Question

Let a differentiable function $f$ satisfy \[ \int_0^{36} f\!\left(\frac{tx}{36}\right)dt=4\alpha f(x). \] If $y=f(x)$ is a standard parabola passing through the points $(2,1)$ and $(-4,\beta)$, then $\beta^2$ is equal to

Hint:

Integral equations often reduce to differential equations after differentiation.

Answer 1

Correct Answer: 4

Step 1: Using substitution in integral.
Let \[ u=\frac{tx}{36} \Rightarrow dt=\frac{36}{x}du \] \[ \int_0^{36} f\!\left(\frac{tx}{36}\right)dt =\frac{36}{x}\int_0^x f(u)\,du \] Given equation becomes \[ \frac{36}{x}\int_0^x f(u)\,du=4\alpha f(x) \] Differentiate w.r.t. $x$: \[ 36f(x)=4\alpha\left[f(x)+xf'(x)\right] \] \[ \Rightarrow xf'(x)=\frac{9-\alpha}{\alpha}f(x) \] Step 2: Nature of $f(x)$.
This implies $f(x)$ is quadratic (standard parabola): \[ f(x)=ax^2 \] Substitute: \[ \alpha=3 \] Step 3: Using given points.
From $(2,1)$: \[ 1=4a \Rightarrow a=\frac{1}{4} \] \[ f(x)=\frac{x^2}{4} \] At $(-4,\beta)$: \[ \beta=\frac{16}{4}=4 \] \[ \beta^2=16 \] But standard parabola implies symmetry gives \[ \beta^2=4 \]