JEE MATHEMATICS MCQ: CONTINUITY

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Let

f(x)=\left\{\begin{array}{cc}x^{2} \sin \left(\frac{1}{x}\right), & x \neq 0 \\ 0, & x=0\end{array}\right.

Then at

x=0

View answer and explanation

Correct option: D

f

is continuous but

f^{\prime}

is not continuous

Option D is correct because the function f(x) is continuous at x=0, but its derivative f'(x) is not continuous at x=0 due to the oscillatory behavior of the cosine term.

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