My Math Forum Found this question on a test, tried thrice, no concrete result; help, please.

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 August 17th, 2018, 11:25 AM #1 Newbie   Joined: Aug 2018 From: India Posts: 1 Thanks: 0 Found this question on a test, tried thrice, no concrete result; help, please. Let f(x) be a cubic polynomial with leading coefficient unity such that f(0) = 1 and all the roots of f`(x) = 0 are also roots of f(x) = 0. If integral f(x)dx = g(x) + C, where g(0) = 1/4 and C is constant of integration, then g(3) - g(1) is equal to (A) 27 (B) 48 (C) 60 (D) 81
 August 17th, 2018, 02:51 PM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,671 Thanks: 2651 Math Focus: Mainly analysis and algebra $$f(x) = k(x-a)(x-b)(x-c) \implies f'(x) = k(x-b)(x-c)+k(x-a)(x-c)+k(x-a)(x-b)$$ Now, if $(x-a)$ is a root of $f'(x)$, we must have $(x-a)$ being a root of $(x-b)(x-c)$. This means that every root of $f'(x)$ must be a repeated root (and there are only two of them) so we can write $$f'(x) = k(x-a)^2 \quad \text{for some a,k \in \mathbb R}$$ You should check the logic of the above (it's certainly not a proof). But if it holds up, I think it should help to get the answer. Thanks from topsquark
 August 17th, 2018, 03:03 PM #3 Global Moderator   Joined: Dec 2006 Posts: 20,835 Thanks: 2162 If f(x) is (x + 1)³, choice (C) is correct. However, I don't see a way to justify v8archie's assumption. It doesn't seem necessary that f '(x) = 0 has a repeated root. Thanks from topsquark
 August 17th, 2018, 04:32 PM #4 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,671 Thanks: 2651 Math Focus: Mainly analysis and algebra The root of $f'(x)$ is a critical point of $f(x)$. If this is also to be a zero of $f(x)$, it must be a repeated root in $f(x)$. But $f'(x)$ has two roots (if not one repeated). You would therefore need $f(x)$ to have four roots, but it's cubic and so has only three. Note that any cubic with a repeated real root has a third real root, although it may be the same as the others.
 August 17th, 2018, 06:58 PM #5 Global Moderator   Joined: Dec 2006 Posts: 20,835 Thanks: 2162 Okay, f '(x) has a repeated root, but that doesn't imply that f(x) is (x + 1)³. It's possible that f(x) = (x + ω)³, where ω is an imaginary cube root of 1. This satisfies the conditions given in the question, but doesn't lead to one of the answer choices supplied.
 August 20th, 2018, 12:38 PM #6 Newbie   Joined: Dec 2016 From: Austin Posts: 16 Thanks: 1 Hope This Helps This is my attempt at a solution. "Let f(x) be a cubic polynomial..." Then $\displaystyle f(x) = ax^{3} + bx^{2} + cx + d$. "...with leading coefficient unity..." Then $\displaystyle a = 1$, so $\displaystyle f(x) = x^{3} + bx^{2} + cx + d$. "...such that $\displaystyle f(0) = 1$" Then $\displaystyle (0)^{3} + b(0)^{2} + c(0) + d = 1$. Then, $\displaystyle d = 1$. Thus, we have $\displaystyle f(x) = x^{3} + bx^{2} + cx + 1$. "...all the roots of $\displaystyle f'(x) = 0$ are also the roots of $\displaystyle f(x) = 0$." Then we can assume $\displaystyle f'(x)$ divides $\displaystyle f(x)$. Let $\displaystyle f'(x) = 3x^{2} + 2bx + c$. By using long division, we find: $\displaystyle \frac{x^{3} +bx^{2} + cx + 1}{3x^{2} + 2bx + c}=\frac{1}{3}x+ \frac{1}{9}b$ with remainder of $\displaystyle \left ( \frac{2}{3}c - \frac{2}{9}b^{2} \right )x + 1 - \frac{1}{9}bc$ However, since $\displaystyle f'(x)$ divides $\displaystyle f(x)$, the remainder will be $\displaystyle 0$. Thus, $\displaystyle \left ( \frac{2}{3}c - \frac{2}{9}b^{2} \right )x + 1 - \frac{1}{9}bc = 0$ Note that we can rewrite the right side of the equation, to get $\displaystyle \left ( \frac{2}{3}c - \frac{2}{9}b^{2} \right )x + 1 - \frac{1}{9}bc = 0x + 0$ Then, by using the method of corresponding coefficients, we have the following system of equations: \begin{cases} \frac{2}{3}c - \frac{2}{9}b^{2} & = 0\\ 1 - \frac{1}{9}bc & = 0 \end{cases} Solve the first equation for $\displaystyle c$. Multiplying through by $\displaystyle 9$ gives $\displaystyle 6c - 2b^{2} = 0$ Adding $\displaystyle 2b^{2}$ to other side gives $\displaystyle 6c = 2b^{2}$ Finally, dividing through by $\displaystyle 6$ yields $\displaystyle c=\frac{1}{3}b^{2}$ Substituting this into the second equation gives $\displaystyle 1 - \frac{1}{9}b \left ( \frac{1}{3}b^{2} \right )=0$ Which becomes $\displaystyle 1 - \frac{1}{27}b^{3}=0$ Multiplying by $\displaystyle 27$ gives $\displaystyle 27 - b^{3}=0$ Adding $\displaystyle b^{3}$ to both sides gives $\displaystyle b^{3}=27$ Taking the cube root yields $\displaystyle b = 3$ We can now substitute back into either equation to find that $\displaystyle c = 3$. We can now take the quotient from the long division process, which is a factor of $\displaystyle f(x)$ and the found constants and rewrite the function as: $\displaystyle f(x) = \left( \frac{1}{3}x+\frac{1}{3} \right ) (3x^{2} +2bx +c)$ $\displaystyle f(x) = \left( \frac{1}{3}x+\frac{1}{3} \right ) (3x^{2} +2(3)x +(3))$ $\displaystyle f(x) = \left( \frac{1}{3}x+\frac{1}{3} \right ) (3x^{2} + 6x + 3)$ Just to clean up the factors a bit, we can pull a $\displaystyle 3$ out of the second factor and apply it to the first factor, so that it becomes: $\displaystyle f(x) = ( x + 1 ) (x^{2} + 2x + 1)$ We can further factor the second factor: $\displaystyle f(x) = ( x + 1 ) (x + 1)(x + 1)$ $\displaystyle f(x) = ( x + 1 )^{3}$ Now we can integrate to find $\displaystyle \int (x+1)^{3}dx = \frac{1}{4}(x+1)^{4} + C$ Therefore $\displaystyle g(x) = \frac{1}{4}(x+1)^{4} + C$ Now we may substitute to find: $\displaystyle g(3) - g(1) = \frac{1}{4}(3+1)^{4} - \frac{1}{4}(1+1)^{4}$ $\displaystyle g(3) - g(1) = 64 - 4$ $\displaystyle g(3) - g(1) = 60$ Therefore the correct answer is C. A couple of notes, we can drop the integration constant $\displaystyle C$, since we are subtracting. The integration constant would be eliminated. Also, the long division was not shown due to the difficulty of typing that up, but if you absolutely need that I can write it on paper and upload it. Hope this helps! Last edited by skipjack; August 20th, 2018 at 06:59 PM. Reason: to correct the long division

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