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

Given the integral condition: \[ \int_0^{36} f\!\left(\frac{tx}{36}\right)dt=4\alpha f(x). \] Changing the variable of integration, let \( u = \frac{tx}{36} \) implying \( du = \frac{x}{36}dt \) or \( dt = \frac{36}{x}du \). When \( t = 0 \), \( u = 0 \) and when \( t = 36 \), \( u = x \). The integral becomes: \[ \int_0^x f(u)\frac{36}{x}du = 4\alpha f(x). \] This simplifies to: \[ 36\int_0^x f(u)du = 4\alpha x f(x). \] Differentiating w.r.t \( x \) gives: \[ 36f(x) = 4\alpha(f(x) + xf'(x)). \] Rearranging: \[ 36f(x) = 4\alpha f(x) + 4\alpha xf'(x) \] \[ 4\alpha xf'(x) = (36 - 4\alpha)f(x). \] Assuming a quadratic for \( f(x) \):\( f(x) = ax^2 + bx + c \). Given points: \( f(2) = 1 \) and \( f(-4) = \beta \). Evaluating: \[ 4 = ax^2 \quad \text{(since parabola's vertex implies symmetry and } \alpha = 3) \] Substituting the points: \[ 4a + 2b + c = 1 \quad \text{(at } x = 2\text{)} \] \[ 16a - 4b + c = \beta \quad \text{(at } x = -4\text{)} \] Substituting into differential equation: \[ 12x(ax^2 + bx + c) = 3[/4(2ax+b) \rightarrow 8a = 2a] \] Calculating \(\beta\): \[ 16a - 4b + c = 1 \Rightarrow \beta = 4 \] Hence, \(\beta^2 = (4)^2 = 16\). Therefore, the value falls within the provided range 4 ≤ 4.