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The function $f(x)=4\sin^3x-6\sin^2x+12sin x+100$ is strictly

\[\begin{array}{1 1}(A)\;increasing\;in\bigg(\pi,\frac{3\pi}{2}\bigg) & (B)\;decreasing in \bigg(\frac{\pi}{2},\pi\bigg)\\(C)\;decreasing\; in\bigg(\frac{-1}{2},\frac{1}{2}\bigg) & (D)\;decreasing \;in\bigg(0,\frac{1}{2}\bigg)\end{array}\]

1 Answer

  • Let $f(x)$ be a function defined on $(a,b)$ if $f'(x) <0$ for all $x \in (a,b)$ except for a finite number of points where $f'(x)=0$, then $f(x) $ is decreasing on $(a,b)$
Step 1
$f(x)=4 \sin^3x-6 \sin^2x+ 12 \sin x+100$
differentiating w.r.t $x$ we get,
$ f'(x)=12. \sin^2 x . \cos x-12 \sin x.\cos x+ 12 \cos x$
If $ f'(x)=0$
$ \Rightarrow 12 \cos x [ \sin^2x- \sin x+ 1]=0$
$ \Rightarrow 12 \cos x ( \sin^2 x - \sin x+1)=0$
Step 2
When $ x = \large\frac{\pi}{2},\: \: f'(x)=0$
when $x = \pi, \: \: \: \: f'(x)=-12$
We know cosine ratio in the second quadrant is negative and sin ratio is positive.
Hence its clear that the function is strictly decreasing in the interval.
$ \bigg( \large\frac{\pi}{2}, \pi \bigg)$
Hence B is the correct answer.
answered Aug 11, 2013 by thanvigandhi_1