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Question:
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If the function $f: R \rightarrow R$ defined by
$$
f(x)=\left\{\begin{array}{cc}
a\left(\frac{1-\cos 2 x}{x^2}\right), & \text { for } x < 0 \\
\frac{b}{x} & \text { for } x=0 \\
\frac{\sqrt{x}}{\sqrt{4+\sqrt{x}}-2}, & \text { for } x>0
\end{array}\right.
$$
Is continuous at $x=0$, then $a+b=$
Options:
$$
f(x)=\left\{\begin{array}{cc}
a\left(\frac{1-\cos 2 x}{x^2}\right), & \text { for } x < 0 \\
\frac{b}{x} & \text { for } x=0 \\
\frac{\sqrt{x}}{\sqrt{4+\sqrt{x}}-2}, & \text { for } x>0
\end{array}\right.
$$
Is continuous at $x=0$, then $a+b=$
Solution:
2461 Upvotes
Verified Answer
The correct answer is:
6
We have,
$$
f(x)=\left\{\begin{array}{cc}
\frac{a(1-\cos 2 x)}{x^2}, & x < 0 \\
b, & x=0 \\
\frac{\sqrt{x}}{\sqrt{4+\sqrt{x}}-2}, & x>0
\end{array}\right.
$$
$f(x)$ is continuous at $x=0$
$$
\therefore \quad \lim _{x \rightarrow 0^{-}} f(x)=f(0)
$$
$$
\lim _{x \rightarrow 0^{-}} \frac{a(1-\cos 2 x)}{x^2}=b
$$
$$
\lim _{x \rightarrow 0} \frac{a\left(2 \sin ^2 x\right)}{x^2}=b
$$
$\Rightarrow \quad 2 a=b$
Also, $\quad \lim _{x \rightarrow 0^{+}} f(x)=b$
$$
\begin{aligned}
& \therefore \quad \lim _{x \rightarrow 0} \frac{\sqrt{x}}{\sqrt{4+\sqrt{x-2}}}=b \\
& \Rightarrow \lim _{x \rightarrow 0} \frac{\sqrt{x}(\sqrt{4+\sqrt{x}}+2)}{\sqrt{x}}=b \\
& \Rightarrow \quad 4=b, a=2 \\
& \therefore \quad a+b=2+4=6
\end{aligned}
$$
$$
f(x)=\left\{\begin{array}{cc}
\frac{a(1-\cos 2 x)}{x^2}, & x < 0 \\
b, & x=0 \\
\frac{\sqrt{x}}{\sqrt{4+\sqrt{x}}-2}, & x>0
\end{array}\right.
$$
$f(x)$ is continuous at $x=0$
$$
\therefore \quad \lim _{x \rightarrow 0^{-}} f(x)=f(0)
$$
$$
\lim _{x \rightarrow 0^{-}} \frac{a(1-\cos 2 x)}{x^2}=b
$$
$$
\lim _{x \rightarrow 0} \frac{a\left(2 \sin ^2 x\right)}{x^2}=b
$$
$\Rightarrow \quad 2 a=b$
Also, $\quad \lim _{x \rightarrow 0^{+}} f(x)=b$
$$
\begin{aligned}
& \therefore \quad \lim _{x \rightarrow 0} \frac{\sqrt{x}}{\sqrt{4+\sqrt{x-2}}}=b \\
& \Rightarrow \lim _{x \rightarrow 0} \frac{\sqrt{x}(\sqrt{4+\sqrt{x}}+2)}{\sqrt{x}}=b \\
& \Rightarrow \quad 4=b, a=2 \\
& \therefore \quad a+b=2+4=6
\end{aligned}
$$
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