Question

Let $f(x)=\{\begin{array}{ll}\frac{1+\cos x}{(\pi -x{)}^2}\cdot \frac{{\sin }^{2}x}{\ln \left(1+{\pi }^2-2\pi x+x^2\right)}& x\ne \pi \\ \lambda & x=\pi \end{array}$ 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 $x=\pi$ ,
$\underset{x\to \pi }{\lim }f\left(x\right)=f\left(\pi \right)$
Now, $\underset{x\to \pi }{\lim }\frac{1+cosx}{{\left(\pi -x\right)}^2}\cdot \frac{{\left(sin\right)}^{2}x}{ln\left\{1+{\left(\pi -x\right)}^2\right\}}=\lambda$
Putting $x=\pi +t$ , we get,
$\therefore \underset{\text{t}\to 0}{\lim }\frac{1-cost}{t^2}\cdot \frac{{\left(sin\right)}^{2}t}{ln\left(1+t^2\right)}=\lambda$
$\underset{\text{t}\to 0}{\lim }\frac{1}{2}\cdot \frac{{\left(sin\right)}^{2}t}{t^2}\cdot \frac{t^2}{ln\left(1+t^2\right)}=\lambda$
$\frac{1}{2}\times 1\times 1=\lambda \Rightarrow \lambda =\frac{1}{2}$