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1

### JEE Main 2021 (Online) 1st September Evening Shift

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
A
both Statement 1 and Statement 2 are true
B
Statement 1 is false and Statement 2 is true
C
both Statement 1 and Statement 2 are false
D
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).
2

### JEE Main 2021 (Online) 27th August Evening Shift

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 :
A
$$- {1 \over 2}$$
B
$$-$$1
C
$${1 \over 2}$$
D
0

## Explanation

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

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

$$y'(x) = {{ - 1} \over 2}$$
3

### JEE Main 2021 (Online) 26th August Evening Shift

The local maximum value of the function $$f(x) = {\left( {{2 \over x}} \right)^{{x^2}}}$$, x > 0, is
A
$${\left( {2\sqrt e } \right)^{{1 \over e}}}$$
B
$${\left( {{4 \over {\sqrt e }}} \right)^{{e \over 4}}}$$
C
$${(e)^{{2 \over e}}}$$
D
1

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

### JEE Main 2021 (Online) 20th July Morning Shift

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:
A
$$- {{20} \over {27}}$$
B
$${{88} \over {27}}$$
C
$${{20} \over {27}}$$
D
$$- {{88} \over {27}}$$

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

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