EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 25 - Continuity and Differentiability>JEE Main - 10th January 2019 Shift 2>Q 55

MEDIUM

JEE Main

IMPORTANT

Earn 100

Let $f : (- 1, 1) \to R$ be a function defined by $f (x) = m a x \{- |x|, - \sqrt{1 - x^{2}}\} .$ If $K$ be the set of all points at which $f$ is not differentiable, then $K$ has exactly

33.33% studentsanswered this correctly

Important Questions on Continuity and Differentiability

EASY

JEE Main

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 25 - Continuity and Differentiability>JEE Main - 12th January 2019 Shift 1>Q 56

Let

S

be the set of all points in

(- π, π)

at which the function,

f (x) = \min \{\sin x, \cos x\}

is not differentiable. Then

S

is a subset of which of the following?

HARD

JEE Main

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 25 - Continuity and Differentiability>JEE Main - 8th April 2018>Q 57

Let

S = \{t \in R : f (x) = |x - π| \cdot (e^{|x|} - 1) \sin |x| is not differentiable at t\} .

Then, the set

S

is equal to:

MEDIUM

JEE Main

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 25 - Continuity and Differentiability>JEE Main - 15th April 2018>Q 58

Let

S = {(λ, μ) \in R \times R : f (t) = (|λ| e^{|t|} - μ) \sin (2 |t|), t \in R

is a differential function}. Then,

S

is a subset of :

MEDIUM

JEE Main

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 25 - Continuity and Differentiability>JEE Main - 16th April 2018>Q 59

If the function

f

defined as

f (x) = \frac{1}{x} - \frac{k - 1}{e^{2 x} - 1}, x \neq 0

is continuous at

x = 0

, then ordered pair

(k, f (0))

is equal to

EASY

JEE Main

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 25 - Continuity and Differentiability>JEE Main - 9th April 2017>Q 60

The value of

k

which the function

f (x) = \{\begin{matrix} {(\frac{4}{5})}^{\frac{\tan 4 x}{\tan 5 x}}, & 0 < x < \frac{π}{2} \\ k + \frac{2}{5}, & x = \frac{π}{2} \end{matrix}

is continuous at

x = \frac{π}{2},

is

HARD

JEE Main

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 25 - Continuity and Differentiability>JEE Main - 9th April 2016>Q 61

If the function

f (x) = \{\begin{matrix} - x, & x < 1 \\ a + \cos^{- 1} (x + b), & 1 \leq x \leq 2 \end{matrix}

is differentiable at

x = 1

, then

\frac{a}{b}

is equal to

HARD

JEE Main

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 25 - Continuity and Differentiability>JEE Main - 3rd April 2016>Q 62

For

x \in R, f (x) = |\log 2 - \sin x|

and

g (x) = f (f (x))

, then

HARD

JEE Main

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 25 - Continuity and Differentiability>JEE Main - 10th April 2016>Q 63

Let

a, b \in R, (a \neq 0) .

If the function

f

, defined as

f (x) = \{\begin{matrix} \frac{2 x^{2}}{a}, 0 \leq x < 1 \\ a, 1 \leq x < \sqrt{2} \\ \frac{2 b^{2} - 4 b}{x^{3}}, \sqrt{2} \leq x < 8 \end{matrix}

,is continuous in the interval

[0, \infty)

, then an ordered pair

(a, b)

can be