Let f:left[-fracpi2,fracpi2right] to mathbbR$f:\left[-\frac{\pi}{2},\frac{\pi}{2}\right] \to \mathbb{R}$ be a differentiable function such that f(0) = frac12$f(0) = \frac{1}{2}$. If the lim_x to 0 fracx int_0^x f(t) dte^x^2 - 1 = alpha$\lim_{x \to 0} \frac{x \int_{0}^{x} f(t) dt}{e^{x^2} - 1} = \alpha$, then 8alpha^2$8\alpha^2$ is equal to:

### Related Formula lim_y to 0 frace^y - 1y = 1$$\lim_{y \to 0} \frac{e^y - 1}{y} = 1$$ Leibniz Integral Rule: fracddx int_0^x f(t) dt = f(x)$$\frac{d}{dx} \int_0^x f(t) dt = f(x)$$ ### Core Logic Given limit is: alpha = lim_x to 0 fracx int_0^x f(t) dte^x^2 - 1$$\alpha = \lim_{x \to 0} \frac{x \int_{0}^{x} f(t) dt}{e^{x^2} - 1}$$ Multiply and divide the denominator by x^2$x^2$ to use standard exponential limit: alpha = lim_x to 0 fracx int_0^x f(t) dtleft(frace^x^2 - 1x^2right) cdot x^2$$\alpha = \lim_{x \to 0} \frac{x \int_{0}^{x} f(t) dt}{\left(\frac{e^{x^2} - 1}{x^2}\right) \cdot x^2}$$ Since lim_xto 0 frace^x^2 - 1x^2 = 1$\lim_{x\to 0} \frac{e^{x^2} - 1}{x^2} = 1$, the expression simplifies to: alpha = lim_x to 0 fracx int_0^x f(t) dt1 cdot x^2 = lim_x to 0 fracint_0^x f(t) dtx$$\alpha = \lim_{x \to 0} \frac{x \int_{0}^{x} f(t) dt}{1 \cdot x^2} = \lim_{x \to 0} \frac{\int_{0}^{x} f(t) dt}{x}$$ ### Step 1: Applying L'Hôpital's Rule This is a 0/0$0/0$ form. Apply L'Hôpital's Rule by differentiating numerator and denominator w.r.t x$x$: alpha = lim_x to 0 fracfracddx int_0^x f(t) dtfracddx(x) = lim_x to 0 fracf(x)1$$\alpha = \lim_{x \to 0} \frac{\frac{d}{dx} \int_{0}^{x} f(t) dt}{\frac{d}{dx}(x)} = \lim_{x \to 0} \frac{f(x)}{1}$$ By continuity of differentiable function f$f$ at 0$0$: alpha = f(0)$\alpha = f(0)$ ### Step 2: Final Calculation We are given f(0) = frac12$f(0) = \frac{1}{2}$, so alpha = frac12$\alpha = \frac{1}{2}$. We need to find 8alpha^2$8\alpha^2$: 8alpha^2 = 8 left(frac12right)^2 = 8 left(frac14right) = 2$$8\alpha^2 = 8 \left(\frac{1}{2}\right)^2 = 8 \left(\frac{1}{4}\right) = 2$$ ### Pattern Recognition Standard expansion/limits on isolated terms in denominators immediately reduce the power of x$x$, setting up a trivial Leibniz derivative application. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Limits and Derivatives Class 12 Maths: Integrals