Question

Let \(f:(0,1)\rightarrow(0,1)\) be a bijective differentiable function such that \(f^{\prime}(x)\ne0 \ \forall x\in(0,1)\) and \(f\left(\frac{1}{2}\right)=\frac{\sqrt{3}}{2}\). Suppose for all \(x\), \[ \lim_{t\rightarrow x}\frac{\int_{0}^{t}\sqrt{1-(f(s))^{2}}\,ds-\int_{0}^{x}\sqrt{1-(f(s))^{2}}\,ds}{f(t)-f(x)}=f(x) \] Then the value of \(f\left(\frac{1}{4}\right)\) belongs to:

• A. $\{\sqrt{7},\sqrt{6}\}$
• B. $\left\{\frac{\sqrt{7}}{2},\frac{\sqrt{15}}{2}\right\}$
• C. $\left\{\frac{\sqrt{7}}{4},\frac{\sqrt{15}}{4}\right\}$
• D. $\left\{\frac{\sqrt{7}}{3},\frac{\sqrt{15}}{3}\right\}$

Hint:

Whenever a limit equation features a variable index integration limit like $\int_0^t g(s)\,ds$ over a factor variable of the form $f(t)-f(x)$, always reach for L'Hôpital's rule. Differentiating with respect to $t$ quickly cancels out the integrals via Leibniz's identity, leaving a straightforward first-order differential equation!

Answer 1

The Correct Option is C

Concept: The given limit is of the indeterminate form $\frac{0}{0}$ as $t \rightarrow x$. Hence, we apply L'Hôpital's Rule by differentiating the expression with respect to the limit variable $t$, using the Leibniz Integral Rule for differentiating under the integral sign. Step 1: Apply L'Hôpital's Rule to eliminate the indeterminate form.
Let us define the integral in the numerator as a function $F(t) = \int_{0}^{t}\sqrt{1-(f(s))^{2}}\,ds$. By the Fundamental Theorem of Calculus (Leibniz Rule): \[ F'(t) = \frac{d}{dt}\left[\int_{0}^{t}\sqrt{1-(f(s))^{2}}\,ds\right] = \sqrt{1-(f(t))^{2}} \] Differentiating the numerator and the denominator with respect to $t$ yields: \[ \lim_{t\rightarrow x} \frac{\sqrt{1-(f(t))^{2}}}{f'(t)} = f(x) \quad \Rightarrow \quad \frac{\sqrt{1-(f(x))^{2}}}{f'(x)} = f(x) \] Rearranging the terms to isolate the derivative $f'(x)$: \[ f'(x) = \frac{\sqrt{1-(f(x))^{2}}}{f(x)} \]

Step 2: Form and solve the separable differential equation.
Let $y = f(x)$, allowing us to rewrite the expression in standard Leibniz differential notation: \[ \frac{dy}{dx} = \frac{\sqrt{1-y^{2}}}{y} \quad \Rightarrow \quad \frac{y}{\sqrt{1-y^{2}}}\,dy = dx \] Integrating both sides of the equation simultaneously: \[ \int \frac{y}{\sqrt{1-y^{2}}}\,dy = \int dx \] Apply the substitution $u = 1-y^2 \implies du = -2y\,dy$: \[ -\sqrt{1-y^{2}} = x + C \quad \cdots (1) \]

Step 3: Evaluate the integration constant $C$.
We are given the initial boundary condition $f\left(\frac{1}{2}\right) = \frac{\sqrt{3}}{2}$, which translates to $y = \frac{\sqrt{3}}{2}$ when $x = \frac{1}{2}$. Substituting these values back into equation (1): \[ -\sqrt{1-\left(\frac{\sqrt{3}}{2}\right)^2} = \frac{1}{2} + C \quad \Rightarrow \quad -\sqrt{1-\frac{3}{4}} = \frac{1}{2} + C \] \[ -\sqrt{\frac{1}{4}} = \frac{1}{2} + C \quad \Rightarrow \quad -\frac{1}{2} = \frac{1}{2} + C \quad \Rightarrow \quad C = -1 \] Substitute $C = -1$ back into equation (1) and rearrange the negative signs: \[ -\sqrt{1-y^2} = x - 1 \quad \Rightarrow \quad \sqrt{1-y^2} = 1 - x \] Square both sides to fully isolate the functional variable $y^2$: \[ 1 - y^2 = (1 - x)^2 \quad \Rightarrow \quad y^2 = 1 - (1 - x)^2 \]

Step 4: Calculate the target value $f\left(\frac{1}{4}\right)$.
Substitute $x = \frac{1}{4}$ into our solved equation: \[ y^2 = 1 - \left(1 - \frac{1}{4}\right)^2 = 1 - \left(\frac{3}{4}\right)^2 \] \[ y^2 = 1 - \frac{9}{16} = \frac{7}{16} \quad \Rightarrow \quad y = \frac{\sqrt{7}}{4} \] Hence, $f\left(\frac{1}{4}\right) = \frac{\sqrt{7}}{4}$. This explicit coordinate value matches and belongs to the element collection set shown in option (C).