Question

the function $f:R\to R$ defined as
$f\left(x\right)=\frac{1}{2}ln{\left(\sqrt{\sqrt{x^2+1}+x}\sqrt{\sqrt{x^2+1}-x}\right)}^2$is

• A. One -one and onto both
• B. One-one but not onto
• C. Onto but not one-one
• D. Niether one-one nor onto

Answer 1

Answer: (D)

Niether one-one nor onto

Given,

$f\left(x\right)=\frac{1}{2}\ln (\sqrt{\sqrt{x^2+1}+x}+\sqrt{\sqrt{x^2+1}-x}$

Now putting

$x\to (-x)$

We get,

$f(-x)=\frac{1}{2}\ln (\sqrt{\sqrt{(-x{)}^2+1}+x}+\sqrt{\sqrt{(-x{)}^2+1}}-f-x)$

$f(-x)=\frac{1}{2}\ln (\sqrt{\sqrt{x^2+1}-x}+\sqrt{\sqrt{x^2+1}+x}$

$\because f\left(x\right)=f(-x)$

Therefore $f\left(x\right)$ is an even function.

and even functions are many on 0 .

Now we know,

$\sqrt{x^2+1}>0;x\in R$

and

$\sqrt{\sqrt{x^2+1}+x}+\sqrt{\sqrt{x^2+1}-x}>1$

$\therefore f\left(x\right)>0$

- Given, $f:R\to R.$

and $co$ domain $\ne R$.

Option D function is neither onto nor one one.