Question

The maximum value of $4{\sin }^2\left(x\right)-12\sin \left(x\right)+7$ is

• A. $25$
• B. $4$
• C. Does not exist
• D. None of these

Answer 1

The correct option is D

None of these

Explanation for the correct option:

We can rewrite the given expression, $4{\sin }^2\left(x\right)-12\sin \left(x\right)+7$ as :

$=4\left[{\sin }^2\left(x\right)–3\sin \left(x\right)\right]+7$

$=4\left[{\sin }^2\left(x\right)–2·\frac{3}{2}\sin \left(x\right)+{\left(\frac{3}{2}\right)}^2-{\left(\frac{3}{2}\right)}^2\right]+7$

$=4\left[{\left(\sin \left(x\right)–\frac{3}{2}\right)}^2–{\left(\frac{3}{2}\right)}^2\right]+7$

$=4\left[{\left(\sin \left(x\right)–\frac{3}{2}\right)}^2\right]–4·\left(\frac{9}{4}\right)+7$

$=4{\left(\sin \left(x\right)–\frac{3}{2}\right)}^2–2$

We know that the range of $\sin \left(x\right)$ is $[-1,1]$.

$-1\le \sin \left(x\right)\le 1$

Subtracting $\frac{3}{2}$ from all sides,

$-1-\frac{3}{2}\le \sin \left(x\right)-\frac{3}{2}\le 1-\frac{3}{2}$

$-\frac{5}{2}\le \sin \left(x\right)–\frac{3}{2}\le -\frac{1}{2}$

Squaring all the sides,

$\frac{1}{4}\le {\left(\sin \left(x\right)–\frac{3}{2}\right)}^2\le \frac{25}{4}$

Multiplying by $4$ on all side,

$1\le 4{\left[\sin \left(x\right)–\frac{3}{2}\right]}^2\le 25$

and subtracting $2$ from all sides,

$-1\le 4{\left[\sin \left(x\right)–\frac{3}{2}\right]}^2-2\le 23$

Thus, the maximum value is $23$.

Hence, $\mathrm{Option}\left(\mathrm{D}\right)$ is the correct answer.