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let $$f(x)=\begin{cases} \dfrac{1}{a-1}(x-1)&x\ge a\\ \dfrac{1}{a-2}(x-2)& x<a \end{cases}$$ There exist $t_{1},t_{2}$ such that $$f(t_{1})=\dfrac{1}{2},f(t_{2})=\dfrac{3}{2}$$ then $$t_{1}-t_{2}$$ the range of values is ?

my book have only have result $$t_{1}-t_{2}\in (-\infty,-\dfrac{1}{2})\cup (\dfrac{1}{2},+\infty)$$ true or false?

zerosofthezeta
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math110
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1 Answers1

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if $a \geq 2$, then :
since $f(x)$ is greater than one only for $x \geq a$ and is less that one only for $x < a$. $f(t_1) = \frac12$ only for $x < a$ and $f(t_2) = \frac32$ only for $x \geq a$. so $ \frac{t_1-2}{a-2} = \frac12 $ and $ \frac{t_2-1}{a-1} = \frac32 $, then $t_1-t_2 = -a + \frac32$ and the rest is left to you. Similarly you can do it for other values of $a$.

Mahdi Khosravi
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    Is there some things lost after "we have:"? +1 – Mikasa Jul 07 '13 at 07:09
  • No, I corrected it! – Mahdi Khosravi Jul 07 '13 at 07:11
  • From the definition of the function, we have $a \not= 2, 1$. So, there are three regions for $a$ to take values. Now, when $ a >= 2 $ and when $ a <= 1 $, it's all fine...I am getting the given interval. But $a$ can take values between 1 and 2. In that case, I am getting $ t1 - t2 $ lies in $ (-0.5, +0.5 ) $ - which doesn't match with the book answer. – Parth Thakkar Jul 07 '13 at 07:34
  • for $1 \leq a \leq 2$ $f(x)$ is always greater that one. and no $t_1$ is available – Mahdi Khosravi Jul 07 '13 at 09:01