Let a differentiable function satisfy the equation If is a standard parabola passing through the points (2, 1) and , then is equal to _____. [2026]

Question

Let a differentiable function $f$ satisfy the equation $\int_{0}^{36} f (\frac{t x}{36}) d t = 4 α f (x) .$ If $y = f (x)$ is a standard parabola passing through the points (2, 1) and $(- 4, β)$ , then $β^{α}$ is equal to _____. [2026]

Smart Achievers · Accepted Answer

(64)

$\int_{0}^{36} f (\frac{t x}{36}) d t = 4 α f (x), Put \frac{t x}{36} = y$

$(- 4, β)$

$\int_{0}^{x} \frac{f (y) 36 d y}{x} = 4 α f (x)$

$\int_{0}^{x} f (y) d y = \frac{α f (x) x}{9}$

$f (x) = \frac{α}{9} (f (x) + x f' (x))$

$(1 - \frac{α}{9}) f (x) = \frac{α x}{9} f' (x)$ $\Rightarrow (9 - α) f (x) = α x f' (x)$

$\frac{f' (x)}{f (x)} = (\frac{9}{α} - 1) \frac{1}{x}$

$\log_{e} f (x) = (\frac{9}{α} - 1) \log_{e} x + \log_{e} c$

$f (x) = c x^{(\frac{9}{α} - 1)}$ For standard parabola,

$\frac{9}{α} - 1 = 2$

$α = 3$

$f (x) = c x^{2}$

$Passing through (2, 1)$

$1 = 4 c \Rightarrow c = \frac{1}{4}$

$y = \frac{x^{2}}{4} passing through (- 4, β)$

$β = 4$

$β^{x} = 4^{3} = 64$

Solution

Q.
Let a differentiable function $f$ satisfy the equation $\int_{0}^{36} f (\frac{t x}{36}) d t = 4 α f (x) .$ If $y = f (x)$ is a standard parabola passing through the points (2, 1) and $(- 4, β)$ , then $β^{α}$ is equal to _____. [2026]

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Solution

Q. Let a differentiable function f satisfy the equation ∫036f(tx36)dt=4αf(x). If y=f(x) is a standard parabola passing through the points (2, 1) and (-4,β), then βα is equal to _____. [2026]

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Q.
Let a differentiable function $f$ satisfy the equation $\int_{0}^{36} f (\frac{t x}{36}) d t = 4 α f (x) .$ If $y = f (x)$ is a standard parabola passing through the points (2, 1) and $(- 4, β)$ , then $β^{α}$ is equal to _____. [2026]