Consider the function f ( x ) = x 3 4 - sin π x + 3

f ( x ) takes on the value 3 1 4 in the interval [ - 2 , 2 ]

f ( x ) does not attain value within the interval [ - 2 , 2 ]

f ( x ) takes no value p , 1 < p < 5 in the interval [ - 2 , 2 ]

Discuss the continuity of the function f x = log ⁡ ( 2 + x ) − log ⁡ ( 2 − x ) tan ⁡ x , for x ≠ 0 1 , for x = 0 at the point x = 0 .

We have, f x = log ⁡ ( 2 + x ) − log ⁡ ( 2 − x ) tan ⁡ x , for x ≠ 0 1 , for x = 0 i.e. f 0 = 1 . . . 1 Also, L = lim x → 0 f ( x ) = lim x → 0 f ( x ) = lim x → 0 log ⁡ ( 2 + x ) − log ⁡ ( 2 − x ) tan ⁡ x = lim x → 0 log ⁡ 2 + x 2 − x tan ⁡ x ∵ log ⁡ m − log ⁡ n = log ⁡ m n = lim x → 0 log ⁡ 1 + x 2 1 − x 2 tan ⁡ x = lim x → 0 log ⁡ 1 + x 2 1 − x 2 x × lim x → 0 x tan ⁡ x = lim x → 0 log ⁡ 1 + x 2 x − lim x → 0 log ⁡ 1 − x 2 x × lim x → 0 x tan ⁡ x = 1 2 − − 1 2 × 1 ∵ lim x → 0 log 1 + x x = 1 , lim x → 0 tan x x = 1 = 1 . . . 2 From 1 and 2 , we have lim x → 0 f ( x ) = f ( 0 ) Thus, the given function is continuous at x = 0 .

If f ( x ) = 1 − 3 tan ⁡ x π − 6 x , for x ≠ π 6 is continuous at x = π 6 , find f π 6 .

It is given that f ( x ) = 1 − 3 tan ⁡ x π − 6 x is continuous at x = π 6 . i.e. lim x → π 6 f ( x ) = lim x → π 6 1 − 3 tan ⁡ x π − 6 x ⇒ lim x → π 6 f ( x ) = 1 − 3 tan ⁡ π 6 π − 6 π 6 ⇒ lim x → π 6 f ( x ) = 1 − 1 π − π = 0 0 form So, we will apply L'Hospital's rule because we are getting indeterminate form. ⇒ lim x → π 6 f ( x ) = lim x → π 6 0 − 3 · sec 2 x - 6 ⇒ lim x → π 6 f ( x ) = lim x → π 6 3 · sec 2 x 6 ⇒ lim x → π 6 f ( x ) = 3 · sec 2 π 6 6 ⇒ lim x → π 6 f ( x ) = 3 6 · 2 3 2 Thus, f π 6 = 2 3 3

Let g : R → R be a differentiable function with g 0 = 0 , g ' 0 = 0 and g ' 1 ≠ 0 . Let f x = x | x | g x , x ≠ 0 0 , x = 0 and h x = e | x | for all x ∈ R . Let f o h ( x ) denote f h x & ( h o f ) ( x ) denote h f x .Then which of the following is (are) true?

h o f is differentiable at x = 0

f o h is differentiable at x = 0

Let f : R → R be a continuous function. Then, f is surjective if

lim x → ∞ f x = - ∞ and lim x → - ∞ f x = ∞

lim x → ∞ f x = ∞ and lim x → - ∞ f x = ∞

lim x → ∞ f x = 0 and lim x → - ∞ f x = ∞

lim x → ∞ f x = 0 and lim x → ∞ f x = - ∞

MEDIUM

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If $f : ℝ \to ℝ$ be defined by $f (x) = x - [x]$ , where $[x]$ denotes the greatest integer less than or equal to $x .$ Also let $S$ denote the range of $f .$ Which of the following is TRUE?

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Important Questions on Continuity and Differentiability

HARD

Mathematics>Differential Calculus>Continuity and Differentiability>Properties of Continuous Functions

Let

k

be a non - zero real number. If

f (x) = \{\begin{cases} \frac{{(e^{x} - 1)}^{2}}{\sin (\frac{x}{k}) \log (1 + \frac{x}{4})} \begin{matrix} , & x \neq 0 \end{matrix} \\ \begin{matrix} 12, & x = 0 \end{matrix} \end{cases}

is a continuous function at

x = 0

, then the value of

k

EASY

Mathematics>Differential Calculus>Continuity and Differentiability>Properties of Continuous Functions

Consider the function

f (x) = \{\begin{matrix} \frac{x + 5}{x - 2} & i f & x \neq 2 \\ 1 & i f & x = 2 \end{matrix}

. Then

f (f (x))

is discontinuous

HARD

Mathematics>Differential Calculus>Continuity and Differentiability>Properties of Continuous Functions

Let

f : [0, 1] \to [0, 1]

, be a non-constant continuous function different from the identity function. Then

EASY

Mathematics>Differential Calculus>Continuity and Differentiability>Properties of Continuous Functions

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},

MEDIUM

Mathematics>Differential Calculus>Continuity and Differentiability>Properties of Continuous Functions

Consider the function

f (x) = \frac{x^{3}}{4} - \sin π x + 3

EASY

Mathematics>Differential Calculus>Continuity and Differentiability>Properties of Continuous Functions

f (x) = \{\begin{matrix} \log {(\sec^{2} x)}^{\cot^{2} x} & f o r & x \neq 0 \\ K & f o r & x = 0 \end{matrix}

is continuous at

x = 0

then

K

EASY

Mathematics>Differential Calculus>Continuity and Differentiability>Properties of Continuous Functions

Prove that the function defined by

f (x) = \cos (x^{2})

is a continuous function.

