Question

If $f(x)=\int_{1}^{x}{\sqrt{4-{{t}^{2}}}}\,\,\,dt,$ then real roots of the equation $x-f'(x)=0$ are

• A. $\pm \,\,1$
• B. $\pm \,\,\sqrt{2}$
• C. $0$ and $1$
• D. $\pm \,\,2$

Answer 1

Answer: (B)

$\pm \,\,\sqrt{2}$

Answer 2

The correct option is(B): ±√2.

Given, $f(x)=\int_{1}^{x}{\sqrt{4-{{t}^{2}}}}\,\,dt$
On differentiating w. r. t. x, we get
$f'(x)=\sqrt{4-{{x}^{2}}}$ (1)
$\therefore$ $x-f'(x)=x-\sqrt{4-{{x}^{2}}}=0$
$\Rightarrow$ $x=\sqrt{4-{{x}^{2}}}$
$\Rightarrow$ ${{x}^{2}}=4-{{x}^{2}}$
$\Rightarrow$ $2{{x}^{2}}=4$
$\Rightarrow$ ${{x}^{2}}=2$
$\Rightarrow$ $x=\pm 2$
Hence, real roots of
$\\{x-f'(x)\\}$ and $\pm \sqrt{2}$ .