July 31st, 2017, 04:52 PM  #1 
Newbie Joined: Jul 2017 From: Brazil Posts: 4 Thanks: 0  CSAT Problem 30 (2015) Problem 30 A continuous function f(x) satisfies the following. For all reals $\displaystyle x ≤ b, f(x) = a(x  b)^2 + c$ For all reals, $\displaystyle f(x) = \int_0^x{ \sqrt{4  f(t)}}dt$ if $\displaystyle \int_0^6{f(x) dt} = \frac{q}{p}$, where p, q are relatively prime positive integers, find p + q. I'm stuck. 
July 31st, 2017, 08:02 PM  #2 
Math Team Joined: Dec 2013 From: Colombia Posts: 6,876 Thanks: 2240 Math Focus: Mainly analysis and algebra 
The question seems somewhat strange in the definition of $f(x) \le b$ without any indication of the value of $b$, but one approach would be to use the Fundamental Theorem of Calculus on the second definition to gather some information on $a$, $b$ and $c$. Following further down that path, attempting to evaluate the first given integral by means of a trigonometric substitution may allow you to use the given integral result to simplify the result and introduce $p$ and $q$. Note that, for the first integral to be valid for all real $x$, we must have that $f(t) \le 4$ for all real $t$. This also constrains $a$, $b$ and $c$. 
July 31st, 2017, 09:10 PM  #3 
Global Moderator Joined: Dec 2006 Posts: 17,725 Thanks: 1359 
In the final equation, cachorroloucobr, is the integration meant to be with respect to $t$ rather than $x$, or is it mistyped?

July 31st, 2017, 09:19 PM  #4 
Newbie Joined: Jul 2017 From: Brazil Posts: 4 Thanks: 0 
Yeah, I mistyped $\displaystyle \int_0^6{f(x) dx} = \frac{q}{p}$ 
July 31st, 2017, 10:22 PM  #5 
Math Team Joined: Jul 2011 From: Texas Posts: 2,606 Thanks: 1290 
$\displaystyle f(x)=\int_0^x \sqrt{4f(t)} \, dt \implies f(0)=0$ and $f'(x)=\sqrt{4f(x)} \implies f(x) \le 4$ $f(x)=a(xb)^2 + c \implies f(0)=ab^2+c$ and $f'(x)=2a(xb)$ from that, $2a(xb) = \sqrt{4f(x)}$ note $x \le b \implies a \le 0$ $4a^2(xb)^2 = 4f(x) \implies f(x) = 44a^2(xb)^2$ matching coefficients, $c=4$ and $a= \dfrac{1}{4}$ finally, $ab^2+c=0 \implies \dfrac{b^2}{4}+4 = 0 \implies b = \pm 4$ note $f'(0)=2 \implies b=4$ Now ... how to deal with a definite integral with an upper limit of integration, $x=6$ when $x \le 4$ ??? 
July 31st, 2017, 11:58 PM  #6 
Global Moderator Joined: Dec 2006 Posts: 17,725 Thanks: 1359 
The value of $x$ can exceed 4.

August 1st, 2017, 05:45 AM  #7  
Newbie Joined: Jul 2017 From: Brazil Posts: 4 Thanks: 0  Quote:
Why do you assume that f (0) = 0?  
August 1st, 2017, 06:29 AM  #8 
Math Team Joined: Jul 2011 From: Texas Posts: 2,606 Thanks: 1290  no "assume" to it ... recall a basic property of definite integrals ... $\displaystyle \int_a^a g(x) \, dx = 0$ If $\displaystyle f(x) = \int_0^x \sqrt{4f(t)} \, dt$, then $\displaystyle f(0) = \int_0^0 \sqrt{4f(t)} \, dt = 0$ 

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