MCQ
The function $f(x)=|x|$ is
- Acontinuous and differentiable everywhere.
- Bcontinuous and differentiable nowhere.
- ✓continuous everywhere, but differentiable everywhere except at $x=0$.
- Dcontinuous everywhere, but differentiable nowhere.
✓
Answer
Correct option: C.
continuous everywhere, but differentiable everywhere except at $x=0$.
$f(x)=|x|$ = $\left\{\begin{aligned} x, & x>0 \\ -x, & x<0\end{aligned}\right.$
The function $f(x)$ is continuous everywhere but not differentiable at $x=0$ as at $x=0$
$L f^{\prime}(0)=\lim _{x \rightarrow 0^{-}} \frac{f(x)-f(0)}{x-0}$
$=\lim _{x \rightarrow 0} \frac{-x-0}{x}=-1$
$R f^{\prime}(0)=\lim _{x \rightarrow 0^{+}} \frac{f(x)-f(0)}{x-0}$
$=\lim _{x \rightarrow 0} \frac{x-0}{x}=1$
$\therefore L f^{\prime}(0) \neq R f^{\prime}(0),$
so $f(x)$ is not differentiable at $x=0$.
The function $f(x)$ is continuous everywhere but not differentiable at $x=0$ as at $x=0$
$L f^{\prime}(0)=\lim _{x \rightarrow 0^{-}} \frac{f(x)-f(0)}{x-0}$
$=\lim _{x \rightarrow 0} \frac{-x-0}{x}=-1$
$R f^{\prime}(0)=\lim _{x \rightarrow 0^{+}} \frac{f(x)-f(0)}{x-0}$
$=\lim _{x \rightarrow 0} \frac{x-0}{x}=1$
$\therefore L f^{\prime}(0) \neq R f^{\prime}(0),$
so $f(x)$ is not differentiable at $x=0$.
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
The area between the curve $y^2 (a + x) = (a - x)^3$ and its vertical asymptote is
→Let $f(x) = tan^{-1}\, (cot\, x - 2\, cot2x)$ then $\left[ {\sum\limits_{r = 1}^7 {f\left( r \right)} } \right]$ is equal to (where [.] represent greatest integer function)
→$\tan ^{-1}\left(\frac{1+\sqrt{3}}{3+\sqrt{3}}\right)+\sec ^{-1}\left(\sqrt{\frac{8+4 \sqrt{3}}{6+3 \sqrt{3}}}\right)$ is equal to $.........$
→Consider $f(x) = \frac{{{x^2}}}{{1 + {x^3}}}$ ; $g(t) = \int {f(t)\,\,dt}$ . If $g(1) = 0$ then $g(x)$ equals
→The minimum number of elements that must be added to the relation $R =\{( a , b ),( b , c )$, (b, d) $\}$ on the set $\{a, b, c, d\}$ so that it is an equivalence relation, is $.........$
→If $ a$ and $ b $ are two non-collinear vectors and $x\,a + y\,b = 0$
→If $f(x) = \left\{ \begin{array}{l}x\frac{{{e^{(1/x)}} - {e^{( - 1/x)}}}}{{{e^{(1/x)}} + {e^{( - 1/x)}}}},\,x \ne 0\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0,\,x = 0\end{array} \right.$ then which of the following is true
→Choose the correct answer in Exercise$:\ $The value of the integral $\int^{1}_{\frac{1}{3}}\frac{(\text{x}-\text{x}^{3})^{\frac{1}{3}}}{\text{x}^{4}}\text{dx}\ \text{is}$
→The sum and product of the mean and variance of a binomial distribution are $82.5$ and $1350$ respectively. They the number of trials in the binomial distribution is.
→$\int\frac{\text{x+sin x}}{1+\text{cos x}}\ dx$ is equal to:
→