bmatrix} 2& -1& -1\\1 &1 &-2 \\3 &3 &-\beta \end{bmatrix}=0\)

= 2(-β + 6) + 1(-β + 6) – 1(3-3) = 0

β = 6

Δ₁ = 0

\(\begin{bmatrix} 3& -1& -1\\ \alpha &1 &-2 \\3 &3 &-\beta \end{bmatrix}=0\)

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

\(\int_6^{16}\frac{ln \ x^2}{ ln \ x^2 + ln(x^2-44x + 484)}dx\)

is;

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

Solution:

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.

Solution:

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

Solution:

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

Solution:

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

Solution:

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

\(\lim_{x \rightarrow \beta}\frac{e^{2(x^2+bx+c)} - 1 - 2(x^2 + bx + c)}{(x-\beta)^2}\)

Solution:

Answer: 2(β - α)² = 2(b² - 4c)

\(\lim_{x \rightarrow \beta}\frac{e^{2(x^2+bx+c)} - 1 - 2(x^2 + bx + c)}{(x-\beta)^2}\)

=
\(\lim_{x \rightarrow \beta}\frac{e^{2(x-\alpha)(x-\beta)} - 1 - 2(x-\alpha)(x-\beta)}{(x-\beta)^2(x-\alpha)^2} \times (x-\alpha)^2\)

Let (x - α)(x - β) = y

then
\(\lim_{y \rightarrow 0} \frac{e^{2y}-1-2y}{y^2} \times (\beta - \alpha)^2\)

Using L hospital rule,

=
\((\beta - \alpha)^2 \lim_{y \rightarrow 0} \frac{2e^{2y}-2}{2y}\)

=
\((\beta - \alpha)^2 \lim_{y \rightarrow 0} \frac{e^{2y}-1}{2y} \times 2\)

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

Solution:

Answer: (b)

(z + i)/(z + 2i) is purely real, so (z + i)/(z + 2i) =
\((\overline{\frac{z + i}{z + 2i}})\)

⇒
\(z\bar{z} - 2iz + i\bar{z} + 2 = z\bar{z} - iz + 2i\bar{z} + 2\)

⇒
\(z(-i) + \bar{z}(-i) = 0\)

⇒
\(z = -\bar{z}\)

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

Solution:

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
b.

Solution:

Answer: (b.

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)

Solution:

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

Solution:

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)

\(d = \sqrt{(4/7)^2 + (6/7)^2 + ( - 12/7)^2 }\)

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)

Solution:

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:

Question 19: Solve

\(\int \frac{1}{(x^2+x+1)^2}dx\)

a.
\(\frac{4}{\sqrt{3}} tan^{-1}(\frac{2x-1}{\sqrt{3}}) + \frac{\sqrt{3}}{4}(\frac{2x-1}{x^2+x+1})+c\)
b.
\(\frac{4}{3\sqrt{3}} tan^{-1}(\frac{2x-1}{\sqrt{3}}) - \frac{\sqrt{3}}{4}(\frac{2x-1}{x^2+x+1})+c\)
c.
\(\frac{4}{3\sqrt{3}} tan^{-1}(\frac{2x+1}{\sqrt{3}}) + \frac{1}{3}(\frac{2x+1}{x^2+x+1})+c\)
d.
\(\frac{4}{3\sqrt{3}} tan^{-1}(\frac{2x+1}{\sqrt{3}}) - \frac{1}{3}(\frac{2x+1}{x^2+x+1})+c\)

Solution: