Question

If A =$\{x|\frac{\pi }{6}\le x\le \frac{\pi }{3}\}$ and $f\left(x\right)=\cos x-x(1+x)$, then $f\left(A\right)$ is equal to

• A. $[\frac{\pi }{6},\frac{\pi }{3}]$
• B. $[-\frac{\pi }{3},-\frac{\pi }{6}]$
• C. $[\frac{1}{2}-\frac{\pi }{3}(1+\frac{\pi }{3}),\frac{\sqrt{3}}{2}-\frac{\pi }{6}(1+\frac{\pi }{6})]$
• D. $[\frac{1}{2}+\frac{\pi }{3}(1-\frac{\pi }{3}),\frac{\sqrt{3}}{2}+\frac{\pi }{6}(1-\frac{\pi }{6})]$

Answer 1

The correct option is C $[\frac{1}{2}-\frac{\pi }{3}(1+\frac{\pi }{3}),\frac{\sqrt{3}}{2}-\frac{\pi }{6}(1+\frac{\pi }{6})]$

Given,$A=x:\frac{\pi }{6}\le \frac{\pi }{3}$ and $f\left(x\right)=cosx-x(1+x)$

Range of f(A) is obtained by substituting its domain values in the f(x) function.

$\Rightarrow f\left(\frac{\pi }{6}\right)=\frac{\sqrt{3}}{2}-\frac{\pi }{6}\ast (1+\frac{\pi }{6})$

$\Rightarrow f\left(\frac{\pi }{3}\right)=\frac{1}{2}-\frac{\pi }{3}\ast (1+\frac{\pi }{3})$

$\Rightarrow \frac{1}{2}-\frac{\pi }{3}\ast (1+\frac{\pi }{3})\le f\left(A\right)\le \frac{\sqrt{3}}{2}-\frac{\pi }{6}\ast (1+\frac{\pi }{6})$

Option C is correct.