NDA 2 2024 Maths, question 86

Topic: Differential calculus

Passage · Q85 to Q86

Let $f(x)=\cos 2x+x$ on $[-\pi/2,\pi/2]$ .

Q86

What is the least value of $f(x)$ ?

(a) $-\left( 1+\frac{\pi}{2} \right)$ Correct answer
(b) $-\left( \frac{1}{2}+\frac{\pi}{2} \right)$
(c) $-\left( 1+\frac{\pi}{4} \right)$
(d) $-2\left( \frac{1}{2}-\frac{\pi}{4} \right)$

Solution.

Answer: (a) $-\left( 1+\frac{\pi}{2} \right)$

Because cos2x≥−1 and x≥−π/2, f(x)=cos2x+x≥−1−π/2.
Both lower bounds are attained simultaneously at x=−π/2, proving the absolute minimum is −(1+π/2).