The function f(x)= cc x^2 & for x<1 2-x & for x 1 . is - Vidyadip

Question

The function $f(x)=\left\{\begin{array}{cc}x^2 & \text { for } x<1 \\ 2-x & \text { for } x \geq 1\end{array}\right.$ is

✓

Answer

At $x=1 \lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1} x^2=1$
And $\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1} 2-x=1$
Also, $f(1)=2-1=1 \because \lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{+}} f(x)=f(x)$
$\therefore f(x)$ is continuous at $x=1$
Now, L.H.D. $=\lim _{x \rightarrow 1^{-}} \frac{f(x)-f(1)}{x-1}=\lim _{x \rightarrow 1} \frac{x^2-1}{x-1}=\lim _{x \rightarrow 1}(x+1)=2$
R.H.D. $=\lim _{x \rightarrow 1^{+}} \frac{f(x)-f(1)}{x-1}=\lim _{x \rightarrow 1} \frac{(2-x)-1}{x-1}=-1$
$\because \quad$ L.H.D. $\neq$ R.H.D. $\therefore f(x)$ is not differentiable at $x=1$.

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "M.C.Q (1 Marks)" from Continuity and Differentiability→Continuity and Differentiability — all question types→MATHS STD 12 Science question papers→Rajasthan Board STD 12 Science question papers→Rajasthan Board question paper generator→

Similar questions

What is the value of $\int_{0}^{\frac{\pi}{2}}\frac{\sin\text{x}-\cos\text{x}}{1+\sin\text{x}\cos\text{x}}\text{dx}?$

$1$
$\frac{\pi}{2}$
$0$
$-\frac{\pi}{2}$

The area of the region bounded by the parabola $y = x^2$ and $y = |x|$ is$:$

Find the cofactor of element -3 in the determinant $\triangle=\begin{bmatrix}1&4&4\\-3&5&9\\2&1&2\end{bmatrix}$ is:

-4
4
-5
-3

If OACB is a parallelogram with $\overrightarrow{\text{OC}}=\vec{\text{a}}$ and $\overrightarrow{\text{AB}}=\vec{\text{b}}$, then $\overrightarrow{\text{OA}}=$

$\big(\vec{\text{a}}+\vec{\text{b}}\big)$
$\big(\vec{\text{a}}-\vec{\text{b}}\big)$
$\frac{1}2\big(\vec{\text{b}}-\vec{\text{a}}\big)$
$\frac{1}2\big(\vec{\text{a}}-\vec{\text{b}}\big)$

The scalar product of two nonzero vectors $\vec{a}$ and $\vec{b}$ is defined as

$\int_{-\pi}^\pi \sin ^5 x d x=?$

$\int\frac{\text{x}^3}{\sqrt{1+\text{x}^2}}\text{ dx}=\text{a}(1+\text{x}^2)^{\frac{3}{2}}+\text{b}\sqrt{1+\text{x}^2}+\text{C},$ then:

$\text{a}=\frac{1}{3},\text{ b}=1$
$\text{a}=-\frac{1}{3},\text{ b}=1$
$\text{a}=-\frac{1}{3},\text{ b}=-1$
$\text{a}=\frac{1}{3},\text{ b}=-1$

$\int\limits^\pi_0\frac{1}{\text{a}+\text{b}\cos\text{x}}\text{ dx}=$

$\frac{\pi}{\sqrt{\text{a}^2-\text{b}^2}}$
$\frac{\pi}{\text{ab}}$
$\frac{\pi}{\text{a}^2-\text{b}^2}$
$({\text{a}+\text{b}})\pi$

If $\text{y}=\frac{2}{\sqrt{\text{a}^2-\text{b}^2}}\tan^{-1}\Big(\frac{\text{a}-\text{b}}{\text{a}+\text{b}}\tan\frac{\text{x}}{2}\Big),\text{a}>\text{b}>0,$then:

$\text{y}_1=\frac{-1}{\text{a}+\text{b}\cos\text{x}}$
$\text{y}_2=\frac{\text{b}\sin\text{x}}{(\text{a}+\text{b}\cos\text{x})^2}$
$\text{y}_1=\frac{1}{\text{a}-\text{b}\cos\text{x}}$
$\text{y}_2=\frac{-\text{b}\sin\text{x}}{(\text{a}-\text{b}\cos\text{x})^2}$

$\int\frac{\text{x}^9}{(4\text{x}^2+1)^6}\text{dx}$ is equal to:

$\frac{1}{5\text{x}}\Big(4+\frac{1}{\text{x}^2}\Big)^{-5}+\text{c}$
$\frac{1}{5}\Big(4+\frac{1}{\text{x}^2}\Big)^{-5}+\text{c}$
$\frac{1}{10\text{x}}\Big(\frac{1}{\text{x}}+4\Big)^{-5}+\text{c}$
$\frac{1}{10\text{x}}\Big(\frac{1}{\text{x}^2}+4\Big)^{-5}+\text{c}$