At the point x = 1 , the function f x = x 3 - 1 ; 1 < x < ∞ x - 1 ; – ∞ < x ≤ 1 is

continuous and not differentiable

discontinuous and differentiable

discontinuous and not differentiable

At the point x = 1 , the function f x = x 3 - 1 ; 1 < x < ∞ x - 1 ; – ∞ < x ≤ 1 is

continuous and not differentiable

For x ∈ R , f x = log ⁡ 2 - sin ⁡ x and g x = f f x , then

g is differentiable at x = 0 and g ' 0 = - sin ⁡ log ⁡ 2

g is not differentiable at x = 0

The function f x = π 4 + tan - 1 x , x ≤ 1 1 2 x - 1 , x > 1 is :

continuous on R - 1 and differentiable on R - - 1 , 1 .

both continuous and differentiable on R - 1

continuous on R - - 1 and differentiable on R - - 1 , 1

both continuous and differentiable on R - - 1

If f x = x e - 1 x + 1 x ; i f x ≠ 0 0 ; i f x = 0 then which of the following is correct?

f x is continuous and f ' 0 does not exist

f x is continuous and f ′ 0 also exists

Let f : R → R be a function such that f x ≤ x 2 , for all x ∈ R . Then, at x = 0 , f is

continuous as well as differentiable

differentiable but not continuous

neither continuous nor differentiable

continuous but not differentiable

Let f : R → R be differentiable at c ∈ R and f c = 0 . If g x = f x , then at x = c , g is:

not differentiable if f ' c = 0

The function f x = e x is

continuous everywhere but not differentiable at x = 0

continuous and differentiable everywhere

EASY

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At the point $x = 1$ , the function $f (x) = \{\begin{cases} x^{3} - 1; & 1 < x < \infty \\ x - 1; & - \infty < x \leq 1 \end{cases}$ is

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Important Questions on Continuity and Differentiability

HARD

Mathematics>Calculus>Continuity and Differentiability>Differentiability of a Function

Let

f (x) = 15 - |x - 10|; x \in R .

Then the set of all values of

x

, at which the function

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

is not differentiable, is:

HARD

Mathematics>Calculus>Continuity and Differentiability>Differentiability of a Function

For

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

and

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

, then

MEDIUM

Mathematics>Calculus>Continuity and Differentiability>Differentiability of a Function

2 y = {(\cot^{- 1} (\frac{\sqrt{3} \cos x + \sin x}{\cos x - \sqrt{3} \sin x}))}^{2} \forall x \in (0, \frac{π}{2})

, then

\frac{d y}{d x}

is equal to

HARD

Mathematics>Calculus>Continuity and Differentiability>Differentiability of a Function

The function

f (x) = \{\begin{cases} \frac{π}{4} + \tan^{- 1} x, & |x| \leq 1 \\ \frac{1}{2} (|x| - 1), & |x| > 1 \end{cases}

is :

MEDIUM

Mathematics>Calculus>Continuity and Differentiability>Differentiability of a Function

Let

f : (- 1, 1) \to R

be a function defined by

f (x) = m a x \{- |x|, - \sqrt{1 - x^{2}}\} .

K

be the set of all points at which

f

is not differentiable, then

K

has exactly

HARD

Mathematics>Calculus>Continuity and Differentiability>Differentiability of a Function

f (x) = [x]

, where

[x]

is the greatest integer not greater than

x

, then

f^{'} (1^{+})

= ….

HARD

Mathematics>Calculus>Continuity and Differentiability>Differentiability of a Function

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

MEDIUM

Mathematics>Calculus>Continuity and Differentiability>Differentiability of a Function

Let a function,

f : (- 1, 3) \to R

be defined as

f (x) = \min {x [x], | x [x] - 2 | + 2},

where

[x]

denotes the greatest integer

\leq x .

Then

f

is :

MEDIUM

Mathematics>Calculus>Continuity and Differentiability>Differentiability of a Function

f (x) = \{\begin{matrix} x e^{- (\frac{1}{|x|} + \frac{1}{x})}; & i f & x \neq 0 \\ 0; & i f & x = 0 \end{matrix}

then which of the following is correct?

HARD

Mathematics>Calculus>Continuity and Differentiability>Differentiability of a Function

Let

f : R \to R

be a function such that

|f (x)| \leq x^{2},

for all

x \in R .

Then, at

x = 0,

f

MEDIUM

Mathematics>Calculus>Continuity and Differentiability>Differentiability of a Function

Let

f : R \to R

be a function defined by

f (x) = \max \{x, x^{2}\}

. Let

S

denotes the set of all points in

R

, where

f

is not differentiable. Then

MEDIUM

Mathematics>Calculus>Continuity and Differentiability>Differentiability of a Function

Let

f (x) = \{\begin{matrix} m a x (|x|, x^{2}), & |x| \leq 2 \\ 8 - 2 |x|, & 2 < |x| \leq 4 \end{matrix} .

Let

S

be the set of points in the interval

(- 4, 4)

at which

f

is not differentiable. Then

S

HARD

Mathematics>Calculus>Continuity and Differentiability>Differentiability of a Function

Let

f : R \to R

be defined as

f (x) = \{\begin{matrix} x^{5} \sin (\frac{1}{x}) + 5 x^{2} & , & x < 0 \\ 0 & , & x = 0 \\ x^{5} \cos (\frac{1}{x}) + λ x^{2} & , & x > 0 \end{matrix}

. The value of

λ

for which

f^{"} (0)

exists, is___.

MEDIUM

Mathematics>Calculus>Continuity and Differentiability>Differentiability of a Function

Let

f : R \to R

be differentiable at

c \in R

and

f (c) = 0

. If

g (x) = |f (x)|

, then at

x = c, g

is:

HARD

Mathematics>Calculus>Continuity and Differentiability>Differentiability of a Function

The function

f (x) = e^{|x|}

MEDIUM

Mathematics>Calculus>Continuity and Differentiability>Differentiability of a Function

f (x) = \{\begin{array}{cl} e^{x}, & x \leq 0 \\ |1 - x|, & x > 0 \end{array}

, then

EASY

Mathematics>Calculus>Continuity and Differentiability>Differentiability of a Function

At how many points between the interval

(- \infty, \infty)

, is the function

f (x) = \sin x

not differentiable

HARD

Mathematics>Calculus>Continuity and Differentiability>Differentiability of a Function

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:

EASY

Mathematics>Calculus>Continuity and Differentiability>Differentiability of a Function

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

Mathematics>Calculus>Continuity and Differentiability>Differentiability of a Function

If the function

g (x) = {\begin{matrix} k \sqrt{x + 1}, & 0 \leq x \leq 3 \\ m x + 2, & 3 < x \leq 5 \end{matrix}

is differentiable, then the value of

k + m

At the point x=1, the function fx=x3-1;1<x<∞x-1;–∞<x≤1 is

Important Questions on Continuity and Differentiability

Learn and Prepare for any exam you want !

At the point $x = 1$ , the function $f (x) = \{\begin{cases} x^{3} - 1; & 1 < x < \infty \\ x - 1; & - \infty < x \leq 1 \end{cases}$ is