Q93 / 120 · NDA 2024 (Phase 2) · Mathematics

Paragraph: Let $f(t)=\ln \left(t+\sqrt{1+t^2}\right)$ and $g(t)= \tan (f(t))$.
Question: Consider the following statements : A. $f(t)$ is an odd function. B. $g(t)$ is an odd function. Which of the statements given above is/are correct?

  • A
    A only
  • B
    B only
  • Correct answer
    C
    Both A and B
  • D
    Neither A nor B

Solution

$\text { let } f(t)=\ln \left(t+\sqrt{1+t^2}\right)$ Put $t=-t$ in $f(t)$ $\begin{aligned} & f(-t)=\ln \left(\sqrt{1+t^2}-t\right)=\ln \left(\frac{1}{t+\sqrt{1+t^2}}\right) \\ & =-\ln \left(t+\sqrt{1+t^2}\right)=-f(t) \\ & \Rightarrow f(-t)=-f(t) \end{aligned}$ $\therefore \quad f(t)$ is odd function and $g(l)=\tan f(t)$ Put $t=-t$ in $g(t)$ $\begin{aligned} & \therefore \quad g(-t)=\tan (f(-t))=\tan (-f(t)) \\ & =-\tan (f(t))=-g(t) \end{aligned}$ So, $g(t)$ is odd function. Both statements I and II are correct.

Hence, the correct answer is option: 'C'.

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