Q87 / 155 · WBJEE 2026 · Mathematics
Let $f: (0, 1) \to (0, 1)$ be a differentiable function such that $f'(x) \neq 0 \, \forall x \in (0, 1)$ and $f\left(\dfrac{1}{2}\right) = \dfrac{\sqrt{3}}{2}$. Suppose for all $x$, $\lim_{t \to x}\dfrac{\displaystyle\int_{0}^{t}\sqrt{1 - (f(s))^2}\,ds - \displaystyle\int_{0}^{x}\sqrt{1 - (f(s))^2}\,ds}{f(t) - f(x)} = f(x)$. Then the value of $f\left(\dfrac{1}{4}\right)$ belongs to
- A$\{\sqrt{7}, \sqrt{6}\}$
- B$\left\{\dfrac{\sqrt{7}}{2}, \dfrac{\sqrt{15}}{2}\right\}$
- Correct answerC$\left\{\dfrac{\sqrt{7}}{4}, \dfrac{\sqrt{15}}{4}\right\}$
- D$\left\{\dfrac{\sqrt{7}}{3}, \dfrac{\sqrt{15}}{3}\right\}$
Solution
The given limit is of the form $\dfrac{0}{0}$ as $t \rightarrow x$.
Applying L'Hopital's rule by differentiating the numerator and the denominator with respect to $t$:
$\lim_{t \rightarrow x} \dfrac{\dfrac{d}{dt} \left( \int_{0}^{t} \sqrt{1 - (f(s))^2} ds - \int_{0}^{x} \sqrt{1 - (f(s))^2} ds \right)}{\dfrac{d}{dt}(f(t) - f(x))} = f(x)$
Using Leibniz's rule for differentiation under the integral sign:
$\lim_{t \rightarrow x} \dfrac{\sqrt{1 - (f(t))^2} - 0}{f'(t) - 0} = f(x)$
$\dfrac{\sqrt{1 - (f(x))^2}}{f'(x)} = f(x)$
$\sqrt{1 - (f(x))^2} = f(x) f'(x)$
Integrating both sides with respect to $x$:
$\int \dfrac{f(x) f'(x)}{\sqrt{1 - (f(x))^2}} dx = \int 1 dx$
Let $1 - (f(x))^2 = u$. Then $-2f(x)f'(x) dx = du$, which means $f(x)f'(x) dx = -\dfrac{du}{2}$.
$\int \dfrac{-1}{2\sqrt{u}} du = x + C$
$-\sqrt{u} = x + C$
$-\sqrt{1 - (f(x))^2} = x + C$
Given $f\left(\dfrac{1}{2}\right) = \dfrac{\sqrt{3}}{2}$. Substituting $x = \dfrac{1}{2}$:
$-\sqrt{1 - \left(\dfrac{\sqrt{3}}{2}\right)^2} = \dfrac{1}{2} + C$
$-\sqrt{1 - \dfrac{3}{4}} = \dfrac{1}{2} + C$
$-\sqrt{\dfrac{1}{4}} = \dfrac{1}{2} + C$
$-\dfrac{1}{2} = \dfrac{1}{2} + C$
$C = -1$
Thus, $-\sqrt{1 - (f(x))^2} = x - 1$
Squaring both sides:
$1 - (f(x))^2 = (x - 1)^2$
$(f(x))^2 = 1 - (x - 1)^2$
To find $f\left(\dfrac{1}{4}\right)$, substitute $x = \dfrac{1}{4}$:
$\left(f\left(\dfrac{1}{4}\right)\right)^2 = 1 - \left(\dfrac{1}{4} - 1\right)^2$
$\left(f\left(\dfrac{1}{4}\right)\right)^2 = 1 - \left(-\dfrac{3}{4}\right)^2$
$\left(f\left(\dfrac{1}{4}\right)\right)^2 = 1 - \dfrac{9}{16}$
$\left(f\left(\dfrac{1}{4}\right)\right)^2 = \dfrac{7}{16}$
$f\left(\dfrac{1}{4}\right) = \pm \dfrac{\sqrt{7}}{4}$
Since the codomain of $f$ is $(0, 1)$, $f(x) > 0$ for all $x \in (0, 1)$.
