Question

If range of the function $f(x) = log_{\frac{1}{\sqrt{3}}}(\sqrt{2}(sinx+cosx)+3)$ is $[a,b]$ then find value of $(a^2+b^2)$.

Answer 1

$4(log_3 5)^2$

Answer 2

The given function is $f(x) = log_{\frac{1}{\sqrt{3}}}(\sqrt{2}(sinx+cosx)+3)$.

Let the argument of the logarithm be $g(x) = \sqrt{2}(sinx+cosx)+3$.

First, we simplify the trigonometric expression $sinx+cosx$. We can write $sinx+cosx = \sqrt{1^2+1^2}\left(\frac{1}{\sqrt{2}}sinx + \frac{1}{\sqrt{2}}cosx\right) = \sqrt{2}\left(sinx \cos\left(\frac{\pi}{4}\right) + cosx \sin\left(\frac{\pi}{4}\right)\right) = \sqrt{2}\sin\left(x+\frac{\pi}{4}\right)$.

So, $\sqrt{2}(sinx+cosx) = \sqrt{2} \cdot \sqrt{2}\sin\left(x+\frac{\pi}{4}\right) = 2\sin\left(x+\frac{\pi}{4}\right)$.

Now, the argument of the logarithm is $g(x) = 2\sin\left(x+\frac{\pi}{4}\right)+3$.

The range of $\sin(\theta)$ for any real $\theta$ is $[-1, 1]$.

Thus, the range of $\sin\left(x+\frac{\pi}{4}\right)$ is $[-1, 1]$.

The range of $2\sin\left(x+\frac{\pi}{4}\right)$ is $[2(-1), 2(1)] = [-2, 2]$.

The range of $g(x) = 2\sin\left(x+\frac{\pi}{4}\right)+3$ is $[-2+3, 2+3] = [1, 5]$.

For the logarithm $f(x) = log_{\frac{1}{\sqrt{3}}}(g(x))$ to be defined, the argument $g(x)$ must be positive. The range of $g(x)$ is $[1, 5]$, and all values in this range are positive, so the logarithm is defined for all real $x$.

The base of the logarithm is $b = \frac{1}{\sqrt{3}}$. Since $0 < \frac{1}{\sqrt{3}} < 1$, the logarithm function $y = log_b(u)$ is a strictly decreasing function of $u$.

This means that the minimum value of $g(x)$ corresponds to the maximum value of $f(x)$, and the maximum value of $g(x)$ corresponds to the minimum value of $f(x)$.

The minimum value of $g(x)$ is $g_{min} = 1$.

The maximum value of $g(x)$ is $g_{max} = 5$.

The maximum value of $f(x)$ is $f_{max} = log_{\frac{1}{\sqrt{3}}}(g_{min}) = log_{\frac{1}{\sqrt{3}}}(1)$.

Since $log_b(1) = 0$ for any valid base $b$, $f_{max} = 0$.

The minimum value of $f(x)$ is $f_{min} = log_{\frac{1}{\sqrt{3}}}(g_{max}) = log_{\frac{1}{\sqrt{3}}}(5)$.

We can write the base $\frac{1}{\sqrt{3}}$ as $3^{-1/2}$.

$f_{min} = log_{3^{-1/2}}(5)$. Using the property $log_{b^k}(m) = \frac{1}{k}log_b(m)$, we get:

$f_{min} = \frac{1}{-1/2} log_3(5) = -2 log_3(5)$.

The range of the function $f(x)$ is $[f_{min}, f_{max}] = [-2 log_3(5), 0]$.

The problem states that the range is $[a,b]$.

Comparing the ranges, we have $a = -2 log_3(5)$ and $b = 0$.

We need to find the value of $a^2+b^2$.

$a^2 = (-2 log_3(5))^2 = (-2)^2 (log_3(5))^2 = 4 (log_3(5))^2$.

$b^2 = 0^2 = 0$.

$a^2+b^2 = 4 (log_3(5))^2 + 0 = 4 (log_3(5))^2$.

The final answer is $4(log_3 5)^2$.

Explanation of the solution:

1. Simplify $\sqrt{2}(sinx+cosx)$ to $2sin(x+\pi/4)$.
2. Find the range of the argument of the logarithm: $g(x) = 2sin(x+\pi/4)+3$. The range of $sin(x+\pi/4)$ is $[-1,1]$, so the range of $g(x)$ is $[2(-1)+3, 2(1)+3] = [1,5]$.
3. The base of the logarithm $1/\sqrt{3}$ is between 0 and 1, so the logarithm is a decreasing function.
4. The range of $f(x) = log_{1/\sqrt{3}}(g(x))$ is $[log_{1/\sqrt{3}}(max(g(x))), log_{1/\sqrt{3}}(min(g(x)))] = [log_{1/\sqrt{3}}(5), log_{1/\sqrt{3}}(1)]$.
5. Calculate the values: $log_{1/\sqrt{3}}(1) = 0$ and $log_{1/\sqrt{3}}(5) = log_{3^{-1/2}}(5) = -2log_3(5)$.
6. The range is $[-2log_3(5), 0]$. Thus, $a = -2log_3(5)$ and $b = 0$.
7. Calculate $a^2+b^2 = (-2log_3(5))^2 + 0^2 = 4(log_3(5))^2$.

Answer: $4(log_3 5)^2$