Question

Let $f(θ)=3\left({\sin }^4\left(\frac{3π}{2}−θ\right)+{\sin }^4(3π+θ)\right)−2\left(1−{\sin }^{2}2θ\right)$ and $S=\left\{θ∈[0,π]:f^{′}(θ)=−\frac{\sqrt{3}}{2}\right\}$. If $4β=\underset{θ∈S}{∑}θ$, then $f(β)$ is equal to

• A. $\frac{11}{8}$
• B. $\frac{3}{2}$
• C. $\frac{9}{8}$
• D. $\frac{5}{4}$

Answer 1

Answer: (D)

$f(θ)=3\left({\sin }^4\left(\frac{3π}{2}−θ\right)+{\sin }^4(3x+θ)\right)−2\left(1−{\sin }^{2}2θ\right)$
$S=\left\{θ∈[0,π]:f^{′}(θ)=−\frac{\sqrt{3}}{2}\right\}$
$⇒f(θ)=3\left({\cos }^4θ+{\sin }^4θ\right)−2{\cos }^{2}2θ$
$⇒f(θ)=3\left(1−\frac{1}{2}{\sin }^{2}2θ\right)−2{\cos }^{2}2θ$
$⇒f(θ)=3−\frac{3}{2}{\sin }^{2}2θ−2{\cos }^2θ$
$=\frac{3}{2}−\frac{1}{2}{\cos }^{2}2θ=\frac{3}{2}−\frac{1}{2}\left(\frac{1+\cos 4θ}{2}\right)$
$f(θ)=\frac{5}{4}−\frac{\cos 4θ}{4}$
$f^{′}(θ)=\sin 4θ$
$⇒f^{′}(θ)=\sin 4θ=−\frac{\sqrt{3}}{2}$
$⇒4θ=nπ+(−1{)}^n\frac{π}{3}$
$⇒θ=\frac{nπ}{4}+(−1{)}^n\frac{π}{12}$
$⇒θ=\frac{π}{12},\left(\frac{π}{4}−\frac{π}{12}\right),\left(\frac{π}{2}+\frac{π}{12}\right),\left(\frac{3π}{4}−\frac{π}{12}\right)$
$⇒4β=\frac{π}{4}+\frac{π}{2}+\frac{3π}{4}=\frac{3π}{2}$
$⇒β=\frac{3π}{8}⇒f(β)=\frac{5}{4}−\frac{\cos \frac{3π}{2}}{4}=\frac{5}{4}$