CUET · MATHS · PYQ PAPER 2025
If \(f(x)=\left\{\begin{array}{ll}\frac{\tan \left(\frac{\pi}{4}-x\right)}{\cot 2 x}, & x \neq \frac{\pi}{4}, \\ 2 K+1, & x=\frac{\pi}{4},\end{array}\right.\) is continuous at \(x=\frac{\pi}{4}\), then the value of \(K\) is equal to
- A \(\frac{1}{4}\)
- B \(\frac{1}{2}\)
- C \(-\frac{1}{2}\)
- D \(-\frac{1}{4}\)
Answer & Solution
Correct Answer
(D) \(-\frac{1}{4}\)
Step-by-step Solution
Detailed explanation
\(f\left(\frac{\pi}{4}\right) = 2K+1\) \(\lim_{x \to \frac{\pi}{4}} f(x) = \lim_{x \to \frac{\pi}{4}} \frac{\tan \left(\frac{\pi}{4}-x\right)}{\cot 2 x}\) Let \(h = \frac{\pi}{4}-x\). As \(x \to \frac{\pi}{4}\), \(h \to 0\). Then \(x = \frac{\pi}{4}-h\).…
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