Discussion Forum

1)

If the function f:R→ r  defined by

$f(x)=\begin{cases}a\left(\frac{1-\cos 2x}{x^{2}}\right), & for x < 0\\b ,& x = 0\\\frac{\sqrt{x}}{\sqrt{4+\sqrt{x}}-2}&for x>0\end{cases}$

 is continuous  at x=0 , then  a+b=

Answer:

Option C

Explanation:

We have, 

  $f(x)=\begin{cases}a\left(\frac{1-\cos 2x}{x^{2}}\right), & for x < 0\\b ,& x = 0\\\frac{\sqrt{x}}{\sqrt{4+\sqrt{x}}-2}&for x>0\end{cases}$

f(x) is continuous at x=0

 $\therefore$   $\lim_{x \rightarrow{0^{-}}}f(x)= f(0)$

$\lim_{x \rightarrow{0^{-}}}\frac{a(1- \cos 2x)}{x^{2}}=b$

$\lim_{x \rightarrow{0^{}}}\frac{a(2 \sin^{2} x)}{x^{2}}=b$

 $\Rightarrow$   2a=b

 Also,    $\lim_{x \rightarrow{0^{+}}}f(x)=b$

 $\therefore$     $\lim_{x \rightarrow{0^{}}}\frac{\sqrt{x}}{\sqrt{4+\sqrt{x-2}}}=b$

 $\Rightarrow\lim_{x \rightarrow{0^{}}}\frac{\sqrt{x}(\sqrt{4+\sqrt{x}}+2)}{\sqrt{x}}=b$

$\Rightarrow$   4=b , a=2

 $\therefore$ a+b=2+4=6