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

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

Let a , b ∈ R and f : R → R be defined by f x = a cos ⁡ x 3 - x + b x sin ⁡ x 3 + x . Then f is -

Differentiable at x = 1 if a = 1 and b = 0

NOT differentiable at x = 0 if a = 1 and b = 0

NOT differentiable at x = 1 if a = 1 and b = 1

Prove that the function f ( x ) = x 2 - 1 ; x ≥ 1 1 - x ; x < 1 is not differentiable at x = 1

To check the differentiability of a function at x = 1 , we need to find L H D & R H D Notice that f ( 1 ) = 1 2 - 1 = 0 R H D = lim h → 0 f ( 1 + h ) - f ( 1 ) h ⇒ R H D = lim h → 0 ( 1 + h ) 2 - 1 - 0 h ⇒ R H D = lim h → 0 h 2 + 2 h h ⇒ R H D = lim h → 0 h + 2 = 2 Now, L H D = lim h → 0 f ( 1 ) - f ( 1 - h ) h ⇒ L H D = lim h → 0 0 - ( 1 - ( 1 - h ) ) h ⇒ L H D = lim h → 0 - h h = - 1 So even though, L H D , R H D are finite, but they are not equal. So the function is not differentiable at x = 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 given by f x = x 5 + 5 x 4 + 10 x 3 + 10 x 2 + 3 x + 1 , x < 0 ; x 2 - x + 1 , 0 ≤ x < 1 ; 2 3 x 3 - 4 x 2 + 7 x - 8 3 , 1 ≤ x < 3 ; x - 2 l o g e x - 2 - x + 10 3 , x ≥ 3 Then which of the following options is/are correct?

f ' has a local maximum at x = 1

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

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

Let f : [ a , b ] → [ 1 , ∞ ) be a continuous function and let g : R → R be defined as g x = 0 i f x < 0 ∫ a x f t d t i f a ≤ x ≤ b ∫ a b f t d t i f x > b . Then :

g x is continuous but not differentiable at b

g x is continuous and differentiable at either a or b but not both

EASY

Earn 100

If the Left Hand Derivative and Right Hand Derivative of a function is not equal but finite then the function must be a continuous function.

50% studentsanswered this correctly

Important Questions on Continuity and Differentiability

MEDIUM

Mathematics>Differential Calculus>Continuity and Differentiability>Differentiability

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>Differential Calculus>Continuity and Differentiability>Differentiability

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

EASY

Mathematics>Differential Calculus>Continuity and Differentiability>Differentiability

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>Differential Calculus>Continuity and Differentiability>Differentiability

For

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

and

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

, then

MEDIUM

Mathematics>Differential Calculus>Continuity and Differentiability>Differentiability

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>Differential Calculus>Continuity and Differentiability>Differentiability

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:

HARD

Mathematics>Differential Calculus>Continuity and Differentiability>Differentiability

Let

a, b \in R

and

f : R \to R

be defined by

f (x) = a \cos (|x^{3} - x|) + b |x| \sin (|x^{3} + x|)

. Then

f

is -

HARD

Mathematics>Differential Calculus>Continuity and Differentiability>Differentiability

f (x) = [x]

, where

[x]

is the greatest integer not greater than

x

, then

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

= ….

MEDIUM

Mathematics>Differential Calculus>Continuity and Differentiability>Differentiability

The greatest integer function defined by

f (x) = [x], 0 < x < 2

is not differentiable at

x =

_____

MEDIUM

Mathematics>Differential Calculus>Continuity and Differentiability>Differentiability

Prove that the function $f (x) = \{\begin{cases} x^{2} - 1 & ; x \geq 1 \\ 1 - x & ; x < 1 \end{cases}$ is not differentiable at $x = 1$

HARD

Mathematics>Differential Calculus>Continuity and Differentiability>Differentiability

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>Differential Calculus>Continuity and Differentiability>Differentiability

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>Differential Calculus>Continuity and Differentiability>Differentiability

Let

f : R \to R

be given by

f (x)

= \{\begin{matrix} x^{5} + 5 x^{4} + 10 x^{3} + 10 x^{2} + 3 x + 1, & x < 0; \\ x^{2} - x + 1, & 0 \leq x < 1; \\ \frac{2}{3} x^{3} - 4 x^{2} + 7 x - \frac{8}{3}, & 1 \leq x < 3; \\ (x - 2) {l o g}_{e} (x - 2) - x + \frac{10}{3}, & x \geq 3 \end{matrix}

Then which of the following options is/are correct?

HARD

Mathematics>Differential Calculus>Continuity and Differentiability>Differentiability

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>Differential Calculus>Continuity and Differentiability>Differentiability

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>Differential Calculus>Continuity and Differentiability>Differentiability

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 :

HARD

Mathematics>Differential Calculus>Continuity and Differentiability>Differentiability

Let $f : [a, b] \to [1, \infty)$ be a continuous function and let $g : R \to R$ be defined as

$g (x) = \{\begin{cases} 0 & i f & x < 0 \\ \int_{a}^{x} f (t) d t & i f & a \leq x \leq b \\ \int_{a}^{b} f (t) d t & i f & x > b \end{cases}$ . Then :

MEDIUM

Mathematics>Differential Calculus>Continuity and Differentiability>Differentiability

Prove that function $f (x) = |x|$ is not differentiable at $x = 0$ .

HARD

Mathematics>Differential Calculus>Continuity and Differentiability>Differentiability

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:

MEDIUM

Mathematics>Differential Calculus>Continuity and Differentiability>Differentiability

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

If the Left Hand Derivative and Right Hand Derivative of a function is not equal but finite then the function must be a continuous function.

Important Questions on Continuity and Differentiability

Learn and Prepare for any exam you want !