JEE Main 2021 (Online) 1st September Evening Shift

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MCQ (Single Correct Answer)

The function $$f(x) = {x^3} - 6{x^2} + ax + b$$ is such that $$f(2) = f(4) = 0$$. Consider two statements :

Statement 1 : there exists x₁, x₂ $$\in$$(2, 4), x₁ < x₂, such that f'(x₁) = $$-$$1 and f'(x₂) = 0.

Statement 2 : there exists x₃, x₄ $$\in$$ (2, 4), x₃ < x₄, such that f is decreasing in (2, x₄), increasing in (x₄, 4) and $$2f'({x_3}) = \sqrt 3 f({x_4})$$.

Then

both Statement 1 and Statement 2 are true

Statement 1 is false and Statement 2 is true

both Statement 1 and Statement 2 are false

Statement 1 is true and Statement 2 is false

Explanation

$$f(x) = {x^3} - 6{x^2} + ax + b$$

$$f(2) = 8 - 24 + 2a + b = 0$$

$$2a + b = 16$$ .... (1)

$$f(4) = 64 - 96 + 4a + b = 0$$

$$4a + b = 32$$ .... (2)

Solving (1) and (2)

a = 8, b = 0

$$f(x) = {x^3} - 6{x^2} + 8x$$

$$f'(x) = 3{x^2} - 12x + 8$$

$$f''(x) = 6x - 12$$

$$\Rightarrow$$ f'(x) is $$ \uparrow $$ for x > 2, and f'(x) is $$ \downarrow $$ for x < 2

$$f'(2) = 12 - 24 + 8 = - 4$$

$$f'(4) = 48 - 48 + 8 = 8$$

$$f'(x) = 3{x^2} - 12x + 8$$

vertex (2, $$-$$4)

f'(2) = $$-$$4, f'(4) = 8, f'(3) = 27 $$-$$ 36 + 8

f'(x₁) = $$-$$1, then x₁ = 3

f'(x₂) = 0

Again

f'(x) < 0 for x $$\in$$ (2, x₄)

f'(x) > 0 for x $$\in$$ (x₄, 4)

x₄ $$\in$$ (3, 4)

f(x) = x³ $$-$$ 6x² + 8x

f(3) = 27 $$-$$ 54 + 24 = $$-$$3

f(4) = 64 $$-$$ 96 + 32 = 0

For x₄(3, 4)

f(x₄) < $$-$$3$$\sqrt 3 $$

and f'(x₃) > $$-$$4

2f'(x₃) > $$-$$8

So, 2f'(x₃) = $$\sqrt 3 $$ f(x₄)

Correct Ans. (a).

JEE Main 2021 (Online) 27th August Evening Shift

MCQ (Single Correct Answer)

If $$y(x) = {\cot ^{ - 1}}\left( {{{\sqrt {1 + \sin x} + \sqrt {1 - \sin x} } \over {\sqrt {1 + \sin x} - \sqrt {1 - \sin x} }}} \right),x \in \left( {{\pi \over 2},\pi } \right)$$, then $${{dy} \over {dx}}$$ at $$x = {{5\pi } \over 6}$$ is :

$$ - {1 \over 2}$$

$$-$$1

$${1 \over 2}$$

Explanation

$$y(x) = {\cot ^{ - 1}}\left[ {{{\cos {x \over 2} + \sin {x \over 2} + \sin {x \over 2} - \cos {x \over 2}} \over {\cos {x \over 2} + \sin {x \over 2} - \sin {x \over 2} + \cos {x \over 2}}}} \right]$$

$$y(x) = {\cot ^{ - 1}}\left( {\tan {x \over 2}} \right) = {\pi \over 2} - {x \over 2}$$

$$y'(x) = {{ - 1} \over 2}$$

JEE Main 2021 (Online) 26th August Evening Shift

MCQ (Single Correct Answer)

The local maximum value of the function $$f(x) = {\left( {{2 \over x}} \right)^{{x^2}}}$$, x > 0, is

$${\left( {2\sqrt e } \right)^{{1 \over e}}}$$

$${\left( {{4 \over {\sqrt e }}} \right)^{{e \over 4}}}$$

$${(e)^{{2 \over e}}}$$

Explanation

$$f(x) = {\left( {{2 \over x}} \right)^{{x^2}}}$$ ; x > 0

$$\ln f(x) = {x^2}(\ln 2 - \ln x)$$

$$f'(x) = f(x)\{ - x + (\ln 2 - \ln x)2x\} $$

$$f'(x) = \underbrace {f(x)}_ + \,.\,\underbrace x_ + \underbrace {(2\ln 2 - 2\ln x - 1)}_{g(x)}$$

$$g(x) = 2{\ln ^2} - 2\ln x - 1$$

$$ = \ln {4 \over {{x^2}}} - 1 = 0 \Rightarrow x = {2 \over {\sqrt e }}$$

$$LM = {2 \over {\sqrt e }}$$

Local maximum value = $${\left( {{2 \over {2/\sqrt e }}} \right)^{{4 \over e}}} = {e^{{2 \over e}}}$$

JEE Main 2021 (Online) 20th July Morning Shift

MCQ (Single Correct Answer)

Let $$A = [{a_{ij}}]$$ be a 3 $$\times$$ 3 matrix, where $${a_{ij}} = \left\{ {\matrix{ 1 & , & {if\,i = j} \cr { - x} & , & {if\,\left| {i - j} \right| = 1} \cr {2x + 1} & , & {otherwise.} \cr } } \right.$$

Let a function f : R $$\to$$ R be defined as f(x) = det(A). Then the sum of maximum and minimum values of f on R is equal to:

$$ - {{20} \over {27}}$$

$${{88} \over {27}}$$

$${{20} \over {27}}$$

$$ - {{88} \over {27}}$$

Explanation

$$A = \left[ {\matrix{ 1 & { - x} & {2x + 1} \cr { - x} & 1 & { - x} \cr {2x + 1} & { - x} & 1 \cr } } \right]$$

$$\left| A \right| = 4{x^3} - 4{x^2} - 4x = f(x)$$

$$f'(x) = 4(3{x^2} - 2x - 1) = 0$$

$$ \Rightarrow x = 1;x = {{ - 1} \over 3}$$

$$\therefore$$ $$f(1) = - 4;f\left( { - {1 \over 3}} \right) = {{20} \over {27}}$$

Sum $$ = - 4 + {{20} \over 7} = - {{88} \over {27}}$$