If f x = x for x ≤ 0 0 for x > 0 then f ( x ) at x = 0 is

Continuous but not differentiable

Not continuous but differentiable

Not continuous and not differentiable

If f x = x - x 4 , x ∈ R , where x denotes the greatest integer function, then:

l i m x → 4 + f x exists but l i m x → 4 - f x does not exist

l i m x → 4 - f x exists but l i m x → 4 + f x does not exist

Both l i m x → 4 - f x and l i m x → 4 + f x exist but are not equal

HARD

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Let $f : R \to R$ be a continuous function such that $f (x) \neq x$ for any $x \in R$ and $f (0) = 1$ . Then

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Important Questions on Continuity

HARD

Mathematics>Differential Calculus>Continuity>Continuous and Discontinuous 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

HARD

Mathematics>Differential Calculus>Continuity>Continuous and Discontinuous 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>Continuous and Discontinuous 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-

EASY

Mathematics>Differential Calculus>Continuity>Continuous and Discontinuous 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:

HARD

Mathematics>Differential Calculus>Continuity>Continuous and Discontinuous Functions

If the function

f (x) = \{\begin{matrix} {[\tan (\frac{π}{4} + x)]}^{\frac{1}{x}} & f o r & x \neq 0 \\ k & f o r & x = 0 \end{matrix}

is continuous at

x = 0,

then

k = ?

HARD

Mathematics>Differential Calculus>Continuity>Continuous and Discontinuous Functions

Let

[x]

be the greatest integer less than or equals to

x

. Then, at which of the following point(s) the function

f (x) = x \cos (π (x + [x]))

is discontinuous?

EASY

Mathematics>Differential Calculus>Continuity>Continuous and Discontinuous 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>Continuous and Discontinuous Functions

If the function

f

defined on

(- \frac{1}{3}, \frac{1}{3})

f (x) = \{\begin{matrix} \frac{1}{x} {l o g}_{e} (\frac{1 + 3 x}{1 - 2 x}), & when x \neq 0 \\ k & , when x = 0 \end{matrix}

, is continuous, then

k

is equal to.

EASY

Mathematics>Differential Calculus>Continuity>Continuous and Discontinuous 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

MEDIUM

Mathematics>Differential Calculus>Continuity>Continuous and Discontinuous Functions

f (x) = \{\begin{matrix} x & for & x \leq 0 \\ 0 & for & x > 0 \end{matrix}

then

f (x)

x = 0

MEDIUM

Mathematics>Differential Calculus>Continuity>Continuous and Discontinuous 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

HARD

Mathematics>Differential Calculus>Continuity>Continuous and Discontinuous 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

MEDIUM

Mathematics>Differential Calculus>Continuity>Continuous and Discontinuous Functions

f (x) = [x] - [\frac{x}{4}], x \in R,

where

[x]

denotes the greatest integer function, then:

HARD

Mathematics>Differential Calculus>Continuity>Continuous and Discontinuous Functions

Let

f : R \to (0, 1)

, be a continuous function then, which of the following function(s), has(have) the values zero at some point in the interval,

(0, 1) ?

EASY

Mathematics>Differential Calculus>Continuity>Continuous and Discontinuous Functions

If the function

f (x) = \{\begin{matrix} a |π - x| + 1, x \leq 5 \\ b |x - π| + 3, x > 5 \end{matrix}

is continuous at

x = 5,

then the value of

a - b

is:

MEDIUM

Mathematics>Differential Calculus>Continuity>Continuous and Discontinuous Functions

If the function

f (x) = \{\begin{matrix} \frac{(e^{k x} - 1) \tan k x}{4 x^{2}}, & x \neq 0 \\ 16, & x = 0 \end{matrix}

is continuous at

x = 0

, then

k

= ….

MEDIUM

Mathematics>Differential Calculus>Continuity>Continuous and Discontinuous Functions

f (x) = \{\begin{matrix} \frac{s i n (p + 1) x + s i n x}{x} & , & x < 0 \\ q & , & x = 0 \\ \frac{\sqrt{x + x^{2}} - \sqrt{x}}{x^{3 / 2}} & , & x > 0 \end{matrix}

is continuous at

x = 0

, then the ordered pair

(p, q)

is equal to:

EASY

Mathematics>Differential Calculus>Continuity>Continuous and Discontinuous 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>Continuous and Discontinuous 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

HARD

Mathematics>Differential Calculus>Continuity>Continuous and Discontinuous Functions

Let

f : [- \frac{1}{2}, 2] \to R

and

g : [- \frac{1}{2}, 2] \to R

be functions defined by

f (x) = [x^{2} - 3]

and

g (x) = |x| f (x) + |4 x - 7| f (x)

, where

[y]

denotes the greatest integer less than or equal to

y

for

y \in R

. Then

Let f:R→R be a continuous function such that f(x)≠x for any x∈R and f(0)=1. Then

Important Questions on Continuity

Learn and Prepare for any exam you want !

Let $f : R \to R$ be a continuous function such that $f (x) \neq x$ for any $x \in R$ and $f (0) = 1$ . Then