Search any question & find its solution
Question:
Answered & Verified by Expert
At $x=1$, the function $f(x)=\left\{\begin{array}{cc}x^{3}-1, & 1 < x < \infty \\ x-1, & -\infty < x \leq 1\end{array}\right.$ is
Options:
Solution:
2165 Upvotes
Verified Answer
The correct answer is:
continuous and non-differentiable.
$f(x)=\left\{\begin{array}{cc}x^{3}-1, & 1 < x < \infty \\ x-1, & -\infty < x \leq 1\end{array}\right.$
We have to check the continuity at $x=1$. $\mathrm{RHL} \Rightarrow \lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1^{+}}\left(x^{3}-1\right)=1-1=0$ $\mathrm{LHL} \Rightarrow \lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{-}}(x-1)=1-1=0$ $f(\mathrm{l})=1-1=0$
Thus the function is continuous at $x=1$.
$f^{\prime}(x)=\left\{\begin{array}{cc}
3 x^{2}, & 1 < x < \infty \\
1, & -\infty < x \leq 1
\end{array}\right.$
Now, check the differentiability at $x=1$.
LHD at $x=1 \Rightarrow 1$
RHD at $x=1 \Rightarrow 3(1)^{2}=3$
$A s, L H D \neq R H D$, function is not differentiable.
We have to check the continuity at $x=1$. $\mathrm{RHL} \Rightarrow \lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1^{+}}\left(x^{3}-1\right)=1-1=0$ $\mathrm{LHL} \Rightarrow \lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{-}}(x-1)=1-1=0$ $f(\mathrm{l})=1-1=0$
Thus the function is continuous at $x=1$.
$f^{\prime}(x)=\left\{\begin{array}{cc}
3 x^{2}, & 1 < x < \infty \\
1, & -\infty < x \leq 1
\end{array}\right.$
Now, check the differentiability at $x=1$.
LHD at $x=1 \Rightarrow 1$
RHD at $x=1 \Rightarrow 3(1)^{2}=3$
$A s, L H D \neq R H D$, function is not differentiable.
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.