Question

If m,M are the minimum and the maximum value of $4+\frac{1}{2}si{n}^{2}2x-2co{s}^{4}x,xϵR,$then M -m is equal to:

• A. $\frac{9}{4}$
• B. $\frac{17}{4}$
• C. $\frac{7}{4}$
• D. $\frac{1}{4}$

Answer 1

The correct option is B $\frac{17}{4}$

Let $y=4+\frac{1}{2}{sin}^{2}2x-2{cos}^{4}x$

$\therefore y=4+\frac{1}{2}{\left(2sinxcosx\right)}^2-2{\left({cos}^{2}x\right)}^2$

$\therefore y=4+\frac{1}{2}\times 4{sin}^{2}x.{cos}^{2}x-2{\left({cos}^{2}x\right)}^2$ (1)

Put ${cos}^{2}x=t$

$\therefore {sin}^{2}x=1-t$

Equation (1) becomes,

$y=4+2(1-t)t-2{\left(t\right)}^2$

$\therefore y=4+2(t-t^2)-2t^2$

$\therefore y=4+2t-2t^2-2t^2$

$\therefore y=4+2t-4t^2$

For y to be maximum or minimum, differentiate y w.r.t. x and equate it to zero.

$\therefore \frac{dy}{dx}=\frac{d}{dx}[4+2t-4t^2]=0$

$\therefore 2-8t=0$

$\therefore 2=8t$

$\therefore t=\frac{2}{8}$

$\therefore {cos}^{2}x=\frac{1}{4}$

$\therefore cosx=\pm \frac{1}{2}$

$\therefore cosx=\frac{1}{2}$ or $cosx=-\frac{1}{2}$

Consider first root $cosx=\frac{1}{2}$

$\therefore x={cos}^{-1}\frac{1}{2}$

$\therefore x=\frac{\pi }{3}$

Consider first root $cosx=-\frac{1}{2}$

$\therefore cosx=-cos\frac{\pi }{3}$

$\therefore cosx=cos(\pi -\frac{\pi }{3})$

$\therefore cosx=cos\left(\frac{2\pi }{3}\right)$

$\therefore x=\frac{2\pi }{3}$

1) For $x=\frac{\pi }{3}$,

$y=4+\frac{1}{2}{\left(sin\frac{2\pi }{3}\right)}^2-2{\left(cos\frac{\pi }{3}\right)}^4$

$\therefore y=4+\frac{1}{2}{\left(\frac{\sqrt{3}}{2}\right)}^2-2{\left(\frac{1}{2}\right)}^4$

$\therefore y=4+\frac{1}{2}\left(\frac{3}{4}\right)-2\left(\frac{1}{16}\right)$

$\therefore y=4+\frac{3}{8}-\frac{1}{8}$

$\therefore y=4+\frac{2}{8}$

$\therefore y=4+\frac{1}{4}=\frac{17}{4}$

2) For $x=\frac{2\pi }{3}$, we get,

$y=4+\frac{1}{2}{\left(sin\frac{4\pi }{3}\right)}^2-2{\left(cos\frac{2\pi }{3}\right)}^4$

$\therefore y=4+\frac{1}{2}{(-\frac{\sqrt{3}}{2})}^2-2{(-\frac{1}{2})}^4$

$\therefore y=4+\frac{1}{2}\left(\frac{3}{4}\right)-2\left(\frac{1}{16}\right)$

$\therefore y=4+\left(\frac{3}{8}\right)-\left(\frac{1}{8}\right)$

$\therefore y=4+\frac{1}{4}$

$\therefore y=\frac{17}{4}$

Now, $\frac{d^{2}y}{{dx}^2}=\frac{d}{dx}\left(\frac{dy}{dx}\right)$

$\therefore \frac{d^{2}y}{{dx}^2}=\frac{d}{dx}(2-8t)$

$\therefore \frac{d^{2}y}{{dx}^2}=-8<0$

Thus, function is maximum and no minimum value exists.

$\therefore M-m=\frac{17}{4}$