Question

# If m and M are the minimum and the maximum values of $4 + \dfrac{1}{2}{\sin ^2}2x - 2{\cos ^4}x$, $x \in R$, then M-m is equal to:$(a){\text{ }}\dfrac{9}{4} \\ (b){\text{ }}\dfrac{{15}}{4} \\ (c){\text{ }}\dfrac{7}{4} \\ (d){\text{ }}\dfrac{1}{4} \\$

Hint – In this problem we have to find the subtraction of minimum and the maximum value of the given expression, first try and convert the given equation into perfect square form all into a single trigonometric ratio either cos or sin using various trigonometric identities and algebraic identities. Then use the range of that remaining single trigonometric ratio to get the answer.

Given equation is
$4 + \dfrac{1}{2}{\sin ^2}2x - 2{\cos ^4}x$
Now as we know $\sin 2x = 2\sin x\cos x,\;{\sin ^2}x = \left( {1 - {{\cos }^2}x} \right)$
So, substitute this value in above equation we have,
$\Rightarrow 4 + \dfrac{1}{2}{\left( {2\sin x\cos x} \right)^2} - 2{\cos ^4}x \\ \Rightarrow 4 + \dfrac{4}{2}{\sin ^2}x{\cos ^2}x - 2{\cos ^4}x \\ \Rightarrow 4 + 2\left( {1 - {{\cos }^2}x} \right){\cos ^2}x - 2{\cos ^4}x \\$
Now simplify the above equation we have,
$\Rightarrow 4 + 2{\cos ^2}x - 4{\cos ^4}x$
Now take (-4) common we have,
$\Rightarrow - 4\left( { - 1 - \dfrac{1}{2}{{\cos }^2}x + {{\cos }^4}x} \right)$
Now in bracket add and subtract by $\dfrac{1}{{16}}$ we have,
$\Rightarrow - 4\left( { - 1 - \dfrac{1}{2}{{\cos }^2}x + {{\cos }^4}x + \dfrac{1}{{16}} - \dfrac{1}{{16}}} \right)$
Now make a complete square we have,
$\Rightarrow - 4\left[ {{{\left( {{{\cos }^2}x - \dfrac{1}{4}} \right)}^2} - \dfrac{{17}}{{16}}} \right]$
Now as we know $0 \leqslant {\cos ^2}x \leqslant 1$
Now subtract by $\dfrac{1}{4}$ in above equation we have,
$\dfrac{{ - 1}}{4} \leqslant {\cos ^2}x - \dfrac{1}{4} \leqslant 1 - \dfrac{1}{4}$
$\dfrac{{ - 1}}{4} \leqslant {\cos ^2}x - \dfrac{1}{4} \leqslant \dfrac{3}{4}$
Now squaring on both sides we have, when we square the extreme L.H.S becomes zero
$0 \leqslant {\left( {{{\cos }^2}x - \dfrac{1}{4}} \right)^2} \leqslant {\left( {\dfrac{3}{4}} \right)^2}$
$0 \leqslant {\left( {{{\cos }^2}x - \dfrac{1}{4}} \right)^2} \leqslant \dfrac{9}{{16}}$
Now subtract by $\dfrac{{ - 17}}{{16}}$ in the above equation we have,
$- \dfrac{{17}}{{16}} \leqslant {\left( {{{\cos }^2}x - \dfrac{1}{4}} \right)^2} - \dfrac{{17}}{{16}} \leqslant \dfrac{9}{{16}} - \dfrac{{17}}{{16}}$
Now simplify the above equation we have,
$- \dfrac{{17}}{{16}} \leqslant {\left( {{{\cos }^2}x - \dfrac{1}{4}} \right)^2} - \dfrac{{17}}{{16}} \leqslant \dfrac{{ - 1}}{2}$
Now multiply by (-4) throughout we have, (when we multiply by negative value the inequality sign changes).
$- 4\left( { - \dfrac{{17}}{{16}}} \right) \geqslant - 4\left[ {{{\left( {{{\cos }^2}x - \dfrac{1}{4}} \right)}^2} - \dfrac{{17}}{{16}}} \right] \geqslant - 4\left( {\dfrac{{ - 1}}{2}} \right)$
Now simplify the above equation we have,
$\left( {\dfrac{{17}}{4}} \right) \geqslant - 4\left[ {{{\left( {{{\cos }^2}x - \dfrac{1}{4}} \right)}^2} - \dfrac{{17}}{{16}}} \right] \geqslant 2$
So, from the above equation it is clear that the minimum value (m) =2, and the maximum value (M) = $\dfrac{{17}}{4}$ of the given equation.
So the value of (M-m) is
$\Rightarrow \left( {M - m} \right) = \dfrac{{17}}{4} - 2 = \dfrac{9}{4}$.
Hence option (a) is correct.

Note – Whenever we face such type of problems there can be two ways first one is being explained above however the another method is a bit lengthy and it involves the concept of maxima and minima by single differentiating first to get the values at which max or minima can occur and then double differentiating to be sure that whether it’s a max or min. Both of these concepts will help you get on the right track to reach the answer.