JEE MATHEMATICS MCQ: CONTINUITY

1 Easy

1 of 1

Analyzing +4 | -1

Let

f:[1,3] \rightarrow R

be a function satisfying

\frac{x}{[x]} \leq f(x) \leq \sqrt{6-x}

, for all

x \neq 2

and

f(2)=1

, where

R

is the set of all real numbers and

[x]

denotes the largest integer less than or equal to

x

.

Statement 1:

\lim \limits_{x \rightarrow 2^{-}} f(x)

exists.
Statement 2:

f

is continuous at

x=2

View answer and explanation

Correct option: D

Statement 1 is true, Statement 2 is false.

Option D (Statement 1 is true, Statement 2 is false) is correct because the left-hand limit exists at x=2, but the function is not continuous at x=2 since the right-hand limit does not exist.

More Questions from CONTINUITY

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If the function

f\left( x \right) = \left\{ {\begin{array}{cccccccccccccccccccc} {{{\left\{ {1 + \left| {\sin x} \right|} \right\}}^{\frac{a}{{\left| {\sin x} \right|}}}}, - \frac{\pi }{6} < x < 0}\\ {b,x = 0}\\ {{e^{\frac{{\tan 2x}}{{\tan 3x}}}},0 < x < \frac{\pi }{6}} \end{array}} \right.

Is continuous at

x = 0

f(x)=\left\{\begin{array}{rr}1+\cos ^{-1}\{\cot x\}, & x<\frac{\pi}{2} \\ \pi[x]+1, & x \geq \frac{\pi}{2}\end{array}\right.

Where [] denotes greatest integer and

\{ \}

denotes fractional part function, Then value of jump of discontinuity is: If

f:R \to R

is defined by

f\left( x \right) = \left\{ {\begin{array}{cccccccccccccccccccc} {\frac{{\cos 3x - \cos x}}{{{x^2}}},\mathrm{for}\,\,x \ne 0}\\ {\lambda ,\mathrm{for}\,\,x = 0} \end{array}} \right.

and if

f

is continuous at

x = 0

, then

\lambda

is equal to Let

f

be a continuous function satisfying

f(x) f(y)=

f(x)+f(y)+f(x y)-2

for all

x, y \in {R}

and

f(2)=5

then

\lim \limits_{{x \rightarrow 4}} f(x)

is If

\mathrm{f}(\mathrm{x})=\left\{\begin{array}{c}\frac{x^2+3 x-10}{x^2+2 x-15} \text {, when } x \neq-5 \\ a, \quad \text { when } x=-5\end{array}\right.

is continuous at

x=-5

, then the value of 'a' will be If

f\left( x \right) = \left\{ {\begin{array}{cccccccccccccccccccc} {x\sin \frac{1}{x},x \ne 0}\\ {k,x = 0} \end{array}} \right.

is continuous at

x = 0

, then the value of

k