Let f:R R be such that f(xy)=f(x)f(y), for all x,y R and f(0) 0.

Question

Let $f:R	o R$ be such that $f(xy)=f(x)f(y)$, for all $x,y\in R$ and $f(0)
e 0$. Let $g:[1,\infty)	o R$ be a differentiable function such that

${x}^{2}g(x)={\int }_{x}^{1}({t}^{2}f(t)-tg(t))dt$

Then $g(2)$ is equal to :

Accepted Answer

We are given

$f(xy)=f(x)f(y)\, x,y\in R,$

and

$f(0)
e 0.$

Also,

${x}^{2}g(x)={\int }_{x}^{1}({t}^{2}f(t)-tg(t))\, dt,\, x\ge 1.$

We have to find $g(2)$.

Use the condition on $f$: Given

$f(xy)=f(x)f(y)$

for all real $x,y$.

Now put $x=0$ and any real $y$. Then

$f(0\cdot y)=f(0)f(y)$

that is,

$f(0)=f(0)f(y).$

Since $f(0)
e 0$, we can divide by $f(0)$:

$f(y)=1\, y\in R.$

So,

$f(t)=1\, t.$

Substitute into the given integral equation: The equation becomes

${x}^{2}g(x)={\int }_{x}^{1}({t}^{2}-tg(t))\, dt.$

Differentiate both sides: Differentiate with respect to $x$.

Left side:

$\frac{d}{dx}[{x}^{2}g(x)]={x}^{2}{g}^{'}(x)+2xg(x).$

Right side, by the Fundamental Theorem of Calculus:

$\frac{d}{dx}{\int }_{x}^{1}({t}^{2}-tg(t))\, dt={x}^{2}-xg(x).$

Hence,

${x}^{2}{g}^{'}(x)+2xg(x)={x}^{2}-xg(x).$

So,

${x}^{2}{g}^{'}(x)+3xg(x)={x}^{2}.$

Divide by $x$ (since $x\ge 1$):

$x{g}^{'}(x)+3g(x)=x.$

Or,

${g}^{'}(x)+\frac{3}{x}g(x)=1.$

Solve the differential equation: This is a linear differential equation:

${g}^{'}(x)+\frac{3}{x}g(x)=1.$

Its integrating factor is

$I.F.={e}^{\int \frac{3}{x}\, dx}={e}^{3\ln x}={x}^{3}.$

Multiply both sides by ${x}^{3}$:

${x}^{3}{g}^{'}(x)+3{x}^{2}g(x)={x}^{3}.$

The left side is

$\frac{d}{dx}({x}^{3}g(x)).$

So,

$\frac{d}{dx}({x}^{3}g(x))={x}^{3}.$

Integrate:

${x}^{3}g(x)=\frac{{x}^{4}}{4}+C.$

Thus,

$g(x)=\frac{x}{4}+\frac{C}{{x}^{3}}.$

Find the constant using the original equation: Put $x=1$ in

${x}^{2}g(x)={\int }_{x}^{1}({t}^{2}-tg(t))\, dt.$

Then

${1}^{2}g(1)={\int }_{1}^{1}({t}^{2}-tg(t))\, dt=0.$

So,

$g(1)=0.$

Now use

$g(1)=\frac{1}{4}+C=0.$

Hence,

$C=-\frac{1}{4}.$

Therefore,

$g(x)=\frac{x}{4}-\frac{1}{4{x}^{3}}.$

Find $g(2)$: $g(2)=\frac{2}{4}-\frac{1}{4({2}^{3})}=\frac{1}{2}-\frac{1}{32}=\frac{16-1}{32}=\frac{15}{32}.$

So the correct answer is

$\frac{15}{32}.$

Hence, Option C.

Solutions · 1