Question:
If the function \(f\) defined on \(\left(\frac{\pi}{6},\frac{\pi}{3}\right)\) by \(f(x)=\begin{cases}\dfrac{\sqrt2\cos x-1}{\cot x-1},&x\ne\dfrac{\pi}{4}\\ k,&x=\dfrac{\pi}{4}\end{cases}\) is continuous, then \(k\) is equal to
If the function \(f\) defined on \(\left(\frac{\pi}{6},\frac{\pi}{3}\right)\) by \(f(x)=\begin{cases}\dfrac{\sqrt2\cos x-1}{\cot x-1},&x\ne\dfrac{\pi}{4}\\ k,&x=\dfrac{\pi}{4}\end{cases}\) is continuous, then \(k\) is equal to
Show Hint
Continuity requires \(k=\lim_{x\to\pi/4}f(x)\), a \(0/0\) form; apply L'Hopital's rule or factor using \(\cos2x\).
Updated On: Nov 5, 2026
- 1
- \(1/\sqrt2\)
- \(\frac12\)
- 2
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The Correct Option is C
Solution and Explanation
Continuity at \(\pi/4\): \(k=\lim_{x\to\pi/4}\frac{\sqrt2\cos x-1}{\cot x-1}\), which is of the form \(\frac00\).
By L'Hopital: numerator derivative \(=-\sqrt2\sin x\to-1\); denominator derivative \(=-\operatorname{cosec}^2x\to-2\).
\(k=\frac{-1}{-2}=\frac12\).
Answer: (3) \(\frac12\)
By L'Hopital: numerator derivative \(=-\sqrt2\sin x\to-1\); denominator derivative \(=-\operatorname{cosec}^2x\to-2\).
\(k=\frac{-1}{-2}=\frac12\).
Answer: (3) \(\frac12\)
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