Question

Arrange the following in descending order:
$A)$ Min.value of $\sin 4x\cos 4x$

$B)$ Min. value of ${\cos }^{2}x$
$C)$ Max. value of $1-8{\sin }^{2}x{\cos }^{2}x$
$D)$ Max.value of $\sqrt{3}\sin x-\cos x$

• A. $D,C,B,A$
• B. $A,B,C,D$
• C. $A,D,B,C$
• D. $C,D,B,A$

Answer 1

Answer: (A)

$D,C,B,A$

A.
We know $\sin 2x=2\sin x\cos x$
$\sin 4x\cos 4x=\frac{\sin 8x}{2}$
minimum of $\sin 8x=-1.$
minimum of $\frac{\sin 8x}{2}=-\frac{1}{2}$
B.
${\cos }^{2}x$
We know $-1\le \cos x\le 1$
Hence minimum of ${\cos }^{2}x=0.$
C.
We know $\sin 2x=2\sin x\cos x$
$f\left(x\right)=1-8{\sin }^{2}x{\cos }^{2}x=1-2(\sin 2x{)}^2$
maximum of $f\left(x\right)=1-0=1$
D.
Maximum value of $a\sin x+b\cos x+c$ is $\sqrt{a^2+b^2}+c$
$f\left(x\right)=\sqrt{3}\sin x-\cos x$
maximum value value of $f\left(x\right)=\sqrt{3+1}=2$
So, decending order of $A,B,C,D$
$\Rightarrow D,C,B,A$.
$\Rightarrow 2,1,0,-\frac{1}{2}$