Question

Let f (x) = $lo{g}_2$ $(\left|sinx\right|+\left|cosx\right|)$,
then range of f(x) is

• A. $0,\frac{1}{2}$
• B. $0,3$
• C. $-\frac{1}{2},0$
• D. $0,1$

Answer 1

The correct option is A $0,\frac{1}{2}$

Here, $f\left(x\right)={\log }_2(|\sin x|+|cosx|)$

Let $t=|\sin x|+|\cos x|$ Then $f\left(x\right)={\log }_{2}t$......(1)

Squaring both sides we get

$t^2=(|\sin x|{)}^2+(|\cos x|{)}^2+|i\sin 2x|$

$={\sin }^{2}x+{\cos }^{2}x+|\sin 2x|$

$t^2=1+|\sin 2x|$ $(\because {\sin }^{2}x+{\cos }^{2}x=1)$

Here, $|\sin 2x|\in [0,1]$ ( Clearly $|\sin x|\in [0,1]$ so )

$\therefore t=\sqrt{1+|\sin 2x|}$ $t_{max}=\sqrt{1+1}=\sqrt{2}$

$\therefore t_{min}m=\sqrt{1+0}=1$

$\therefore t\in [1,\sqrt{2}]$.........(2)

Now, Using (2) in eqn (1), we shall find the range of $f\left(x\right)$ :-

$f\left(x\right)={\log }_{2}t$ and $t\in [1,\sqrt{2}]$

i.e. $1\le t\le \sqrt{2}$

Take ${\log }_2$ we get.

$\Rightarrow {\log }_{2}1\le {\log }_{2}t\le {\log }_2\sqrt{2}$

$\Rightarrow 0\le f\left(x\right)\le {\log }_2{2}^{1/2}$ $(\because \log 1=0{\log }_{2}2=1$

$\Rightarrow 0\le f\left(x\right)\le \frac{1}{2}{\log }_{2}2$

$\therefore 0\le f\left(x\right)\le \frac{1}{2}$

$\therefore$ option $A$ is correct

ie $[0,\frac{1}{2}]$.