Let f :R R be a function defined as f ( x ) = l 5, , , , , , , , … - Vidyadip

MCQ

Let $f :R \to R$ be a function defined as $f\left( x \right) = \left\{ \begin{array}{l}
5,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,if\,\,\,\,\,\,\,x \le 1\,\,\,\,\,\,\,\\
a + bx,\,\,\,\,if\,\,\,\,\,\,1 < x < 3\\
b + 5x,\,\,\,\,if\,\,\,\,\,\,3 \le x < 5\\
30,\,\,\,\,\,\,\,\,\,\,if\,\,\,\,\,\,\,x \ge 5
\end{array} \right.\,\,\,\,$ Then $f$ is

A
continuous if $a = 5$ and $b = 5$
B
continuous if $a = 5$ and $b = 10$
C
continuous if $a = 0$ and $b = 5$
✓
not continuous for any values of $a$ and $b$

✓

Answer

Correct option: D.

not continuous for any values of $a$ and $b$

d
For $x=1$

$R.H.L=a+b$

$L.H.L=5$

So to be continuous at $x=1$

$a+b=5$ ..........$(i)$

for $x=3$

$R.H.L=b+15$

$L.H.L=a+3b$

$b+15=a+3b$

$a+2b=15$ ........$(ii)$

for $x=5$

$R.H.L=30$

$L.H.L=b+25$

$b+25=30$

$b=5$.

From equation $(ii)$

$a=10$

but $a=10$ and $b=5$ does not satisfied equation $(i)$

So $f(x)$ is discontinuous for $a \in R$ and $b \in R$

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 5. continuity and differentiation→5. continuity and differentiation — all question types→Maths STD 12 Science question papers→Gujarat Board STD 12 Science question papers→Gujarat Board question paper generator→

Similar questions

Function $f(x) = \frac{{{x^2} - 2}}{{\sqrt {1 + {x^2}} }}$

Evaluate: $\int\left(5 x^3+2 x^{-5}-7 x+\frac{1}{\sqrt{x}}+\frac{5}{x}\right) d x$

If $\text{AB}=\text{A}$ and $\text{BA = B}$ then $\text{B}^2 $ is equal to:

$\text{B}$
$\text{A}$
$\text{-B}$
$\text{B}^2$

Integration of $\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)$ with respect to $x$ :

If $y \frac{d y}{d x}=x\left[\frac{y^{2}}{x^{2}}+\frac{\phi\left(\frac{y^{2}}{x^{2}}\right)}{\phi^{\prime}\left(\frac{y^{2}}{x^{2}}\right)}\right], x>0, \phi>0$, and $y(1)=-1$ then $\phi\left(\frac{\mathrm{y}^{2}}{4}\right)$ is equal to :

The differential coefficient of the function $|x - 1| + |x - 3|$ at the point $x = 2$ is

Solve:$\sin { \left( { \tan }^{ -1 }\text{x} \right) } ,\left| \text{x} \right| <1$ is equal to:

$\frac { \text{x} }{ \sqrt { 1-{ \text{x} }^{ 2 } } }$
$\frac { \text{x} }{ \sqrt { 1-{ \text{x} }^{ 2 } } }$
$\frac { \text{x} }{ \sqrt { 1-{ \text{x} }^{ 2 } } }$
$\frac { \text{x} }{ \sqrt { 1+{ \text{x} }^{ 2 } } }$

If $\frac{\text{d}}{\text{dx}}[\text{x}^\text{n}-\text{a}_1\text{x}^{\text{n}-1}+\text{a}_2\text{x}^{\text{n}-2}+...+(-1)^\text{n}\text{a}_\text{n }]\text{e}^\text{x}=\text{x}^\text{n}\ \text{e}^\text{x},$ then the value of a, 0 < r < is equals to:

$\frac{\text{n}!}{\text{r}!}$
$\frac{(\text{n}-\text{r})!}{\text{r}!}$
$\frac{\text{n}!}{(\text{n}-\text{r})!}$
$\text{None of these}$

If $\left| {\begin{array}{*{20}{c}}
{^9{C_4}}&{^9{C_5}}&{^{10}{C_r}} \\
{^{10}{C_6}}&{^{10}{C_7}}&{^{11}{C_{r + 2}}} \\
{^{11}{C_8}}&{^{11}{C_9}}&{^{12}{C_{r + 4}}}
\end{array}} \right| = 0$ then $r$ is equal to

$\int\limits_0^\infty $ $\frac{x}{{(1\,\, + \,\,x)\,\,(1\,\, + \,\,{x^2})}}$ $d x$ :