1

### JEE Main 2021 (Online) 31st August Evening Shift

Let f be any continuous function on [0, 2] and twice differentiable on (0, 2). If f(0) = 0, f(1) = 1 and f(2) = 2, then
A
f''(x) = 0 for all x $\in$ (0, 2)
B
f''(x) = 0 for some x $\in$ (0, 2)
C
f'(x) = 0 for some x $\in$ [0, 2]
D
f''(x) > 0 for all x $\in$ (0, 2)

## Explanation

f(0) = 0, f(1) = 1 and f(2) = 2

Let h(x) = f(x) $-$ x has three roots

By Rolle's theorem h'(x) = f'(x) $-$ 1 has at least two roots

h''(x) = f''(x) = 0 has at least one roots
2

### JEE Main 2021 (Online) 31st August Evening Shift

If $\alpha = \mathop {\lim }\limits_{x \to {\pi \over 4}} {{{{\tan }^3}x - \tan x} \over {\cos \left( {x + {\pi \over 4}} \right)}}$ and $\beta = \mathop {\lim }\limits_{x \to 0 } {(\cos x)^{\cot x}}$ are the roots of the equation, ax2 + bx $-$ 4 = 0, then the ordered pair (a, b) is :
A
(1, $-$3)
B
($-$1, 3)
C
($-$1, $-$3)
D
(1, 3)

## Explanation

$\alpha = \mathop {\lim }\limits_{x \to {\pi \over 4}} {{{{\tan }^3}x - \tan x} \over {\cos \left( {x + {\pi \over 4}} \right)}};{0 \over 0}$ form

Using L Hospital rule

$\alpha = \mathop {\lim }\limits_{x \to {\pi \over 4}} {{3{{\tan }^2}x{{\sec }^2}x - {{\sec }^2}x} \over { - \sin \left( {x + {\pi \over 4}} \right)}}$

$\alpha$ = $-$4

$\beta = \mathop {\lim }\limits_{x \to 0} {(\cos x)^{\cot x}} = {e^{\mathop {\lim }\limits_{x \to 0} {{(\cos x - 1)} \over {\tan x}}}}$

$\beta = {e^{\mathop {\lim }\limits_{x \to 0} {{ - (1 - \cos x)} \over {{x^2}}}.{{{x^2}} \over {{{\left( {{{\tan x} \over x}} \right)}^x}}}}}$

$\beta = {e^{\mathop {\lim }\limits_{x \to 0} \left( {{{ - 1} \over 2}} \right).{x \over 1}}} = {e^0} \Rightarrow \beta = 1$

$\alpha$ = $-$4; $\beta$ = 1

If ax2 + bx $-$ 4 = 0 are the roots then

16a $-$ 4b $-$ 4 = 0 & a + b $-$ 4 = 0

$\Rightarrow$ a = 1 & b = 3
3

### JEE Main 2021 (Online) 31st August Morning Shift

$\mathop {\lim }\limits_{x \to 0} {{{{\sin }^2}\left( {\pi {{\cos }^4}x} \right)} \over {{x^4}}}$ is equal to :
A
${\pi ^2}$
B
$2{\pi ^2}$
C
$4{\pi ^2}$
D
$4\pi$

## Explanation

$\mathop {\lim }\limits_{x \to 0} {{{{\sin }^2}\left( {\pi {{\cos }^4}x} \right)} \over {{x^4}}}$

= $\mathop {\lim }\limits_{x \to 0} {{1 - \cos \left( {2\pi {{\cos }^4}x} \right)} \over {2{x^4}}}$

= $\mathop {\lim }\limits_{x \to 0} {{1 - \cos \left( {2\pi - 2\pi {{\cos }^4}x} \right)} \over {{{\left[ {2\pi (1 - {{\cos }^4}x)} \right]}^2}}}4{\pi ^2}.{{{{\sin }^4}x} \over {2{x^4}}}{\left( {1 + {{\cos }^2}x} \right)^2}$

$= {1 \over 2}.4{\pi ^2}.{1 \over 2}{(2)^2} = 4{\pi ^2}$
4

### JEE Main 2021 (Online) 31st August Morning Shift

If the function
$f(x) = \left\{ {\matrix{ {{1 \over x}{{\log }_e}\left( {{{1 + {x \over a}} \over {1 - {x \over b}}}} \right)} & , & {x < 0} \cr k & , & {x = 0} \cr {{{{{\cos }^2}x - {{\sin }^2}x - 1} \over {\sqrt {{x^2} + 1} - 1}}} & , & {x > 0} \cr } } \right.$ is continuous

at x = 0, then ${1 \over a} + {1 \over b} + {4 \over k}$ is equal to :
A
$-$5
B
5
C
$-$4
D
4

## Explanation

If f(x) is continuous at x = 0, RHL = LHL = f(0)

$\mathop {\lim }\limits_{x \to {0^ + }} f(x) = \mathop {\lim }\limits_{x \to {0^ + }} {{{{\cos }^2}x - {{\sin }^2}x - 1} \over {\sqrt {{x^2} + 1} - 1}}.{{\sqrt {{x^2} + 1} + 1} \over {\sqrt {{x^2} + 1} + 1}}$ (Rationalisation)

$\mathop {\lim }\limits_{x \to {0^ + }} - {{2{{\sin }^2}x} \over {{x^2}}}.\left( {\sqrt {{x^2} + 1} + 1} \right) = - 4$

$\mathop {\lim }\limits_{x \to {0^ - }} f(x) = \mathop {\lim }\limits_{x \to {0^ - }} {1 \over x}\ln \left( {{{1 + {x \over a}} \over {1 - {x \over b}}}} \right)$

$\mathop {\lim }\limits_{x \to {0^ - }} {{\ln \left( {1{x \over a}} \right)} \over {\left( {{x \over a}} \right).\,a}} + {{\ln \left( {1 - {x \over b}} \right)} \over {\left( { - {x \over b}} \right)\,.\,b}}$$= {1 \over a} + {1 \over b}$

So, ${1 \over a} + {1 \over b} = - 4 = k$

$\Rightarrow {1 \over a} + {1 \over b} + {4 \over k} = - 4 - 1 = - 5$