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Suppose $f(x)=x^3+ ax^2 + bx +c$ . Now a,b,c are chosen respectively by throwing a dice 3 times. Now find the Probability that $f(x)$ is a increasing function ?


MY APPROACH :

i really have given a lot thought to it but i have no clue. i cant find even the first step towards solving the problem . Didnt understand what they meant by saying f(x) is a increasing function. when a function becomes an increasing function? what are the conditions of that ? and what is the solution of this question?

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    $\begingroup$ So are $a,b,c$ numbers from $1$ to $6$ or numbers from $3$ to $18$? $\endgroup$ Mar 2, 2014 at 14:17
  • $\begingroup$ this is the question given .. no explanation was available ... only this much information(i.e. only the question)....so whats going on your mind???what it should be??? for me i think its 1 to 6 $\endgroup$ Mar 2, 2014 at 14:19
  • $\begingroup$ The word "respectively" is the odd part, but I think you are correct. $\endgroup$ Mar 2, 2014 at 14:20
  • $\begingroup$ now what is the solution??? $\endgroup$ Mar 2, 2014 at 14:21
  • $\begingroup$ Are you demanding a solution from he??? $\endgroup$ Mar 2, 2014 at 14:22

1 Answer 1

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HINT : $$\begin{align}f'(x) &= 3x^2 + 2ax + b \\ &= 3(x + \frac{a}{3})^2 + b - \frac{a^2}{3}\\ &\ge b - \frac{a^2}{3}\end{align}$$

Now, $f'(x) > 0 \implies f(x)$ is an increasing function (an unrelated but good question to think about : does $f(x)$ increasing $\implies f'(x) > 0$?).

What can you now say about the probability for which this condition is satisfied? Or more directly, what is the probability that $b - \frac{a^2}{3} > 0$? (Interestingly, it does not depend on $c$).

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  • $\begingroup$ sorry i didnt understand the last line...3(x+a/3)^2 + b is always greater than 0 right??? so t be $f'(x) > 0$ we need (a^2/3) to be less than 3(x+a/3)^2 + b right??? then why you are not considering that case $\endgroup$ Mar 2, 2014 at 14:41
  • $\begingroup$ Nope. If we set $b = 1$ and $a = 6$, then it is possible to find $x$ such that $f'(x) < 0$. What the last line meant was that since $(x + \frac{a^2}{3})^2$ is a square, and since all (real) squares are at least $0$, then $f'(x) \ge 3(0) + b - \frac{a^2}{3} = b - \frac{a^2}{3}$. So we only need to consider the cases for which $b - \frac{a^2}{3} > 0 $ so that $f'(x) > 0$. $\endgroup$
    – Yiyuan Lee
    Mar 2, 2014 at 14:50
  • $\begingroup$ No problem. Very glad to be able to help you! $\endgroup$
    – Yiyuan Lee
    Mar 2, 2014 at 15:00

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