Let a differentiable function f satisfy the equation _36^0f((tx)/(36))dt=4 f(x).

Question

Let a differentiable function $f$ satisfy the equation ${\int }_{36}^{0}f(\frac{tx}{36})dt=4\alpha f(x)$. If $y=f(x)$ is a standard parabola passing through the points $(2,1)$ and $(-4,\beta )$, then ${\beta }^{\alpha }$ is equal to $____$ .

Answer: 64

Accepted Answer

Given, ${\int }_{36}^{0}f(\frac{tx}{36})dt=4\alpha f(x)$

Let $\frac{tx}{36}=u\Rightarrow \frac{x}{36}dt=du$

${\int }_{x}^{0}f(u)\frac{36}{x}du=4\alpha f(x)$

Integration is w.r.t $u$, so treat $x$ as a constant

${\int }_{x}^{0}f(u)du=\frac{4x}{36}\alpha f(x)=\frac{x}{9}\alpha f(x)$

Now, differentiate both sides w.r.t $x$, use Leibniz theorem in LHS:

$1\cdot f(x)=\frac{\alpha }{9}[1\cdot f(x)+x{f}^{'}(x)]$

$\Rightarrow$ $9f(x)=\alpha f(x)+\alpha x{f}^{'}(x)$

$\Rightarrow$ $\alpha x{f}^{'}(x)=f(x)(9-\alpha )$

$\Rightarrow$ $\frac{{f}^{'}(x)}{f(x)}=\frac{9-\alpha }{\alpha x}$

Integrating both sides

$\int \frac{{f}^{'}(x)}{f(x)}dx=\int \frac{9-\alpha }{\alpha }(\frac{1}{x})dx$

$\Rightarrow$ $\ln f(x)=\frac{9-\alpha }{\alpha }\ln x+c$

$\Rightarrow$ $f(x)={e}^{\frac{9-\alpha }{\alpha }\ln x}\cdot {e}^{c}$

$\Rightarrow$ $f(x)={x}^{\frac{9-\alpha }{\alpha }}\cdot K\, {K={e}^{c}}$

It is given that $f(x)$ is a standard parabola passing through $(2,1)$ and $(-4,\beta )$.

So $f(x)=K{x}^{2}$ type.

$\frac{9-\alpha }{\alpha }=2\implies 9-\alpha =2\alpha \implies 3\alpha =9$

$\Rightarrow$ $\alpha =3$

$f(x)=K{x}^{2}$ passing through $(2,1)$

$f(2)=1$

$1=K(2{)}^{2}\implies K=\frac{1}{4}$

$f(x)=\frac{1}{4}{x}^{2}$

$f(x)$ is also passing through $(-4,\beta ),f(-4)=\beta $

$\beta =\frac{1}{4}(-4{)}^{2}=\frac{16}{4}=4$

then ${\beta }^{\alpha }={4}^{3}=64$

Solutions · 1