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July 31st, 2017, 05:52 PM   #1
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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.
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July 31st, 2017, 09:02 PM   #2
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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 $c$.
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July 31st, 2017, 10:10 PM   #3
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In the final equation, cachorroloucobr, is the integration meant to be with respect to $t$ rather than $x$, or is it mistyped?
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July 31st, 2017, 10:19 PM   #4
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Yeah, I mistyped
$\displaystyle \int_0^6{f(x) dx} = \frac{q}{p}$
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July 31st, 2017, 11:22 PM   #5
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$\displaystyle f(x)=\int_0^x \sqrt{4-f(t)} \, dt \implies f(0)=0$ 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 $x \le 4$ ???
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August 1st, 2017, 12:58 AM   #6
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The value of $x$ can exceed 4.
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August 1st, 2017, 06:45 AM   #7
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Quote:
Originally Posted by skeeter View Post
$\displaystyle f(x)=\int_0^x \sqrt{4-f(t)} \, dt \implies f(0)=0$ 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?
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August 1st, 2017, 07:29 AM   #8
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Quote:
Originally Posted by cachorroloucobr View Post
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 $\displaystyle f(0) = \int_0^0 \sqrt{4-f(t)} \, dt = 0$
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