Question

The function f(x) = 4sin³x - 6sin²x + 12sin x + 100 is strictly
(a) increasing in $\left[\mathrm{\pi },\frac{3\mathrm{\pi }}{2}\right]$ (b) decreasing in $\left(\frac{\mathrm{\pi }}{2},\mathrm{\pi }\right)$
(c) decreasing in $\left[-\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right]$ (d) decreasing in $\left[0,\frac{\mathrm{\pi }}{2}\right]$

Answer 1

The given function is f(x) = 4sin³x − 6sin²x + 12sinx + 100.

f(x) = 4sin³x − 6sin²x + 12sinx + 100

Differentiating both sides with respect to x, we get

$f\text{'}\left(x\right)=4\times 3{\sin }^{2}x\times \cos x-6\times 2\sin x\times \cos x+12\cos x$

$f\text{'}\left(x\right)=12{\sin }^{2}x\cos x-12\sin x\cos x+12\cos x$

$\Rightarrow f\text{'}\left(x\right)=12\cos x\left({\sin }^{2}x-\sin x+1\right)$

$\Rightarrow f\text{'}\left(x\right)=12\cos x\left[{\left(\sin x-\frac{1}{2}\right)}^2+\frac{3}{4}\right]$

Now,

${\left(\sin x-\frac{1}{2}\right)}^2+\frac{3}{4}>0$ ∀ x ∈ R

When $x\in \left[-\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right]$, cosx ≥ 0

$\therefore f\text{'}\left(x\right)\ge 0$

So, f(x) is increasing in $\left[-\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right]$.

When $x\in \left[0,\frac{\mathrm{\pi }}{2}\right]$, cosx ≥ 0

$\therefore f\text{'}\left(x\right)\ge 0$

So, f(x) is increasing in $\left[0,\frac{\mathrm{\pi }}{2}\right]$.

When $x\in \left[\mathrm{\pi },\frac{3\mathrm{\pi }}{2}\right]$, cosx ≤ 0

$\therefore f\text{'}\left(x\right)\le 0$

So, f(x) is decreasing in $\left[\mathrm{\pi },\frac{3\mathrm{\pi }}{2}\right]$.

When $x\in \left(\frac{\mathrm{\pi }}{2},\mathrm{\pi }\right)$, cosx < 0

$\therefore f\text{'}\left(x\right)<0$

So, f(x) is strictly decreasing in $\left(\frac{\mathrm{\pi }}{2},\mathrm{\pi }\right)$.

Thus, the function f(x) = 4sin³x − 6sin²x + 12sin x + 100 is strictly decreasing in $\left(\frac{\mathrm{\pi }}{2},\mathrm{\pi }\right)$.

Hence, the correct answer is option (b).