Question

Range of the function $f(x) = \log_2(2-\log_{\sqrt{2}}(16\sin^2 x + 1))$ is :

Answer 1

$(-\infty, 1]$

Answer 2

Let the given function be $f(x) = \log_2(2-\log_{\sqrt{2}}(16\sin^2 x + 1))$.

For the function to be defined, the following conditions must be met:

1. The argument of the inner logarithm must be positive: $16\sin^2 x + 1 > 0$. Since $\sin^2 x \ge 0$ for any real $x$, $16\sin^2 x \ge 0$, and thus $16\sin^2 x + 1 \ge 1$. This condition is always satisfied.

2. The argument of the outer logarithm must be positive: $2-\log_{\sqrt{2}}(16\sin^2 x + 1) > 0$. $2 > \log_{\sqrt{2}}(16\sin^2 x + 1)$. Since the base of the logarithm $\sqrt{2} = 2^{1/2} > 1$, the inequality can be rewritten as: $(\sqrt{2})^2 > 16\sin^2 x + 1$ $2 > 16\sin^2 x + 1$ $1 > 16\sin^2 x$ $\sin^2 x < \frac{1}{16}$.

This condition $\sin^2 x < \frac{1}{16}$ defines the domain of the function $f(x)$.

Now, let's determine the range of the expression $16\sin^2 x + 1$ for the values of $x$ satisfying $\sin^2 x < \frac{1}{16}$. We know that $0 \le \sin^2 x < \frac{1}{16}$. Multiplying by 16, we get $0 \le 16\sin^2 x < 16 \times \frac{1}{16} = 1$. Adding 1 to all parts of the inequality, we get $0 + 1 \le 16\sin^2 x + 1 < 1 + 1$. So, $1 \le 16\sin^2 x + 1 < 2$.

Let $u = 16\sin^2 x + 1$. The range of $u$ for the domain of $f(x)$ is $[1, 2)$.

Now consider the term $\log_{\sqrt{2}}(u)$. The base of the logarithm is $\sqrt{2} = 2^{1/2}$. $\log_{\sqrt{2}}(u) = \log_{2^{1/2}}(u) = \frac{\log_2 u}{\log_2 2^{1/2}} = \frac{\log_2 u}{1/2} = 2\log_2 u$.

Since $u \in [1, 2)$ and the base of the logarithm (2) is greater than 1, $\log_2 u$ is an increasing function. When $u=1$, $2\log_2(1) = 2 \times 0 = 0$. As $u$ approaches 2 from the left ($u \to 2^-$), $2\log_2 u$ approaches $2\log_2(2) = 2 \times 1 = 2$. So, the range of $\log_{\sqrt{2}}(16\sin^2 x + 1)$ is $[0, 2)$.

Let $v = \log_{\sqrt{2}}(16\sin^2 x + 1)$. The range of $v$ is $[0, 2)$. The argument of the outer logarithm is $2 - v$. Since $v \in [0, 2)$, the range of $-v$ is $(-2, 0]$. The range of $2 - v$ is $2 + (-2, 0] = (0, 2]$.

Let $w = 2 - \log_{\sqrt{2}}(16\sin^2 x + 1)$. The range of $w$ is $(0, 2]$.

Finally, the function is $f(x) = \log_2(w)$. Since $w \in (0, 2]$ and the base of the logarithm (2) is greater than 1, $\log_2 w$ is an increasing function. As $w$ approaches 0 from the right ($w \to 0^+$), $\log_2 w$ approaches $-\infty$. When $w = 2$, $\log_2 w = \log_2 2 = 1$. So, the range of $\log_2(w)$ for $w \in (0, 2]$ is $(-\infty, 1]$.

Therefore, the range of the function $f(x)$ is $(-\infty, 1]$.