The function $ f(x)= \begin{cases}\frac{1-\cos 4 x}{x^2} & \text { if } x 0\end{cases} $ is continuous at $x=0$ for

JEE MATHEMATICS MCQ: CONTINUITY

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The function

f(x)= \begin{cases}\frac{1-\cos 4 x}{x^2} & \text { if } x < 0 \\ \frac{2 \sqrt{x}}{\sqrt{16+\sqrt{x}}-4} & \text { if } x=0 \\ \text { if } x > 0\end{cases}

is continuous at

x=0

for

View answer and explanation

Correct option: D

no value of

A

Option D (no value of

A

) is correct because the left-hand limit and right-hand limit at x=0 are not equal, so the function cannot be continuous at x=0 for any value of A.

Practice All Chapter MCQs →

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.

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.

\{ \}

f:R \to R

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.

f

x = 0

\lambda

f

f(x) f(y)=

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

x, y \in {R}

f(2)=5

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

\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.

x=-5

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

x = 0

k