Download App

STD 12 - 5. continuity and differentiation — Mathematics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceMathematicsSTD 12 - 5. continuity and differentiation1 Mark
MCQ
At the point $x = 1$, the given function $f(x) = \left\{ \begin{array}{l}{x^3} - 1;\,\,1 < x < \infty \\x - 1;\,\, - \infty < x \le 1\end{array} \right.$ is
  • A
    Continuous and differentiable
  • ✓
    Continuous and not differentiable
  • C
    Discontinuous and differentiable
  • D
    Discontinuous and not differentiable
✓

Answer

Correct option: B.
Continuous and not differentiable
b
(b) We have $Rf'(1) = \mathop {\lim }\limits_{h \to 0} \,\,\frac{{f(1 + h) - f(1)}}{h}$

$ = \mathop {\lim }\limits_{h \to 0} \,\,\frac{{\left\{ {{{(1 + h)}^3} - 1} \right\} - 0}}{h} = 3$

$Lf'(1) = \mathop {\lim }\limits_{h \to 0} \,\,\frac{{f(1 - h) - f(1)}}{{ - h}}$

$ = \mathop {\lim }\limits_{h \to 0} \,\,\frac{{\left\{ {(1 - h) - 1} \right\} - 0}}{{ - h}} = 1$

$\therefore \,\,\,Rf'(1) \ne Lf'(1)$ $ \Rightarrow \,\,f(x)$ is not differentiable at $x = 1.$

Now, $f(1 + 0) = \mathop {\lim }\limits_{h \to 0} \,f(1 + h) = 0$

and $f(1 - 0) = \mathop {\lim }\limits_{h \to 0} \,f(1 - h) = 0$

$\therefore \,\,\,f(1 + 0) = f(1 - 0) = f(0)$

$ \Rightarrow \,\,f(x)$ is continuous at $x = 1.$

Hence at $x = 1,\,\,\,f(x)$ is continuous and not differentiable.

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 STD 12 - 5. continuity and differentiation→STD 12 - 5. continuity and differentiation — all question types→Mathematics STD 12 Science question papers→CBSE Board STD 12 Science question papers→CBSE Board question paper generator→

Similar questions

If $A^5 = 0$ Such that $\text{A}^{\text{n}}\neq\text{I for }1\leq\text{n}\leq4,\text{ then}(\text{I}-\text{A})^{-1}$ equals:
→
The corner points of the feasible region determined by the system of linear constraints are (0, 10),(5, 5),(15, 15),(0, 20). Let z = px + qy where p, q > 0. Condition on p and q so that the maximum of z occurs at both the points (15, 15) and (0, 20) is __________:
→
If $f(x) = {(a - {x^n})^{1/n}},$ where $a > 0$ and $n$ is a positive integer, then $f[f(x)] = $
→
Let $y=y(x)$ satisfies the equation $\frac{d y}{d x}-|A|=0$, for all $x>0$, where $A=\left[\begin{array}{ccc}y & \sin x & 1 \\ 0 & -1 & 1 \\ 2 & 0 & \frac{1}{x}\end{array}\right] .$

If $y(\pi)=\pi+2$, then the value of $y\left(\frac{\pi}{2}\right)$ is:

→
$\int \limits_{\frac{3 \sqrt{2}}{4}}^{\frac{3 \sqrt{3}}{4}} \frac{48}{\sqrt{9-4 x^2}} d x$ is equal to
→
If the law of motion in a straight line is $s = {1 \over 2}v\,t,$ then acceleration is
→
If $\text{x}=\text{a}\cos^3\theta,\text{y}=\text{a}\sin^3,$ then $\sqrt{1+\Big(\frac{\text{dy}}{\text{dx}}\Big)^2}=$
→
The solution of the differential equation $(1 + {y^2}) + \left( {x - {e^{{{\tan }^{ - 1}}y}}} \right)\frac{{dy}}{{dx}} = 0,$ is
→
Maximum area of a parallelogram inscirbed in a triangle $ABC$ is (given $A$ is also one of the vertex of parallelogram) equal to

(where symbols have their usual meaning in $\Delta\,ABC$ )

→
The solution of ${e^{dy/dx}} = (x + 1)$, $y(0) = 3$ is
→