1
MCQ (Single Correct Answer)

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
MCQ (Single Correct Answer)

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
MCQ (Single Correct Answer)

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
MCQ (Single Correct Answer)

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

Questions Asked from Differentiation

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