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 July 31st, 2017, 05:52 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, 09:02 PM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,700 Thanks: 2682 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 $c$. Thanks from cachorroloucobr July 31st, 2017, 10:10 PM #3 Global Moderator   Joined: Dec 2006 Posts: 21,128 Thanks: 2337 In the final equation, cachorroloucobr, is the integration meant to be with respect to $t$ rather than $x$, or is it mistyped? Thanks from cachorroloucobr July 31st, 2017, 10:19 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, 11:22 PM #5 Math Team   Joined: Jul 2011 From: Texas Posts: 3,101 Thanks: 1677 $\displaystyle f(x)=\int_0^x \sqrt{4-f(t)} \, dt \implies f(0)=0$ and $f'(x)=\sqrt{4-f(x)} \implies f(x) \le 4$ $f(x)=a(x-b)^2 + c \implies f(0)=ab^2+c$ and $f'(x)=2a(x-b)$ from that, $2a(x-b) = \sqrt{4-f(x)}$ note $x \le b \implies a \le 0$ $4a^2(x-b)^2 = 4-f(x) \implies f(x) = 4-4a^2(x-b)^2$ 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 $x \le 4$ ??? Thanks from cachorroloucobr August 1st, 2017, 12:58 AM #6 Global Moderator   Joined: Dec 2006 Posts: 21,128 Thanks: 2337 The value of $x$ can exceed 4. August 1st, 2017, 06:45 AM   #7
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 Originally Posted by skeeter $\displaystyle f(x)=\int_0^x \sqrt{4-f(t)} \, dt \implies f(0)=0$ and $f'(x)=\sqrt{4-f(x)} \implies f(x) \le 4$ $f(x)=a(x-b)^2 + c \implies f(0)=ab^2+c$ and $f'(x)=2a(x-b)$ from that, $2a(x-b) = \sqrt{4-f(x)}$ note $x \le b \implies a \le 0$ $4a^2(x-b)^2 = 4-f(x) \implies f(x) = 4-4a^2(x-b)^2$ 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 $x \le 4$ ???
I don't understand:
Why do you assume that f (0) = 0? August 1st, 2017, 07:29 AM   #8
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 Originally Posted by cachorroloucobr I don't understand: Why do you assume that f (0) = 0?
no "assume" to it ...

recall a basic property of definite integrals ...

$\displaystyle \int_a^a g(x) \, dx = 0$

If $\displaystyle f(x) = \int_0^x \sqrt{4-f(t)} \, dt$, then $\displaystyle f(0) = \int_0^0 \sqrt{4-f(t)} \, dt = 0$ Tags 2015, csat, problem Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post MathOpen Math Events 1 October 11th, 2015 09:41 AM Cindy Computer Science 0 July 25th, 2015 07:50 AM mobel Number Theory 33 December 30th, 2014 12:03 PM

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