Therefore, $f\left(\dfrac{1}{4}\right) = \dfrac{\sqrt{7}}{4}$.
This value belongs to the set $\left\{ \dfrac{\sqrt{7}}{4}, \dfrac{\sqrt{15}}{4} \right\}$.
Answer: $\left\{\dfrac{\sqrt{7}}{4}, \dfrac{\sqrt{15}}{4}\right\}$
Applying L'Hopital's rule by differentiating the numerator and the denominator with respect to $t$:
$\lim_{t \rightarrow x} \dfrac{\dfrac{d}{dt} \left( \int_{0}^{t} \sqrt{1 - (f(s))^2} ds - \int_{0}^{x} \sqrt{1 - (f(s))^2} ds \right)}{\dfrac{d}{dt}(f(t) - f(x))} = f(x)$
Using Leibniz's rule for differentiation under the integral sign:
$\lim_{t \rightarrow x} \dfrac{\sqrt{1 - (f(t))^2} - 0}{f'(t) - 0} = f(x)$
$\dfrac{\sqrt{1 - (f(x))^2}}{f'(x)} = f(x)$
$\sqrt{1 - (f(x))^2} = f(x) f'(x)$
Integrating both sides with respect to $x$:
$\int \dfrac{f(x) f'(x)}{\sqrt{1 - (f(x))^2}} dx = \int 1 dx$
Let $1 - (f(x))^2 = u$. Then $-2f(x)f'(x) dx = du$, which means $f(x)f'(x) dx = -\dfrac{du}{2}$.
$\int \dfrac{-1}{2\sqrt{u}} du = x + C$
$-\sqrt{u} = x + C$
$-\sqrt{1 - (f(x))^2} = x + C$
Given $f\left(\dfrac{1}{2}\right) = \dfrac{\sqrt{3}}{2}$. Substituting $x = \dfrac{1}{2}$:
$-\sqrt{1 - \left(\dfrac{\sqrt{3}}{2}\right)^2} = \dfrac{1}{2} + C$
$-\sqrt{1 - \dfrac{3}{4}} = \dfrac{1}{2} + C$
$-\sqrt{\dfrac{1}{4}} = \dfrac{1}{2} + C$
$-\dfrac{1}{2} = \dfrac{1}{2} + C$
$C = -1$
Thus, $-\sqrt{1 - (f(x))^2} = x - 1$
Squaring both sides:
$1 - (f(x))^2 = (x - 1)^2$
$(f(x))^2 = 1 - (x - 1)^2$
To find $f\left(\dfrac{1}{4}\right)$, substitute $x = \dfrac{1}{4}$:
$\left(f\left(\dfrac{1}{4}\right)\right)^2 = 1 - \left(\dfrac{1}{4} - 1\right)^2$
$\left(f\left(\dfrac{1}{4}\right)\right)^2 = 1 - \left(-\dfrac{3}{4}\right)^2$
$\left(f\left(\dfrac{1}{4}\right)\right)^2 = 1 - \dfrac{9}{16}$
$\left(f\left(\dfrac{1}{4}\right)\right)^2 = \dfrac{7}{16}$
$f\left(\dfrac{1}{4}\right) = \pm \dfrac{\sqrt{7}}{4}$
Since the codomain of $f$ is $(0, 1)$, $f(x) > 0$ for all $x \in (0, 1)$.
Therefore, $f\left(\dfrac{1}{4}\right) = \dfrac{\sqrt{7}}{4}$.
This value belongs to the set $\left\{ \dfrac{\sqrt{7}}{4}, \dfrac{\sqrt{15}}{4} \right\}$.
Answer: $\left\{\dfrac{\sqrt{7}}{4}, \dfrac{\sqrt{15}}{4}\right\}$
Hence, the correct answer is option: 'C'.