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1

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

Explanation

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

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

$$y'(x) = {{ - 1} \over 2}$$
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
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 = $${\left( {{2 \over {2/\sqrt e }}} \right)^{{4 \over e}}} = {e^{{2 \over e}}}$$
3

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

JEE Main 2021 (Online) 25th February Morning Slot

MCQ (Single Correct Answer)
If Rolle's theorem holds for the function $$f(x) = {x^3} - a{x^2} + bx - 4$$, $$x \in [1,2]$$ with $$f'\left( {{4 \over 3}} \right) = 0$$, then ordered pair (a, b) is equal to :
A
($$-$$5, $$-$$8)
B
(5, $$-$$8)
C
($$-$$5, 8)
D
(5, 8)

Explanation

$$f(1) = f(2)$$

$$ \Rightarrow 1 - a + b - 4 = 8 - 4a + 2b - 4$$

$$3a - b = 7$$ ..... (1)

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

$$ \Rightarrow f'\left( {{4 \over 3}} \right) = 0 \Rightarrow 3 \times {{16} \over 9} - {8 \over 3}a + b = 0$$

$$ \Rightarrow - 8a + 3b = - 16$$ ..... (2)

$$ \therefore $$ $$a = 5,b = 8$$

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