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The least value of $4x^2-4ax +a^2-2a+2$ on $[0,2]$ is $3$. What is the integer part of $a$?

We know that minimum value of a quadratic is $-\cfrac{b}{2a}$.

We will get one condition from here and $-\cfrac{b}{2a}$ should be equal to $3$.

But the problem is that this limit is for the whole function, not for an interval, and it might not apply to the interval we have been given.

Adriano
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maths lover
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  • I did a major edit to clarify the question - do check it says what you want. Also it would help to know where the question came from. – Mark Bennet Jul 13 '13 at 10:39

1 Answers1

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Hint: there are three possibilities, given the shape of the graph of a quadratic

Let $f(x)=4x^2-4ax +a^2-2a+2$, then

either $f(x)$ is increasing on $[0,2]$, in which case the minimum value occurs at $x=0$

or $f(x)$ is decreasing on $[0,2]$, in which case the minimum value occurs at $x=2$

or $f(x)$ has minimum value within $[0,2]$

Test each of the three possibilities and see which gives you a consistent solution.

Added later

If $x=0$ we need to solve $a^2-2a+2=3$, so $a=1\pm\sqrt2$. But then we have to check those two values of $a$ to see whether $f$ is decreasing (which is the condition for using $x=0$).

If $x=2$ we have $16-8a+a^2-2a+2=a^2-10a+18=3$ whence $a=5\pm\sqrt{10}$, and we need to test the deacreasing condition.

Mark Bennet
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