Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
Oh I see....... –  B. 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

2 Answers 2

up vote 6 down vote accepted

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.

share|improve this answer

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 $(2,-2,0)$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.