Question

If $(1 + |\cos x|)^{\frac{|\tan x|}{b}} = (1 + |\cos x|)^{b(|\tan x|^2 - |\tan x|)}$, b > 0 where $\pi < x < \frac{3\pi}{2}$, then least integral value of btanx is

Answer 1

2

Answer 2

For $\pi < x < \frac{3\pi}{2}$, $x$ is in the third quadrant. In this quadrant, $\tan x > 0$ and $\cos x < 0$. Therefore, $|\tan x| = \tan x$ and $|\cos x| = -\cos x$. The given equation is $(1 + |\cos x|)^{\frac{|\tan x|}{b}} = (1 + |\cos x|)^{b(|\tan x|^2 - |\tan x|)}$. Since $\cos x \in (-1, 0)$, $|\cos x| \in (0, 1)$. Thus, the base $(1 + |\cos x|)$ is in the range $(1, 2)$, which is greater than 1. We can equate the exponents: $\frac{|\tan x|}{b} = b(|\tan x|^2 - |\tan x|)$ Substitute $|\tan x| = \tan x$: $\frac{\tan x}{b} = b(\tan^2 x - \tan x)$ Since $b > 0$ and $\tan x > 0$ in the given interval, we can divide both sides by $b \tan x$: $\frac{1}{b^2} = \tan x - 1$ $\tan x = 1 + \frac{1}{b^2}$ We need to find the least integral value of $b \tan x$. Substitute the expression for $\tan x$: $b \tan x = b \left(1 + \frac{1}{b^2}\right) = b + \frac{1}{b}$ By the AM-GM inequality, for $b > 0$, we have: $\frac{b + \frac{1}{b}}{2} \ge \sqrt{b \cdot \frac{1}{b}}$ $\frac{b + \frac{1}{b}}{2} \ge 1$ $b + \frac{1}{b} \ge 2$ The minimum value of $b + \frac{1}{b}$ is 2, which occurs when $b = \frac{1}{b}$, i.e., $b^2 = 1$. Since $b > 0$, this means $b=1$. Thus, the minimum value of $b \tan x$ is 2. The expression $b \tan x$ can take any value greater than or equal to 2. The least integral value in this range is 2.