Q. If $A = \left\{ x:\frac{\pi}{6} \leq x \leq \frac{\pi}{3} \right\}$ and $f(x) = \cos{x - x(1 + x)}$, then $f(A)$ is equal to
A) $\left\lbrack - \frac{\pi}{3},\ - \frac{\pi}{6} \right\rbrack$
B) $\left\lbrack \frac{\pi}{6},\ \frac{\pi}{3} \right\rbrack$
C) $\left\lbrack \frac{1}{2} - \frac{\pi}{3}\left( 1 + \frac{\pi}{3} \right),\frac{\sqrt{3}}{2} - \frac{\pi}{6}\left( 1 + \frac{\pi}{6} \right) \right\rbrack$
D) $\left\lbrack \frac{1}{2} + \frac{\pi}{3}\left( 1 - \frac{\pi}{3} \right),\frac{\sqrt{3}}{2} + \frac{\pi}{6}\left( 1 - \frac{\pi}{6} \right) \right\rbrack$
Explanation
(c)
Since, $f(x)$ is a continuous decreasing function on $\left\lbrack \frac{\pi}{6},\ \frac{\pi}{3} \right\rbrack$
$\therefore\ \ f(x)$ attains every value between $\left\lbrack \frac{\pi}{6},\ \frac{\pi}{3} \right\rbrack$
Its minimum value
$f\left( \frac{\pi}{3} \right) = \frac{1}{2} - \frac{\pi}{3}\left( 1 + \frac{\pi}{3} \right)$
And maximum value
$f\left( \frac{\pi}{6} \right) = \frac{\sqrt{3}}{2} - \frac{\pi}{6}\left( 1 + \frac{\pi}{6} \right)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.