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CONTINUITY AND DIFFERENTIABILITY — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSCONTINUITY AND DIFFERENTIABILITY1 Mark
Question
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:
  1. $\text{a}=1,\text{ b}=-1$
  2. $\text{a}=-1,\text{ b}=1+\sqrt{2}$
  3. $\text{a}=-1,\text{ b}=1$
  4. $\text{None os these}.$
✓

Answer

  1. a = -1, b = 1
Solution:
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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