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NDA 2024
Exam
NDA
Year
2024
Subject
Mathematics
Chapter
Functions

NDA 2024 · Mathematics · Functions

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

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: Consider the following statements : A. f(t)f(t) is an odd function. B. g(t) is an odd function. Which of the statements given above is/are correct?

Options

  1. AA only
  2. BB only
  3. CBoth A and B
  4. DNeither A nor B

Correct answer

C. Both A and B

Step-by-step solution

\text{ let } f(t)=\ln \left(t+\sqrt{1+t^2}\right) Put t=−tint=-t inf(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}\therefore \quad f(t)isoddfunctionandis odd function andg(l)=\tan f(t)PutPutt=-t in g(t)g(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}$So,So,g(t)$ is odd function. Both statements I and II are correct.
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