Question

If $y=2+\frac{2}{x+1}+\frac{1}{x-1},$ then

• A. domain of the function is $R-\left\{1\right\}$
• B. domain of the function is $R-\{-1,1\}$
• C. range of the function is $R-\left\{2\right\}$
• D. range of the function is $R$

Answer 1

Answer: (B), (D)

The correct options are
B domain of the function is $R-\{-1,1\}$
D range of the function is $R$

$y=2+\frac{2}{x+1}+\frac{1}{x-1}$
Clearly, function is not defined at $x=\pm 1$
Hence, domain of the function is $R-\{-1,1\}$

Now, $y-2=\frac{2}{x+1}+\frac{1}{x-1}$
$\Rightarrow y-2=\frac{3x-1}{x^2-1}...\left(1\right)$
$\Rightarrow (y-2)x^2-3x-(y-3)=0$

For real solutions, $\Delta \ge 0$ assuming $y-2\ne 0$
$\Rightarrow 9+4(y-2)(y-3)\ge 0$
$\Rightarrow 4y^2-20y+33\ge 0$
which is always true as $\Delta <0$

Now, put $y=2$ in equation $\left(1\right),$ we get
$x=\frac{1}{3}\in R-\{-1,1\}$
Therefore, $y=2$ also lies in the range.
Hence, range is $R.$