f(x) = log21/2(2-log2(16sin2x+1)) : Find the range of the function - Maths - Relations and Functions

Question

f(x) = log21/2
(2-log2(16sin2x+1)) : Find the range of the
function

Ajanta Trivedi · Accepted Answer

the given function is: $f (x) = \log_{\sqrt{2}} (2 - \log_{2} (16 \sin^{2} x + 1))$

let us first find the domain of the function:

$2 - \log_{2} (16 \sin^{2} x + 1) > 0 \log_{2} (16 \sin^{2} x + 1) < 2 16 \sin^{2} x + 1 < 2^{2} 16 \sin^{2} x < 4 - 1 16 \sin^{2} x < 3$

$\sin^{2} x < \frac{3}{16} ............ (1) - \frac{\sqrt{3}}{4} < \sin x < \frac{\sqrt{3}}{4}$

by the eq(1)

$1 \leq 16 \sin^{2} x + 1 < 4 \log_{2} 1 \leq \log_{2} (16 \sin^{2} x + 1) < \log_{2} 4 0 \leq \log_{2} (16 \sin^{2} x + 1) < 2 - 2 < - \log_{2} (16 \sin^{2} x + 1) \leq 0$

$0 < 2 - \log_{2} (16 \sin^{2} x + 1) \leq 2 - \infty < \log_{2} [2 - \log_{2} (16 \sin^{2} x + 1)] \leq 1 .............. (2) f (x) = \log_{\sqrt{2}} [2 - \log_{2} (16 \sin^{2} x + 1)] = 2 \log_{2} [2 - \log_{2} (16 \sin^{2} x + 1)] \Rightarrow - \infty < f (x) \leq 2$

thus range of the given function is: $f (x) \in (- \infty, 2]$

hope this helps you.

cheers!!

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