Question

If $A=\{x:\pi /6\le x\le \pi /3\}$ and $f\left(x\right)=cosx-x(1+x)$ then $f\left(A\right)=\{y:\frac{\sqrt{c}}{2}-\frac{\pi }{b}(1+\frac{\pi }{3})\le y\le \frac{1}{a}-\frac{\pi }{6}(1+\frac{\pi }{d}\left)\right\}$Find $a+b+c+d$

Answer 1

Given

$f\left(x\right)=\cos x-x(1+x)$

$\frac{\pi }{6}\le x\le \frac{\pi }{3}$

$⟹1+\frac{\pi }{6}\le (1+x)\le 1+\frac{\pi }{3}$

$⟹\frac{\pi }{6}(1+\frac{\pi }{6})\le x(1+x)\le \frac{\pi }{3}(1+\frac{\pi }{3})$

$⟹-\frac{\pi }{3}(1+\frac{\pi }{3})\le -x(1+x)\le -\frac{\pi }{6}(1+\frac{\pi }{6})\cdots \left(1\right)$

$⟹\cos \frac{\pi }{6}\le \cos x\le \cos \frac{\pi }{3}$

$⟹\frac{\sqrt{3}}{2}\le \cos x\le \frac{1}{2}\cdots \left(2\right)$

$\left(1\right)+\left(2\right)$

$⟹\frac{\sqrt{3}}{2}-\frac{\pi }{3}(1+\frac{\pi }{3})\le \cos x-x(1+x)\le \frac{1}{2}-\frac{\pi }{6}(1+\frac{\pi }{6})\cdots \left(1\right)$

$⟹a=2,b=3,c=3,d=6$

$⟹a+b+c+d=14$