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Continuity — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsContinuity1 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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