Let $f:(0,1) \in(0,1)$ be a differentiable function such that $f(x) \neq 0$ for all $x \in(0,1)$ and $f\left(\frac{1}{2}\right)=\frac{\sqrt{3}}{2}$. Suppose for all $x$, $\mathop {\lim }\limits_{t \to x} \,\,\,\frac{{\int_0^t {\sqrt {1 - {{(f(s))}^2}} } ds - \int_0^x {\sqrt {1 - {{(f(s))}^2}} } ds}}{{f(t) - f(x)}} = f(x)$ Then, the value of $f\left(\frac{1}{4}\right)$ belongs to

$\left\{\frac{\sqrt{7}}{2}, \frac{\sqrt{15}}{2}\right\}$

$\left\{\frac{\sqrt{7}}{3}, \frac{\sqrt{15}}{3}\right\}$

JEE MATHEMATICS MCQ: DEFINITE INTEGRATION

1 Easy

1 of 1

Analyzing +4 | -1

Let

f:(0,1) \in(0,1)

be a differentiable function such that

f(x) \neq 0

for all

x \in(0,1)

and

f\left(\frac{1}{2}\right)=\frac{\sqrt{3}}{2}

. Suppose for all

x

\lim\limits_{t \to x} \,\,\,\frac{{\int_0^t {\sqrt {1 - {{(f(s))}^2}} } ds - \int_0^x {\sqrt {1 - {{(f(s))}^2}} } ds}}{{f(t) - f(x)}} = f(x)

Then, the value of

f\left(\frac{1}{4}\right)

belongs to

View answer and explanation

Correct option: D

\left\{\frac{\sqrt{7}}{4}, \frac{\sqrt{15}}{4}\right\}

The correct answer is

\left{\frac{\sqrt{7}}{4}, \frac{\sqrt{15}}{4}\right\}

because after solving the differential equation derived from the given limit and using the initial condition, we find

f(x) = \sqrt{2x - x^2}

, leading to

f(\frac{1}{4}) = \frac{\sqrt{7}}{4}