HARD

Mathematics>Differential Calculus>Continuity and Differentiability>Properties of Continuous Functions

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

MEDIUM

Mathematics>Differential Calculus>Continuity and Differentiability>Properties of Continuous Functions

Discuss the continuity of the function

f (x) = \{\begin{cases} \frac{\log (2 + x) - \log (2 - x)}{\tan x} & , for x \neq 0 \\ 1 & , for x = 0 \end{cases}

at the point

x = 0

MEDIUM

Mathematics>Differential Calculus>Continuity and Differentiability>Properties of Continuous Functions

f (x) = \frac{1 - \sqrt{3} \tan x}{π - 6 x}, for x \neq \frac{π}{6}

is continuous at

x = \frac{π}{6}

, find

f (\frac{π}{6})

HARD

Mathematics>Differential Calculus>Continuity and Differentiability>Properties of Continuous Functions

Let

f : [0, π] \to R

be defined as

f (x) = \{\begin{matrix} \sin x, & if & x is irrational and x \in [0, π] \\ \tan^{2} x, & if & x is rational and x \in [0, π] \end{matrix}

,

The number of points in

[0, π]

at which the function

f

is continuous is

EASY

Mathematics>Differential Calculus>Continuity and Differentiability>Properties of Continuous Functions

f (x) = 1 - x, for 0 < x \leq 1 = k

, for

x = 0

is continuous at

x = 0

, then

k =

_____

EASY

Mathematics>Differential Calculus>Continuity and Differentiability>Properties of Continuous Functions

If the function

f (x) = \{\begin{matrix} \frac{\sqrt{2 + cos x} - 1}{{(π - x)}^{2}}, & x \neq π \\ k, & x = π \end{matrix}

is continuous at

x = π

, then

k

equals

MEDIUM

Mathematics>Differential Calculus>Continuity and Differentiability>Properties of Continuous Functions

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

Mathematics>Differential Calculus>Continuity and Differentiability>Properties of Continuous Functions

f (x) = \{\begin{cases} \frac{1}{|x|}; & |x| \geq 1 \\ a x^{2} + b; & - 1 < x < 1 \end{cases}

is differentiable

\forall x

, then one of the value of

a

and

b

is-

MEDIUM

Mathematics>Differential Calculus>Continuity and Differentiability>Properties of Continuous Functions

If the function

f

defined on

(\frac{π}{6}, \frac{π}{3})

f (x) = \{\begin{matrix} \frac{\sqrt{2} c o s x - 1}{c o t x - 1}, x \neq \frac{π}{4} \\ k, x = \frac{π}{4} \end{matrix}

is continuous, then

k

is equal to

EASY

Mathematics>Differential Calculus>Continuity and Differentiability>Properties of Continuous Functions

Let

f : [- 1,3] \to R

be defined as

f (x) = \{\begin{matrix} |x| + [x], \\ x + |x|, \\ x + [x], \end{matrix} \begin{matrix} - 1 \leq x < 1 \\ 1 \leq x < 2 \\ 2 \leq x \leq 3, \end{matrix}

Where

[t]

denotes the greatest integer less than or equal to

t

. Then,

f

is discontinuous at:

MEDIUM

Mathematics>Differential Calculus>Continuity and Differentiability>Properties of Continuous Functions

Function $f (x)$ is continuous on its domain $[- 2, 2]$ , where

$f (x) = \{\begin{cases} \frac{\sin a x}{x} + 2, & for - 2 \leq x < 0 \\ 3 x + 5, & for 0 \leq x \leq 1 \\ \sqrt{x^{2} + 8} - b, & for 1 < x \leq 2 \end{cases}$

Find the value of $a + b + 2$ .

HARD

Mathematics>Differential Calculus>Continuity and Differentiability>Properties of Continuous Functions

Let

g : R \to R

be a differentiable function with

g (0) = 0, g^{'} (0) = 0

and

g^{'} (1) \neq 0 .

Let

f (x) = \{\begin{matrix} \frac{x}{| x |} g (x), & x \neq 0 \\ 0, & x = 0 \end{matrix}

and

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

for all

x \in R .

Let

(f o h) (x)

denote

f (h (x)) & (h o f) (x)

denote

h (f (x))

.Then which of the following is (are) true?

EASY

Mathematics>Differential Calculus>Continuity and Differentiability>Properties of Continuous Functions

Let

f : R \to R

be a continuous function. Then,

f

is surjective if

If f:ℝ→ℝ be defined by f(x)=x-[x], where [x] denotes the greatest integer less than or equal to x. Also let S denote the range of f. Which of the following is TRUE?

Important Questions on Continuity and Differentiability

Learn and Prepare for any exam you want !

If $f : ℝ \to ℝ$ be defined by $f (x) = x - [x]$ , where $[x]$ denotes the greatest integer less than or equal to $x .$ Also let $S$ denote the range of $f .$ Which of the following is TRUE?