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Continuity and Differentiability — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsContinuity and Differentiability1 Mark
MCQ
The function $\text{f(x)}=\begin{cases}\frac{\text{x}^2}{\text{a}},&0\leq\text{x}<1\\\text{a},&1\leq\text{x}<\sqrt{2}\\\frac{2\text{b}^2-4\text{b}}{\text{x}^2},&\sqrt{2}\leq\text{x}<\infty\end{cases}$ is continuous for $0\leq\text{x}<\infty,$ then the most suitable values of $a$ and $b$ are:
  • A
    $\text{a}=1,\text{ b}=-1$
  • B
    $\text{a}=-1,\text{ b}=1+\sqrt{2}$
  • ✓
    $\text{a}=-1,\text{ b}=1$
  • D
    None of these.
✓

Answer

Correct option: C.
$\text{a}=-1,\text{ b}=1$
Given$, f(x)$ is continuous for $0\leq\text{x}<\infty.$
This means that $f(x)$ is continuous for $\text{x}=1,\sqrt{2.}$
Now,
If $(x)$ is continuons at $x = 1,$ then
$\lim\limits_{\text{x}\rightarrow1^-}\text{f(x)}=\text{f}(1)$
$\Rightarrow\lim\limits_{\text{h}\rightarrow0}\text{f}(1-\text{h})=\text{a}$
$\Rightarrow\frac{(1-\text{h})^2}{\text{a}}=\text{a}$
$\Rightarrow\frac{1}{\text{a}}=\text{a}$
$\Rightarrow\text{a}^2=1$
$\Rightarrow\text{a}=\pm1$
If $f(x)$ is continuous at $\text{x}={\sqrt{2}},$ then
$\lim\limits_{\text{x}\rightarrow\sqrt{2}^-}\text{f(x)}=\text{f}(\sqrt{2})$
$\Rightarrow\lim\limits_{\text{h}\rightarrow0}\text{f}(\sqrt{2}-\text{h})=\frac{2\text{b}^2-4\text{b}}{2}$
$\Rightarrow\lim\limits_{\text{h}\rightarrow0}\text{a=b}^2-2\text{b}$
$\Rightarrow\text{a = b}^2-2\text{b}$
$\Rightarrow\text{b}^2-2\text{b - a}=0$
$\therefore$ For $a = -1,$ We have
$\text{b}^2-2\text{b}+1=0$
$\Rightarrow(\text{b}-1)^2=0$
$\Rightarrow\text{b}=1$
Thus$, a = -1$ and $b=1$

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