Question

$\largef(x)$ is a quadratic function. Another function $\largeg(x)$ satisfies $g(x)=\int\{x^{2}+f(x)\}dx$ and $f(x)g(x)=-2x^{4}+8x^{3}.$
Find $\largeg(1).$

Answer 1

$\Huge\text{ \cdots\longrightarrow\boxed{\bold{g(1)=2.}} }$

$\Large\text{\boxed{\bold{[Topic: Integration]}}}$

Let's compare the degrees of different functions. Let us denote the degree of $\largef(x)$ by $\largedeg(f)$. Then, -

$\large\cdots\longrightarrow deg(f)=2.\ \cdots\text{[Eqn.1]}$

$\large\cdots\longrightarrow deg(fg)=4.\ \cdots\text{[Eqn.2]}$

$\large\text{ \displaystyle\cdots\longrightarrow deg(g)=deg(\int \{x^{2}+f(x)\}\ dx).\ \cdots\text{[Eqn.3]} }$

We know that -

$\large\cdots\longrightarrow deg(fg)=deg(f)+deg(g).$

Hence, in [Eqn.2], -

$\large\cdots\longrightarrow\boxed{deg(g)=2.}$

Then, in [Eqn.3], -

$\large\text{ \cdots\longrightarrow\boxed{deg(\int \{x^{2}+f(x)\}\ dx)=2.} }$

Integrated functions have degrees 1 higher. Hence, -

$\large\text{ \cdots\longrightarrow\boxed{deg(\{x^{2}+f(x)\})=1.} }$

$\Large\text{\boxed{\bold{[Explanation]}}}$

We know that -

$\largef(x)=-x^{2}+ax+b,$

because of -

$\large\text{ \cdots\longrightarrow\boxed{deg(\{x^{2}+f(x)\})=1} }$

We can refer to the product of two functions.

$\largef(x)g(x)$ factorizes to $\large-2x^{3}(x-4)$.

Since we know the leading term of $\largef(x)$, the remaining cases are -

$\cdots\longrightarrow\begin{cases} & f(x)=-x^{2} \\ & g(x)=2x(x-4) \end{cases}\text{ or }\begin{cases} & f(x)=-x(x-4) \\ & g(x)=2x^{2}. \end{cases}$

$\Large\text{\boxed{\bold{[Case\ A.]}}}$

Considering the indefinite integral $\large\text{ \int \{x^{2}+f(x)\}\ dx }$, -

$\large\text{ \cdots\longrightarrow\displaystyle\int \{x^{2}+f(x)\}\ dx=\int0\ dx=C. }$

But, -

$\large\cdots\longrightarrow g(x)\neq C.$

Hence Case A is false.

$\Large\text{\boxed{\bold{[Case\ B.]}}}$

Considering the indefinite integral $\large\text{ \int \{x^{2}+f(x)\}\ dx }$, -

$\large\text{ \cdots\longrightarrow\displaystyle\int \{x^{2}+f(x)\}\ dx=\int4x\ dx=2x^{2}+C. }$

And, -

$\large\cdots\longrightarrow g(x)=2x^{2}.$

Hence Case B is confirmed.

$\Large\text{\boxed{\bold{[Final\ answer]}}}$

We know that, -

$\large\text{ \cdots\longrightarrow\boxed{g(x)=2x^{2}.} }$

Hence, -

$\Large\text{ \cdots\longrightarrow \boxed{\bold{g(1)=2.}} }$