Question ID: #873
Let a differentiable function f satisfy the equation $\int_{0}^{36}f(\frac{tx}{36})dt=4\alpha f(x)$. If $y=f(x)$ is a standard parabola passing through the points $(2, 1)$ and $(-4,\beta)$, Then $\beta^{\alpha}$ is equal to
Solution:
Given equation:
$$\int_{0}^{36}f\left(\frac{tx}{36}\right)dt = 4\alpha f(x)$$
Let $u = \frac{tx}{36} \Rightarrow du = \frac{x}{36} dt \Rightarrow dt = \frac{36}{x} du$.
Limits: When $t=0, u=0$. When $t=36, u=x$.
Substitute in the integral:
$$\int_{0}^{x} f(u) \frac{36}{x} du = 4\alpha f(x)$$
$$\int_{0}^{x} f(u) du = \frac{4\alpha x}{36} f(x) = \frac{\alpha x}{9} f(x)$$
Differentiating both sides with respect to $x$ using Newton-Leibniz rule:
$$f(x) = \frac{\alpha}{9} [f(x) + x f'(x)]$$
$$9f(x) = \alpha f(x) + \alpha x f'(x)$$
$$(9-\alpha)f(x) = \alpha x f'(x)$$
$$\frac{f'(x)}{f(x)} = \frac{9-\alpha}{\alpha x}$$
Integrating both sides:
$$\ln f(x) = \frac{9-\alpha}{\alpha} \ln x + \ln c$$
$$f(x) = c x^{\frac{9-\alpha}{\alpha}}$$
Since $y=f(x)$ is a standard parabola, it is of the form $y=cx^2$.
Comparing exponents:
$$\frac{9-\alpha}{\alpha} = 2 \Rightarrow 9-\alpha = 2\alpha \Rightarrow 3\alpha = 9 \Rightarrow \alpha = 3$$
So, $f(x) = cx^2$.
Passes through $(2, 1)$:
$$1 = c(2)^2 \Rightarrow 4c = 1 \Rightarrow c = \frac{1}{4}$$
Equation is $y = \frac{x^2}{4}$.
Passes through $(-4, \beta)$:
$$\beta = \frac{(-4)^2}{4} = \frac{16}{4} = 4$$
We need to find $\beta^\alpha$:
$$\beta^\alpha = 4^3 = 64$$
Ans. (64)
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