Functions 2 Question 16
17.
If $f(x)=\sin ^{2} x+\sin ^{2} x+\frac{\pi}{3}+\cos x \cos x+\frac{\pi}{3} \quad$ and g $\frac{5}{4}=1$, then $(g \circ f)(x)=\ldots$.
(1996, 2M)
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Answer:
Correct Answer: 17. (1)
Solution:
- $f(x)=\sin ^{2} x+\sin ^{2}(x+\pi / 3)+\cos x \cos (x+\pi / 3)$
$\Rightarrow f(x)=\sin ^{2} x+(\sin x \cos \pi / 3+\cos x \sin \pi / 3)^{2}$ $+\cos x \cos (x+\pi / 3)$
$+\cos x(\cos x \cos \pi / 3-\sin x \sin \pi / 3)$
$\Rightarrow f(x)=\sin ^{2} x+\frac{\sin ^{2} x}{4}+\frac{3 \cos ^{2} x}{4}+\frac{2 \cdot \sqrt{3}}{4} \sin x \cos x$ $+\frac{\cos ^{2} x}{2}-\cos x \sin x \cdot \frac{\sqrt{3}}{2}$
$=\sin ^{2} x+\frac{\sin ^{2} x}{4}+\frac{3 \cos ^{2} x}{4}+\frac{\cos ^{2} x}{2}$
$ =\frac{5}{4} \sin ^{2} x+\frac{5}{4} \cos ^{2} x=\frac{5}{4} $
and $\quad g \circ f(x)=g{f(x)}=g(5 / 4)=1$
Alternate Solution
$f(x)=\sin ^{2} x+\sin ^{2}(x+\pi / 3)+\cos x \cos (x+\pi / 3)$
$\Rightarrow f^{\prime}(x)=2 \sin x \cos x+2 \sin (x+\pi / 3) \cos (x+\pi / 3)$
$-\sin x \cos (x+\pi / 3)-\cos x \sin (x+\pi / 3)$
$=\sin 2 x+\sin (2 x+2 \pi / 3)-[\sin (x+x+\pi / 3)]$
$=2 \sin \frac{2 x+2 x+2 \pi / 3}{2} \cdot \cos \frac{2 x-2 x-2 \pi / 3}{2}$
$-\sin (2 x+\pi / 3)$
$=2[\sin (2 x+\pi / 3) \cdot \cos \pi / 3]-\sin (2 x+\pi / 3)$
$=2 \sin (2 x+\pi / 3) \cdot \frac{1}{2}-\sin 2 x+\frac{\pi}{3}=0$
$\Rightarrow f(x)=c$, where $c$ is a constant.
But $f(0)=\sin ^{2} 0+\sin ^{2}(\pi / 3)+\cos 0 \cos \pi / 3$
Therefore, $(g \circ f)(x)=g[f(x)]=g(5 / 4)=1$
