1

### JEE Main 2019 (Online) 9th January Evening Slot

If   x $=$ 3 tan t and y $=$ 3 sec t, then the value of ${{{d^2}y} \over {d{x^2}}}$ at t $= {\pi \over 4},$ is :
A
${1 \over {3\sqrt 2 }}$
B
${1 \over {6\sqrt 2 }}$
C
${3 \over {2\sqrt 2 }}$
D
${1 \over 6}$

## Explanation

x = 3 tan t and y = 3 sec t

So that ${{dx} \over {dt}}$ = 3sec2t and ${{dy} \over {dt}}$ = 3 sec t tan t

${{dy} \over {dx}}$ = ${{dy/dt} \over {dx/dt}}$ = sin t

${{{d^2}y} \over {d{x^2}}}$ = (cos t)$.{{dt} \over {dx}}$

${{{d^2}y} \over {d{x^2}}} = \left( {\cos t} \right).{1 \over {3{{\sec }^2}t}}$

${{{d^2}y} \over {d{x^2}}}$ = ${1 \over 3}$(cos3 t)

${\left( {{{{d^2}y} \over {d{x^2}}}} \right)_{t = \pi /4}}$ = ${1 \over 3} \times {\left( {{1 \over {\sqrt 2 }}} \right)^3} = {1 \over {6\sqrt 2 }}$
2

### JEE Main 2019 (Online) 12th January Morning Slot

For x > 1, if (2x)2y = 4e2x$-$2y,

then (1 + loge 2x)2 ${{dy} \over {dx}}$ is equal to :
A
${{x\,{{\log }_e}2x - {{\log }_e}2} \over x}$
B
loge 2x
C
x loge 2x
D
${{x\,{{\log }_e}2x + {{\log }_e}2} \over x}$

## Explanation

(2x)2y = 4e2x-2y

2y$\ell$n2x = $\ell$n4 + 2x $-$ 2y

y = ${{x + \ell n2} \over {1 + \ell n2x}}$

y ' = ${{\left( {1 + \ell n2x} \right) - \left( {x + \ell n2} \right){1 \over x}} \over {{{\left( {1 + \ell n2x} \right)}^2}}}$

y '${\left( {1 + \ell n2x} \right)^2} = \left[ {{{x\ell n2x - \ell n2} \over x}} \right]$
3

### JEE Main 2019 (Online) 8th April Morning Slot

If $2y = {\left( {{{\cot }^{ - 1}}\left( {{{\sqrt 3 \cos x + \sin x} \over {\cos x - \sqrt 3 \sin x}}} \right)} \right)^2}$,

x $\in$ $\left( {0,{\pi \over 2}} \right)$ then $dy \over dx$ is equal to:
A
$2x - {\pi \over 3}$
B
${\pi \over 6} - x$
C
${\pi \over 3} - x$
D
$x - {\pi \over 6}$

## Explanation

$2y = {\left( {{{\cot }^{ - 1}}\left( {{{\sqrt 3 \cos x + \sin x} \over {\cos x - \sqrt 3 \sin x}}} \right)} \right)^2}$

$\Rightarrow$ 2y = ${\left( {{{\cot }^{ - 1}}\left( {{{\sqrt 3 + \tan x} \over {1 - \sqrt 3 \tan x}}} \right)} \right)^2}$

$\Rightarrow$ 2y = ${\left( {{{\cot }^{ - 1}}\left( {{{\tan {\pi \over 3} + \tan x} \over {1 - \tan {\pi \over 3}\tan x}}} \right)} \right)^2}$

$\Rightarrow$ 2y = ${\left( {{{\cot }^{ - 1}}\tan \left( {{\pi \over 3} + x} \right)} \right)^2}$

$\Rightarrow$ 2y = ${\left( {{\pi \over 2} - {{\tan }^{ - 1}}\tan \left( {{\pi \over 3} + x} \right)} \right)^2}$

As x $\in$ $\left( {0,{\pi \over 2}} \right)$ then

${{{\tan }^{ - 1}}\tan \left( {{\pi \over 3} + x} \right)}$ = ${\left( {{\pi \over 3} + x} \right)}$

$\Rightarrow$ 2y = ${\left( {{\pi \over 2} - \left( {{\pi \over 3} + x} \right)} \right)^2}$

$\Rightarrow$ 2y = ${\left( {{\pi \over 6} - x} \right)^2}$

$\therefore$ $2{{dy} \over {dx}} = 2\left( {{\pi \over 6} - x} \right)\left( { - 1} \right)$

$\Rightarrow$ ${{dy} \over {dx}} = \left( {x - {\pi \over 6}} \right)$
4

### JEE Main 2019 (Online) 8th April Evening Slot

If ƒ(1) = 1, ƒ'(1) = 3, then the derivative of ƒ(ƒ(ƒ(x))) + (ƒ(x))2 at x = 1 is :
A
33
B
12
C
9
D
15

## Explanation

Given ƒ(1) = 1, ƒ'(1) = 3

Let y = ƒ(ƒ(ƒ(x))) + (ƒ(x))2

On differentiating both sides with respect to x we get,

${{dy} \over {dx}}$ = ƒ'(ƒ(ƒ(x))).ƒ'(ƒ(x)).ƒ'(x) + 2ƒ(x).ƒ'(x)

Now at x = 1,

${{dy} \over {dx}}$ = ƒ'(ƒ(ƒ(1))).ƒ'(ƒ(1)).ƒ'(1) + 2ƒ(1).ƒ'(1)

= ƒ'(ƒ(1)).ƒ'(1).ƒ'(1) + 2.1.ƒ'(1)

= ƒ'(1).ƒ'(1).ƒ'(1) + 2.1.ƒ'(1)

= 3$\times$3$\times$3 + 2$\times$3

= 33

NEET