JEE Mains · Maths · STD 12 - 7.2 definite integral
Let \(g(t)=\int \limits_{-\pi / 2}^{\pi / 2} \cos \left(\frac{\pi}{4} t+f(x)\right) \,d x\), where \(f(x)=\log _{e}\left(x+\sqrt{x^{2}+1}\right), x \in R\). Then which one of the following is correct?
- A \(g(1)+g(0)=0\)
- B \(g(1)=\sqrt{2} g(0)\)
- C \(g(1)=g(0)\)
- D \(\sqrt{2} g(1)=g(0)\)
Answer & Solution
Correct Answer
(D) \(\sqrt{2} g(1)=g(0)\)
Step-by-step Solution
Detailed explanation
\(g(t)=\int_{-\pi / 2}^{\pi / 2} \cos \left(\frac{\pi}{4} t+f(x)\right) \,d x\) \(g(t)=\pi \cos \frac{\pi}{4} t+\int_{-\pi / 2}^{\pi / 2} f(x) \,d x\) \(g(t)=\pi \cos \frac{\pi}{4} t\) \(g(1)=\frac{\pi}{\sqrt{2}}, g(0)=\pi\)
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