NDA 2 2024 Maths, question 94

Topic: Integral calculus

Passage · Q93 to Q94

Let $f(t)=\ln \left( t+\sqrt{1+t^{2}} \right)$ and $g(t)=\tan (f(t))$ .

Q94

What is $\int _{-\pi}^{\pi}g(t) dt$ equal to?

(a) $-1$
(b) $0$ Correct answer
(c) $1/2$
(d) $1$

Solution.

Answer: (b) $0$

$f(-t) = \ln\left(-t + \sqrt{1 + t^{2}}\right) = -f(t)$ , so $f$ is odd, and so is $g(t) = \tan(f(t))$ .
The integral of an odd function over $[-\pi, \pi]$ is 0.

Trap: $g$ has breaks where $f(t) = \pm\frac{\pi}{2}$ , inside the interval, so strictly the integral does not exist; the expected answer uses the odd-function rule.