Question

The Maximum value of $4{\sin ^2}x - 12\sin x + 7$ is
(A) 25
(B) 4
(C) Does not exist
(D) None of the above

Answer 1

First we have to change the equation in ${(a + b)^2}$ and we know that the value of $\sin x$ is varies from $- 1$ to $1$ by using this concept we can solve this question.

Complete step-by-step answer:
It is given that we have to find the maximum value of $4{\sin ^2}x - 12\sin x + 7$
Now we take common $4$ from $4{\sin ^2}x + 12\sin x$
we get : $4({\sin ^2}x - 3\sin x) + 7$
Now we have to make this equation in ${(a - b)^2}$ form
So we have multiple and divide $2$ in the term $3\sin x$ and
Add and subtract ${\left( {\dfrac{3}{2}} \right)^2}$ in the equation as :
$4\left( {{{\sin }^2}x - 2 \times \dfrac{3}{2} \times \sin x + {{\left( {\dfrac{3}{2}} \right)}^2} - \left( {\dfrac{3}{2}} \right)}^2 \right) + 7$

We know that ${(a - b)^2}$ $= {a^2} - 2ab + {b^2}$
So it is changes as $\left[ {4\left( {{{\left( {\sin x - \dfrac{3}{2}} \right)}^2} - \dfrac{9}{4}} \right) + 7} \right]$
Now solving further we get $\left[ {\left( {4{{\left( {\sin x + \dfrac{3}{2}} \right)}^2} - 9} \right) + 7} \right]$
And finally we get
$\left[ {4{{\left( {\sin x - \dfrac{3}{2}} \right)}^2} - 2} \right]$
Now we have to find out the maximum value of this
We know that
$- 1 \leqslant \sin x \leqslant 1$ for any value of $x$
Subtract $\dfrac{3}{2}$ in it
$- 1 - \dfrac{3}{2} \leqslant \sin x - \dfrac{3}{2} \leqslant 1 - \dfrac{3}{2}$
or
$- \dfrac{5}{2} \leqslant \sin x - \dfrac{3}{2} \leqslant - \dfrac{1}{2}$
On squaring all the terms the sign of fraction will change ;
$\dfrac{1}{4} \leqslant {\left( {\sin x - \dfrac{3}{2}} \right)^2} \leqslant \dfrac{{25}}{4}$
On multiplying by $4$ we get
$1 \leqslant 4{\left( {\sin x - \dfrac{3}{2}} \right)^2} \leqslant 25$
Now subtract $2$ from all sides
$- 1 \leqslant 4{\left( {\sin x - \dfrac{3}{2}} \right)^2} - 2 \leqslant 23$
Hence the maximum value of $\left[ {4{{\left( {\sin x - \dfrac{3}{2}} \right)}^2} - 2} \right]$ is $23$
or maximum value of $4{\sin ^2}x - 12\sin x + 7$ is $23$

So, the correct answer is “Option D”.

Note: Always be careful when you multiply negative sign in inequality the sign of equation is changed .Same method will be used if we replace the $\sin x$ to $\cos x$ or If question is given in any other form like $\tan x,\cot x,\cos ecx,$ try to convert it into in $\sin x$ and $\cos x$ form .