If $y=f(x)$ is a linear function satisfying the relation $f(xy)=f(x)f(y)$, then the curve $P(x,y)=\alpha$ cuts $y=f^{-1}x$ at? 
If $y=f(x)$ is  a linear function satisfying the relation $f(xy)=f(x)f(y)\forall x,y\in\mathbb R$, then the curve $$y^2+\int_0^x(\sin t+a^2t^3+bt)dt=\alpha,\alpha\in\mathbb R^+$$
  cuts $y=f^{-1}x$ at?

My try:


*

*$f(x)=x$ and $f^{-1}(x)=x$

*$y^2-\cos x+\frac14a^2x^4+\frac12bx^2=\alpha-C=C'$

*So $x^2+C'=\cos x-\frac14a^2x^4-\frac12bx^2$



Edit: 
Options are:


*

*No point.

*exactly one point.

*at least two points.(Correct Option)

*infinite points.

 A: The only $f$ that is linear and satisfies $f(xy)=f(x)f(y)$ and has an inverse is the identity function. So you are asking to solve the equation
$$
\begin{align}
x^2+\int_0^x(\sin(t)+a^2t^3+bt)\,dt&=\alpha\\
x^2+\left[-\cos(t)+\frac14a^2t^4+\frac12bt^2\right]_0^x&=\alpha\\
x^2-\cos(x)+1+\frac14a^2x^4+\frac12bx^2&=\alpha\\
1+\left(\frac12b+1\right)x^2+\frac14a^2x^4&=\cos(x)+\alpha\\
\left(\frac{b+2}{2a}+\frac12ax^2\right)^2&=\cos(x)+\alpha-1+\left(\frac{b+2}{2a}\right)^2\\
\end{align}$$
On the right, we have cosine function shifted up/down by some constant. On the left, we have a relatively easy to understand quartic polynomial. There can only be a finite number of solutions.
With specific values for $a$, $b$, and $\alpha$, we could solve for the solutions in $x$ using numerical methods, like Newton-Raphson.

We only need decide if there are $0$ solutions, $1$ solution, finitely many but at least $2$ solutions, or infinitely many solutions.
Note that at $x=0$, the left side is less than the right side since $\alpha>0$. But as $x\to\infty$, the left side approaches $\infty$ while the right side remains bounded. So there must be a point where the left side overtakes the right side, and therefore there is at least $1$ solution.
Note that $0$ is not a solution since $\alpha>0$. Because the two sides of the equation are even functions, whenever solutions exist, they come in pairs. $x$ will be a solution if and only if $-x$ is. So there are at least $2$ solutions.
The range of the right side is a compact interval $[A,B]$. Since the left side is an even polynomial of degree $4$, the preimage of $[A,B]$ under the left function is either a compact interval $[-C,C]$ or a union of two, three, or four compact intervals. Call this compact preimage $P$. Any solutions must lie in $P$, since $P$ is precisely where outputs from the left side have a possibility to match outputs of the right side.
Over a connected component of a compact set $P$, it's not possible for two distinct analytic functions to have the same output infinitely many times. Otherwise, consider the function that you get from their difference. This analytic function would have infinitely many zeros on that connected component, and the only analytic function on a compact connected set with infinitely many zeros is the zero function.
Therefore there are a finite, even number of solutions, and the third multiple choice option is correct. If $\alpha$ is large enough, it is easy to imagine conditions where there are precisely two solutions. If $b=-2$, $\alpha=1$, and $a$ is very small, it is easy to imagine a large number of solutions. I'll assert without real proof that depending on $a$, $b$, and $\alpha$, any even number at least $2$ is possible for the solution count. 
