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Question

Mathematics Question on Trigonometric Functions

Let f(?)=(1+sin2?)(2sin2?)f(?) = (1+sin^2?)(2-sin^2?). Then for all values of ??

A

f(?)>94f\left(?\right) > \frac{9}{4}

B

f(?)<2f\left(?\right) < 2

C

f(?)>114f\left(?\right) > \frac{11}{4}

D

2f(?)942\le f\left(?\right)\le \frac{9}{4}

Answer

2f(?)942\le f\left(?\right)\le \frac{9}{4}

Explanation

Solution

Given f(θ)=(1+sin2θ)(2sin2θ)f(\theta) =\left(1+\sin ^{2} \theta\right)\left(2-\sin ^{2} \theta\right)
=2+2sin2θsin2θsin4θ=2+2 \sin ^{2} \theta-\sin ^{2} \theta-\sin ^{4} \theta
=sin4θ+sin2θ+2=-\sin ^{4} \theta+\sin ^{2} \theta+2
=(sin4θsin2θ2)=-\left(\sin ^{4} \theta-\sin ^{2} \theta-2\right)
=-\left\\{\sin ^{4} \theta-\sin ^{2} \theta+\frac{1}{4}-\frac{9}{4}\right\\}
=+94(sin2θ12)2=+\frac{9}{4}-\left(\sin ^{2} \theta-\frac{1}{2}\right)^{2} ....(i)
1sinθ1\because -1 \leq \sin \theta \leq 1
0sin2θ1\Rightarrow 0 \leq \sin ^{2} \theta \leq 1
12sin2θ1212\Rightarrow -\frac{1}{2} \leq \sin ^{2} \theta-\frac{1}{2} \leq \frac{1}{2}
0(sin2θ12)214\Rightarrow 0 \leq\left(\sin ^{2} \theta-\frac{1}{2}\right)^{2} \leq \frac{1}{4}
0(sin2θ12)214\Rightarrow 0 \geq-\left(\sin ^{2} \theta-\frac{1}{2}\right)^{2} \geq-\frac{1}{4}
9494(sin2θ12)29414\Rightarrow \frac{9}{4} \geq \frac{9}{4}-\left(\sin ^{2} \theta-\frac{1}{2}\right)^{2} \geq \frac{9}{4}-\frac{1}{4}
2f(θ)94\Rightarrow 2 \leq f(\theta) \leq \frac{9}{4} [from E(i)]