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Question

Let f (x) = $$log_2$$ $$\left (\left | sin x  \right | +\left | cos x \right |\right )$$,
then range of f(x) is


A
0,12
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B
0,3
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C
12,0
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D
0,1
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Solution

The correct option is A $$0, \dfrac{1}{2}$$
Here, $$f(x)=\log_2(|\sin x|+|cos x|)$$
Let $$t=|\sin x|+|\cos x|$$ Then $$f(x)=\log_2t$$......(1)
Squaring both sides we get
$$t^2=(|\sin x|)^2+(|\cos x|)^2+|i\sin 2x|$$
$$=\sin^2x+\cos^2x+|\sin 2x|$$
$$t^2=1+|\sin 2x|$$      $$(\because \sin^2x+\cos^2x=1)$$
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 $$f(x)$$ :-
$$f(x)=\log_2t$$ and $$t\in [1, \sqrt 2]$$
i.e. $$1\le t\le \sqrt 2$$
Take $$\log_2$$ we get.
$$\Rightarrow \log_21\le \log_2 t\le \log_2 \sqrt 2$$
$$\Rightarrow 0\le f(x) \le \log_2 2^{1/2}$$      $$\left(\because \log 1=0\\ \log_2 2=1\right.$$
$$\Rightarrow 0\le f(x) \le \dfrac{1}{2}\log_2 2$$
$$\therefore 0\le f(x) \le \dfrac{1}{2}$$
$$\therefore$$ option $$A$$ is correct
ie $$\left[0, \dfrac{1}{2}\right]$$.

Mathematics

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