Question:
If the function \(f(x)=\begin{cases}\dfrac{\sqrt{2+\cos x}-1}{(\pi-x)^2}, & x\ne\pi\\ k, & x=\pi\end{cases}\) is continuous at \(x=\pi\), then k equals :
If the function \(f(x)=\begin{cases}\dfrac{\sqrt{2+\cos x}-1}{(\pi-x)^2}, & x\ne\pi\\ k, & x=\pi\end{cases}\) is continuous at \(x=\pi\), then k equals :
Show Hint
Continuity needs \(k=\lim_{x\to\pi}f(x)\); put \(t=\pi-x\) and rationalise.
Updated On: Nov 11, 2026
- \(0\)
- \(\frac12\)
- \(2\)
- \(\frac14\)
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The Correct Option is D
Solution and Explanation
Let \(t=\pi-x\to 0\). Then \(\cos x=-\cos t\) and \(f=\frac{\sqrt{2-\cos t}-1}{t^2}\).
Rationalise: \(\frac{(2-\cos t)-1}{t^2\left(\sqrt{2-\cos t}+1\right)}=\frac{1-\cos t}{t^2}\cdot\frac{1}{\sqrt{2-\cos t}+1}\).
As \(t\to0\): \(\frac{1-\cos t}{t^2}\to\frac12\) and \(\sqrt{2-\cos t}+1\to 2\).
So \(k=\frac12\cdot\frac12=\frac14\).
Answer: (4) \(\frac14\)
Rationalise: \(\frac{(2-\cos t)-1}{t^2\left(\sqrt{2-\cos t}+1\right)}=\frac{1-\cos t}{t^2}\cdot\frac{1}{\sqrt{2-\cos t}+1}\).
As \(t\to0\): \(\frac{1-\cos t}{t^2}\to\frac12\) and \(\sqrt{2-\cos t}+1\to 2\).
So \(k=\frac12\cdot\frac12=\frac14\).
Answer: (4) \(\frac14\)
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