JEE Mains · Maths · STD 12 - 7.2 definite integral
If \(a=\lim _{n \rightarrow \infty} \sum_{k=1}^{n} \frac{2 n}{n^{2}+k^{2}}\) and \(f(x)=\) \(\sqrt{\frac{1-\cos x}{1+\cos x}}, x \in(0,1)\), then.
- A \(2 \sqrt{2} f \left(\frac{ a }{2}\right)= f ^{\prime}\left(\frac{ a }{2}\right)\)
- B \(f \left(\frac{ a }{2}\right) f ^{\prime}\left(\frac{ a }{2}\right)=\sqrt{2}\)
- C \(\sqrt{2} f \left(\frac{ a }{2}\right)= f ^{\prime}\left(\frac{ a }{2}\right)\)
- D \(f \left(\frac{ a }{2}\right)=\sqrt{2} f ^{\prime}\left(\frac{ a }{2}\right)\)
Answer & Solution
Correct Answer
(C) \(\sqrt{2} f \left(\frac{ a }{2}\right)= f ^{\prime}\left(\frac{ a }{2}\right)\)
Step-by-step Solution
Detailed explanation
\(a =\frac{1}{ n } \sum_{ k =1}^{n} \frac{2}{1+\left(\frac{ k }{ n }\right)^{2}}=\int_{0}^{1} \frac{2}{1+ x ^{2}} dx =\frac{\pi}{2}\) \(f ( x )=\tan \left(\frac{ x }{2}\right) ; x \in(0,1)\) \(f \left(\frac{\pi}{4}\right)=\sqrt{2}-1\)…
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