EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 26 - Differential Coefficient>JEE Main - 4 September 2020 Shift 1>Q 5

MEDIUM

JEE Main

IMPORTANT

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If $(a + \sqrt{2} b \cos x) (a - \sqrt{2} b \cos y) = a^{2} - b^{2}$ , where $a > b > 0,$ then $\frac{d x}{d y}$ at $(\frac{π}{4}, \frac{π}{4})$ is:

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Important Questions on Differential Coefficient

MEDIUM

JEE Main

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 26 - Differential Coefficient>JEE Main - 4 September 2020 Shift 1>Q 6

Suppose a differentiable function

f (x)

satisfies the identity

f (x + y) = f (x) + f (y) + x y^{2} + x^{2} y,

for all real

x

and

y

. If

\lim_{x \to 0} \frac{f (x)}{x} = 1,

then

f^{'} (3)

is equal to :

HARD

JEE Main

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 26 - Differential Coefficient>JEE Main - 5 September 2020 Shift 2>Q 7

The derivative of

\tan^{- 1} (\frac{\sqrt{1 + x^{2}} - 1}{x})

with respect to

\tan^{- 1} (\frac{2 x \sqrt{1 - x^{2}}}{1 - 2 x^{2}})

at

x = \frac{1}{2}

is :

EASY

JEE Main

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 26 - Differential Coefficient>JEE Main - 6 September 2020 Shift 2>Q 8

If

y = (\frac{2}{π} x - 1) cosec x

is the solution of the differential equation,

\frac{d y}{d x} + p (x) y = - \frac{2}{π} cosec x, 0 < x < \frac{π}{2}

, then the function

p (x)

is equal to :

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JEE Main

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 26 - Differential Coefficient>JEE Main - 7th January 2020 Shift 1>Q 9

If

y (α) = \sqrt{2 (\frac{t a n α + c o t α}{1 + t a n^{2} α}) + \frac{1}{s i n^{2} α}}, α \in (\frac{3 π}{4}, π)

, then

\frac{d y}{d α}

at

α = \frac{5 π}{6}

is

MEDIUM

JEE Main

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 26 - Differential Coefficient>JEE Main - 7th January 2020 Shift 1>Q 10

Let

x^{k} + y^{k} = a^{k}, (a, k > 0)

and

\frac{d y}{d x} + {(\frac{y}{x})}^{\frac{1}{3}} = 0,

then

k

is

MEDIUM

JEE Main

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 26 - Differential Coefficient>JEE Main - 7th January 2020 Shift 2>Q 11

Let

y = y (x)

be a function of

x

satisfying

y \sqrt{1 - x^{2}} = k - x \sqrt{1 - y^{2}}

, where

k

is a constant and

y (\frac{1}{2}) = - \frac{1}{4}

. Then

\frac{d y}{d x}

at

x = \frac{1}{2}

, is equal to

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JEE Main

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 26 - Differential Coefficient>JEE Main - 8th January 2020 Shift 1>Q 12

Let

f (x) = {(\sin (\tan^{- 1} x) + \sin (\cot^{- 1} x))}^{2} - 1

,

|x| > 1

. If

\frac{d y}{d x} = \frac{1}{2} \frac{d}{d x} (\sin^{- 1} (f (x)))

and

y (\sqrt{3}) = \frac{π}{6},

then

y (- \sqrt{3})

is equal to:

MEDIUM

JEE Main

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 26 - Differential Coefficient>JEE Main - 9th January 2020 Shift 2>Q 13

If

x = 2 \sin θ - \sin 2 θ

and

y = 2 \cos θ - \cos 2 θ,

θ \in [0, 2 π],

then

\frac{d^{2} y}{d x^{2}}

at

θ = π

is: