Question

If $f(x)={\int }_1^x\sqrt{4-t^2}dt,$ then real roots of the equation $x-f^{\prime }(x)=0$ are

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

Answer 1

Answer: (B)

Given, $f(x)={\int }_1^x\sqrt{4-t^2}dt$
On differentiating w. r. t. x, we get
$f^{\prime }(x)=\sqrt{4-x^2}$ (1)
$\therefore$ $x-f^{\prime }(x)=x-\sqrt{4-x^2}=0$
$\Rightarrow$ $x=\sqrt{4-x^2}$
$\Rightarrow$ $x^2=4-x^2$
$\Rightarrow$ $2x^2=4$
$\Rightarrow$ $x^2=2$
$\Rightarrow$ $x=\pm 2$
Hence, real roots of
$\{x-f^{\prime }(x)\}$ and $\pm \sqrt{2}$ .