Question

The value of the f(0), so that the function f(x) = $\frac{\sqrt{(a^2-ax+x^2)-\sqrt{(a^2+ax+x^2)}}}{\sqrt{(a+x)-\sqrt{(a-x)}}}$ becomes continuous for all x, is given by:

• A. $a\sqrt{a}$
• B. $\sqrt{a}$
• C. $-\sqrt{a}$
• D. $a-\sqrt{a}$

Answer 1

The correct option is A $a\sqrt{a}$

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

$f\left(0\right)=\sqrt{\frac{a^2-\sqrt{a^2}}{\sqrt{a+\sqrt{a}}}}\Rightarrow \frac{\sqrt{a^2\pm a}}{\sqrt{a+\sqrt{a}}}$

$a^2\pm \sqrt{a^2}\ge 0\Rightarrow a^4\pm 2a^2\sqrt{a^2}+a^2\ge 0$

$a^2[a^2\pm 2\sqrt{a^2}+1]\ge 0$

$\sqrt{a+\sqrt{a}}\ge 0$

$a+\sqrt{a}>0$

$a^2+a+2a\sqrt{a}>0$

$a[1+a+2\sqrt{a}]>0$

$f\left(0\right)=\frac{a^2\left(2\sqrt{a^2}\right)}{a2\sqrt{a}}=a\sqrt{a}$