# A Calculus Question on onto functions with a specified range.

The following question was from a mock test of a competitive exam.

Suppose $f:\mathbb{R} \to [-8,8]$ is an onto function and $f(x) = \dfrac{bx}{(a-3)x^3 + x^2 + 4}$ where $a,b \in \mathbb{R}^+$. If the set of all values of $m$ for which the equation $f(x) = mx$ has three distinct real solutions in the open interval $(p,q)$, then find the value of $a+b+p+q$.

I tried the problem and obtained an answer. According to me, it is $43$. I will post the answer later. My solution was quite involved and long. I want to know if there is a quick solution to this problem.

Thank you.

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Oh I see....... –  Babak S. May 27 '13 at 7:22
Every real cubic has a (real) zero, so the denominator of $f$ goes to zero somewhere. $x = 0$ is not a possible root of the denominator, and so $f$ must become unbounded near the zero. Hence $f$ cannot have the specified range. –  Zach L. May 27 '13 at 7:24

First off, the function $f(x)=\dfrac{bx}{(a-3)x^3 + x^2 + 4}$ has a cubic polynomial in the denominator (assuming $a\ne3$), and since we assume $a\in\Bbb{R}$, this equation must have a real root, which cannot be canceled by the $x$ in the numerator, because $x=0$ is not a root of the cubic (which instead evaluates to $4$ at $x=0$). In the vicinity of this root, $f(x)$ is unbounded above and below, so clearly it doesn't satisfy the criterion that $\operatorname{ran}f=[-8,8]$. Thus $a=3$ (in order to make the denominator have no real roots).

The new function $f(x)=\dfrac{bx}{x^2 + 4}$ has no poles, and goes to $0$ at $x\to\pm\infty$, and it's self-evidently continuous, so it takes on a maximum and minimum somewhere, which can be found using the first-derivative test. $f'(x)=-\dfrac{b(x^2-4)}{(x^2+4)^2}=0$ when $x=\pm2$, and at these points $f(x)=\pm\frac b4$. Thus $\operatorname{ran}f=\big[\!-\frac b4,\frac b4\!\big]=[-8,8]$ implies $b=32$.

The function is now $f(x)=\dfrac{32x}{x^2 + 4}$. Note that $f$ is odd. Thus $f(0)=0=m\cdot0$ is always a solution to $f(x)=mx$, and nontrivial solutions come in pairs $\pm x$, since $mx$ is also an odd function. Having discarded $x=0$, we can divide by $x$ and rearrange the equation to get $x^2 + 4=32/m$. This equation has a solution when $32/m > 4$ (noting that $32/m=4$ merely yields a triple root at $0$ which is stated to be inadmissible), which is equivalent to $0<m<8$. The case $m=0$ must be analyzed separately, but $\dfrac{32x}{x^2 + 4}=0$ only when $x=0$, so it is not in the case of interest. Thus $m\in(0,8)=(p,q)$ implies $p=0$ and $q=8$.

Putting it all together, we have $a+b+p+q=3+32+0+8=43$, so it looks like you got the answer right.

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Since $f:\mathbb{R} \to [-8,8]$,

$\max f(x)=\dfrac{bx}{(a-3)x^3+x^2+4}=8$ , we need to have $a=3$ so as to make Polynomial have real roots. Here $x=+2, b=32$ .

And also $\min f(x) \dfrac{bx}{(a-3)x^3+x^2+4}=-8$, you get $x=-2$ and $b=32$

$f(x)=mx$ gives a straight line with $m$ as a slope. Now the function $f(x)=\dfrac{bx}{x^2+4}$ needs to have slope that is equal to $m$. So, $f'(x)=-\dfrac{32(x^2-4)}{(x^2+4)^2}=m$, when $x= \pm 2 \implies m=0$ and when $x=0$ you get $m=8$, therefore $m \in [0,8]$ claiming the roots to be $(2,-2,0)$

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