KCET · Maths · Continuity and Differentiability
If the function \(f(x)=\left\{\begin{array}{cl}\frac{1-\cos x}{x^{2}}, & \text { for } x \neq 0 \\ k, & \text { for } x=0\end{array}\right.\) continuous at \(x=0\), then the value of \(k\) is
- A 1
- B 0
- C \(\frac{1}{2}\)
- D \(-1\)
Answer & Solution
Correct Answer
(C) \(\frac{1}{2}\)
Step-by-step Solution
Detailed explanation
Given, \(f(x)=\left\{\begin{array}{cc}\frac{1-\cos x}{x^{2}}, & x \neq 0 \\ k & , x=0\end{array}\right.\)
Since, \(x\) is continuous
\[
\begin{array}{ll}
\Rightarrow & \lim _{x \rightarrow 0} f(x)=f(0) \\
\therefore & \lim _{x \rightarrow 0} \frac{1-\cos x}{x^{2}}=k \\
\Rightarrow \quad & \lim _{x \rightarrow 0} \frac{-(-\sin x)}{2 x}=k \\
\Rightarrow \quad & \frac{1}{2} \lim _{x \rightarrow 0} \frac{\sin x}{x}=k \\
\Rightarrow \quad & \frac{1}{2} \cdot 1=k \Rightarrow k=\frac{1}{2}
\end{array}
\]
Since, \(x\) is continuous
\[
\begin{array}{ll}
\Rightarrow & \lim _{x \rightarrow 0} f(x)=f(0) \\
\therefore & \lim _{x \rightarrow 0} \frac{1-\cos x}{x^{2}}=k \\
\Rightarrow \quad & \lim _{x \rightarrow 0} \frac{-(-\sin x)}{2 x}=k \\
\Rightarrow \quad & \frac{1}{2} \lim _{x \rightarrow 0} \frac{\sin x}{x}=k \\
\Rightarrow \quad & \frac{1}{2} \cdot 1=k \Rightarrow k=\frac{1}{2}
\end{array}
\]
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