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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY1 Mark
MCQ
The value of f(0), so that the function $\text{f(x)}=\frac{\sqrt{\text{a}^2+\text{ax+x}^2}-\sqrt{\text{a}^2+\text{ax+x}^2}{}}{\sqrt{\text{a+x}}-\sqrt{\text{a-x}}}$ becomes continuous for all x, given by:
  • A
    $\text{a}^{\frac{3}{2}}$
  • B
    $\text{a}^{\frac{1}{2}}$
  • ✓
    $-\text{a}^{\frac{1}{2}}$
  • D
    $-\text{a}^{\frac{3}{2}}$
✓

Answer

Correct option: C.
$-\text{a}^{\frac{1}{2}}$
Given, $\text{f(x)}=\frac{\sqrt{\text{a}^2+\text{ax+x}^2}-\sqrt{\text{a}^2+\text{ax+x}^2}{}}{\sqrt{\text{a+x}}-\sqrt{\text{a-x}}}$

$\Rightarrow\text{f(x)}=\frac{\big(\sqrt{\text{a}^2-\text{ax+x}^2}-\sqrt{\text{a}^2+\text{ax+x}^2}\big)\big(\sqrt{\text{a}^2-\text{ax+x}^2}+\sqrt{\text{a}^2+\text{ax+x}^2}\big)}{\big(\sqrt{\text{a+x}}-\sqrt{\text{a-x}}\big)\big(\sqrt{\text{a}^2-\text{ax+x}^2}+\sqrt{\text{a}^2+\text{ax+x}^2}\big)}$

$\Rightarrow\text{f(x)}=\frac{\big(\text{a}^2-\text{ax+x}^2\big)-\big(\text{a}^2+\text{ax+x}^2\big)}{\big(\sqrt{\text{a+x}}-\sqrt{\text{a-x}}\big)\big(\sqrt{\text{a}^2-\text{ax+x}^2}+\sqrt{\text{a}^2+\text{ax+x}^2}\big)}$

$\Rightarrow\text{f(x)}=\frac{(-2\text{ax})\big(\sqrt{\text{a+x}}+\sqrt{\text{a-x}}\big)}{\big(\sqrt{\text{a+x}}-\sqrt{\text{a-x}}\big)\big(\sqrt{\text{a}^2-\text{ax+x}^2}+\sqrt{\text{a}^2+\text{ax+x}^2}\big)\big(\sqrt{\text{a+x}}+\sqrt{\text{a-x}}\big)}$

$\Rightarrow\text{f(x)}=\frac{(-2\text{ax})\big(\sqrt{\text{a+x}}+\sqrt{\text{a-x}}\big)}{\big(\text{a+x-a+x}\big)\big(\sqrt{\text{a}^2-\text{ax+x}^2}+\sqrt{\text{a}^2}+\text{ax+x}^2\big)}$

$\Rightarrow\text{f(x)}=\frac{(-2\text{ax})\big(\sqrt{\text{a+x}}+\sqrt{\text{a-x}}\big)}{(2\text{x})\big(\sqrt{\text{a}^2-\text{ax+x}^2}+\sqrt{\text{a}^2+\text{ax+x}^2}\big)}$

$\Rightarrow\text{f(x)}=\frac{-\text{a}\big(\sqrt{\text{a+x}}+\sqrt{\text{a-x}}\big)}{\big(\sqrt{\text{a}^2-\text{ax+x}^2}+\sqrt{\text{a}^2+\text{ax+x}^2}\big)}$

So, if f(x) is continuous at x = 0, then

$\lim\limits_{\text{x}\rightarrow0}\text{f(x)}=\text{f(0)}$

$\Rightarrow\lim\limits_{\text{x}\rightarrow0}\Bigg[\frac{-\text{a}\big(\sqrt{\text{a+x}}+\sqrt{\text{a-x}}\big)}{\big(\sqrt{\text{a}^2-\text{ax+x}}+\sqrt{\text{a}^2+\text{ax+x}^2}\big)} \Bigg]$

$\Rightarrow\bigg[\frac{-2\text{a}(\sqrt{\text{a}})}{(\sqrt{a}^2+\sqrt{\text{a}^2})}\bigg]=\text{f(0)}$

$\Rightarrow\begin{bmatrix}\frac{-2\text{a}(\sqrt{a})}{(\text{a+a})} \end{bmatrix}=\text{f(0)}$

$\Rightarrow\text{f}(0)$

$=-\sqrt{a}$

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