Question

The interval of $a$ for which the equation ${\tan }^{2}x-(a-4)\tan x+4-2a=0$ has at least one solution such that $x\in [0,\pi /4]$ is

• A. $a\in (2,3)$
• B. $a\in [2,3]$
• C. $a\in (1,4)$
• D. $a\in [1,4]$

Answer 1

The correct option is B $a\in [2,3]$

Given ${\tan }^{2}x-(a-4)\tan x+4-2a=0$

we know roots of a quadratic equation $a{x}^2+bx+c=0$

$x=\frac{-b\pm \left(\sqrt{b^2-4ac}\right)}{2a}$

similarly for given equation

$tanx=\frac{a-4-\sqrt{(a-4{)}^2-4(4-2a)}}{2}$

$tanx=\frac{a-4-a}{2}=a-2,$

$\therefore tanx=a-2$ and $tanx\ne -2$ $\dots [\because x\in [0,\frac{\pi }{4}]\Rightarrow \tan \theta ispositive]$

$\because x\in [0,\frac{\pi }{4}]$

$\therefore 0\le a-2\le 1$

$\Rightarrow 2\le a\le 3$