Need solution for RD Sharma maths class 12 chapter 8 Continuity exercise multiple choice question 11

The correct option is (b)

Hint:

Use the given formula:

\begin{aligned} &\text { (i) } \lim _{x \rightarrow 0} f(x) g(x)=e \lim _{x \rightarrow 0}(f(x)-1) \cdot g(x)\\ &\text { Where } \lim _{x \rightarrow 0} f(x)=1 \end{aligned}

\begin{aligned} &\text { And } \lim _{x \rightarrow 0} g(x)=0\\ &\text { (ii) } \lim _{x \rightarrow 0}\left\{\frac{\cos x-1}{x}\right\}=-1 \end{aligned}

(iii) A function f(x) is said to be continuous at a point x = a of its domain, if

$\lim _{x \rightarrow a^{+}} f(a+h)=\lim _{x \rightarrow a^{-}} f(a-h)=f(a)$

Given:

$f(x)=\left\{\begin{array}{l} (\cos x)^{\frac{1}{x}} \\ k \end{array}\right.$    \begin{aligned} , x & \neq 0 \\ , x &=0 \end{aligned}

And f(x) is continuous at x = 0

Solution:

$\lim _{x \rightarrow 0} f(x)=f(0)$

Calculate the value of k

$\lim _{x \rightarrow 0}(\cos x)^{\frac{1}{x}}=k$

Using formula (i)

$\lim _{x \rightarrow 0} f(x)^{g(x)}=e \lim _{x \rightarrow 0}\left\{\frac{\cos x-1}{x}\right\}=k$

Apply formula (ii)

\begin{aligned} &e^{0}=k \\ & k=1 \end{aligned}

So, option (b) is correct.