Question

Let $(1+k){\tan }^{2}x-4\tan x+k-1=0$ have real roots $\tan x_1,\tan x_2(\tan x_1>\tan x_2).$ Then

• A. $k^2<5$
• B. $\tan (x_1+x_2)=2$
• C. for $k=2,x_1=\frac{\pi }{4}$
• D. for $k=1,x_2=0$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A $k^2<5$
B $\tan (x_1+x_2)=2$
C for $k=2,x_1=\frac{\pi }{4}$
D for $k=1,x_2=0$

$(1+k){\tan }^{2}x-4\tan x+k-1=0$
For real and distinct roots, $D>0$
$\Rightarrow k^2<5$

$\tan x_1+\tan x_2=\frac{4}{k+1}\cdots \left(1\right)$
$\tan x_1\cdot \tan x_2=\frac{k-1}{k+1}\cdots \left(2\right)$

$\tan (x_1+x_2)=\frac{\tan x_1+\tan x_2}{1-\tan x_1\cdot \tan x_2}$
$\Rightarrow \tan (x_1+x_2)=2$

Now, $(\tan x_1-\tan x_2{)}^2=(\tan x_1+\tan x_2{)}^2-4\tan x_1\cdot \tan x_2$
$\Rightarrow \tan x_1-\tan x_2=\frac{2\sqrt{5-k^2}}{k+1}$
If $k=2,\tan x_1-\tan x_2=\frac{2}{3}\cdots \left(3\right)$
$\left(1\right)+\left(3\right)\Rightarrow \tan x_1=1$
$\therefore x_1=\frac{\pi }{4}$

If $k=1,\tan x_1-\tan x_2=2\cdots \left(4\right)$
$\left(1\right)-\left(4\right)\Rightarrow \tan x_2=0$
$\therefore x_2=0$