Question

Let $f(x)=\left\{\begin{array}{ll}\frac{1+\cos x}{(\pi-x)^{2}} \cdot \frac{\sin ^{2} x}{\ln \left(1+\pi^{2}-2 \pi x+x^{2}\right)} & x \neq \pi \\ \lambda & x=\pi\end{array}\right.$ is continuous at $x=\pi $ , then $\lambda $ is equal to

• A. $1$
• B. $-1$
• C. $\frac{1}{2}$
• D. $\frac{1}{4}$

Answer 1

Answer: (C)

For $f\left(x\right)$ to be continuous at $x=\pi $ ,
$\underset{x \rightarrow \pi }{lim} f \left(x\right)=f\left(\pi \right)$
Now, $\underset{x \rightarrow \pi }{lim} \frac{1 + cos ⁡ x}{\left(\pi - x\right)^{2}}\cdot \frac{\left(sin\right)^{2} ⁡ x}{ln ⁡ \left\{1 + \left(\pi - x\right)^{2}\right\}}=\lambda $
Putting $x=\pi +t$ , we get,
$\therefore \underset{\text{t} \rightarrow 0}{lim} \frac{1 - cos t}{t^{2}} \cdot \frac{\left(sin\right)^{2} t}{ln \left(1 + t^{2}\right)} = \lambda $
$\underset{\text{t} \rightarrow 0}{lim} \frac{1}{2} \cdot \frac{\left(sin\right)^{2} t}{t^{2}} \cdot \frac{t^{2}}{ln \left(1 + t^{2}\right)} = \lambda $
$\frac{1}{2}\times 1\times 1=\lambda \Rightarrow \lambda =\frac{1}{2}$