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JEE Main 2024
Trigonometry
Trigonometric Ratio and Identites
Medium

Question

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

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Solution

f(θ)=3(sin4(3π2θ)+sin4(3x+θ))2(1sin22θ)f(\theta)=3\left(\sin ^{4}\left(\frac{3 \pi}{2}-\theta\right)+\sin ^{4}(3 x+\theta)\right)-2\left(1-\sin ^{2} 2 \theta\right) S={θ[0,π]:f(θ)=32}S=\left\{\theta \in[0, \pi]: f^{\prime}(\theta)=-\frac{\sqrt{3}}{2}\right\} f(θ)=3(cos4θ+sin4θ)2cos22θ\Rightarrow \mathrm{f}(\theta)=3\left(\cos ^{4} \theta+\sin ^{4} \theta\right)-2 \cos ^{2} 2 \theta f(θ)=3(112sin22θ)2cos22θ\Rightarrow \mathrm{f}(\theta)=3\left(1-\frac{1}{2} \sin ^{2} 2 \theta\right)-2 \cos ^{2} 2 \theta f(θ)=332sin22θ2cos2θ\Rightarrow \mathrm{f}(\theta)=3-\frac{3}{2} \sin ^{2} 2 \theta-2 \cos ^{2} \theta =3212cos22θ=3212(1+cos4θ2)=\frac{3}{2}-\frac{1}{2} \cos ^{2} 2 \theta=\frac{3}{2}-\frac{1}{2}\left(\frac{1+\cos 4 \theta}{2}\right) f(θ)=54cos4θ4f(\theta)=\frac{5}{4}-\frac{\cos 4 \theta}{4} f(θ)=sin4θf^{\prime}(\theta)=\sin 4 \theta f(θ)=sin4θ=32\Rightarrow f^{\prime}(\theta)=\sin 4 \theta=-\frac{\sqrt{3}}{2} 4θ=nπ+(1)nπ3\Rightarrow 4 \theta=\mathrm{n} \pi+(-1)^{\mathrm{n}} \frac{\pi}{3} θ=nπ4+(1)nπ12\Rightarrow \theta=\frac{\mathrm{n} \pi}{4}+(-1)^{\mathrm{n}} \frac{\pi}{12}

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