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NDA 2024
Exam
NDA
Year
2024
Subject
Mathematics
Chapter
Definite Integration

NDA 2024 · Mathematics · Definite Integration

Paragraph: Let f(t)= (t+ 1+t^2 ) and g(t)= (f(t)) . Question: What...

Paragraph: Let f(t)=ln⁡f(t)=\ln\left(t+\sqrt{1+t^2}\right)andg(t)=tan⁡(f(t))and g(t)= \tan (f(t)).
Question: What is ∫−ππ\int_{-\pi}^\pi g(t) d t equal to?

Options

  1. A-1
  2. B0
  3. C12\frac{1}{2}
  4. D1

Correct answer

B. 0

Step-by-step solution

 let f(t)=ln⁡(t+1+t2)Put\text{ let } f(t)=\ln \left(t+\sqrt{1+t^2}\right) Put t=-t in f(t)f(t)f(−t)=ln⁡(1+t2−t)=ln⁡(1t+1+t2)=−ln⁡(t+1+t2)=−f(t)⇒f(−t)=−f(t)\begin{aligned}& f(-t)=\ln \left(\sqrt{1+t^2}-t\right)=\ln \left(\frac{1}{t+\sqrt{1+t^2}}\right)\\ & =-\ln\left(t+\sqrt{1+t^2}\right)=-f(t) \\ & \Rightarrow f(-t)=-f(t) \end{aligned}∴f(t)\therefore \quad f(t) is odd function and g(l)=tan⁡f(t)g(l)=\tan f(t) Put t=−tint=-t ing(t)∴g(−t)=tan⁡(f(−t))=tan⁡(−f(t))=−tan⁡(f(t))=−g(t)\begin{aligned}& \therefore \quad g(-t)=\tan(f(-t))=\tan(-f(t)) \\ & =-\tan(f(t))=-g(t) \end{aligned}So,So,g(t)isoddfunction.weknowthatis odd function. we know thatg(t)isoddfunctionthen∫−ππis odd function then \int_{-\pi}^\pi g(t) d t=0
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