Question

Let $f:(-\infty ,\infty )\to [2,\infty )$ be a function defined by $f\left(x\right)=\sqrt{x^2-2a+a^2},a\in R,$ then the value of a for which $f$ is onto

• A. $1$
• B. $\sqrt{3}$
• C. $1\pm \sqrt{3}$
• D. $1\pm \sqrt{5}$

Answer 1

The correct option is D $1\pm \sqrt{5}$

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

$\therefore f\left(x\right)\in \left[2\infty \right)$

$\Rightarrow x^2-2a+a^2⩾4$

$\Rightarrow x^2+a^2-2a-4⩾0$

$\Rightarrow x^2+(a^2-2a-4)⩾0$

It is possible iff $a^2-2a-4⩾0$ because

$x^2$ is already non negative.

$a^2-2a-4⩾0$

$\Rightarrow a^2-2a+1-5⩾0$

$\Rightarrow (a-1{)}^2-(\sqrt{5}{)}^2⩾0\Rightarrow (a-1-\sqrt{5})(a-1+\sqrt{5})⩾0$

$a\in (-\infty ,-1-\sqrt{5}]\cup [1+\sqrt{5}\infty )$

Values of 'a' for function to be onto are $1\pm \sqrt{5}$

Answer: option (D).