Question

If range of $f\left(x\right)=\cos x,x\in (\frac{-\pi }{3},\frac{\pi }{6})$ is $(a,b)$, then

• A. $a+b=\frac{3}{2}$
• B. $b-a=1-\frac{\sqrt{3}}{2}$
• C. $a^2+b^2=\frac{5}{6}$
• D. $a^2+b^2=\frac{7}{4}$

Answer 1

The correct option is A $a+b=\frac{3}{2}$

Given $f\left(x\right)=\cos x,x\in (\frac{-\pi }{3},\frac{\pi }{6})$

For $x\in (-\frac{\pi }{3},0)⟹\cos x\in (1/2,1)$

For $x\in (\frac{\pi }{6},0)⟹\cos x\in (\sqrt{3}/2,1)$

So for $x\in (\frac{-\pi }{3},\frac{\pi }{6})$ ,Range of $f\left(x\right)$ is $(\frac{1}{2},1)$

So $a=\frac{1}{2},b=1$