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Mathematics Question on General and Particular Solutions of a Differential Equation

If f(x) is differentiable and ∫0t2xf(x)dx=25t5,\int_0^{t^2}x f(x) dx =\frac{2}{5}t^5, then f(425)f\bigg(\frac{4}{25}\bigg) equals

A

25\frac{2}{5}

B

−52-\frac{5}{2}

C

1

D

52\frac{5}{2}

Answer

25\frac{2}{5}

Explanation

Solution

Here, ∫0t2xf(x)dx=25t5,\int_0^{t^2}x f(x) dx =\frac{2}{5}t^5,
Using Newton Leibnitz's formula, differentiating both
sides, we get
t^2 \\{ f(t^2)\\} \bigg \\{\frac{d}{dt}(t^2)\bigg\\}-0.f(0) \bigg\\{\frac{d}{dt}(0)\bigg\\}=2t^4
⇒    t2f(t2)2t=2t4   ⇒  f(t2)=t\Rightarrow \, \, \, \, t^2f(t^2)2t=2t^4 \, \, \, \Rightarrow \, \, f(t^2)=t
∴       f(425)=−25          [putting t=25]\therefore \, \, \, \, \, \, \, f\bigg(\frac{4}{25}\bigg)=-\frac{2}{5} \, \, \, \, \, \, \, \, \, \, \bigg[ putting \, t=\frac{2}{5}\bigg]
⇒    f(425)=25\Rightarrow \, \, \, \, f\bigg(\frac{4}{25}\bigg)=\frac{2}{5}