If $0<a<\sqrt{2}$ and $x$ is an integer,

$S$ is the sum of all the possible values of $a$ that allows the below statement to be true.

$\boxed{\text{There is only one } x \text{ that satisfies the above set of inequalities.}}$

Find the value of $6000S$ .

*
This problem is a part of
<Grade 10 CSAT Mock test> series
.
*

The answer is 9000.

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From $x^2-a^2x\geq0$ , we get

$x\leq0$ or $x\geq a^2$ .

.

And from $x^2-4ax+4a^2-1<0$ , we get

$2a-1<x<2a+1$ .

$(i)\quad0<a<\dfrac{1}{2}$

$-1<2a-1<x\leq0$ or $a^2\leq x<2a+1<2$ .

Since $0\leq a^2\leq\dfrac{1}{4}$ ,

There are two values for $x$ , $0$ and $1$ , which doesn't satisfy the problem.

$(ii)\quad a=\dfrac{1}{2}$

.

$\dfrac{1}{4}\leq x<2$ .

There is one value for $x$ , which is $1$ .

$(iii)\quad \dfrac{1}{2}<a<1$

$a^2\leq x<2a+1$ .

Since $\dfrac{1}{4}<a<1$ and $2<2a+1<3$ ,

There are two values for $x$ , $1$ and $2$ , which doesn't satisfy the problem.

$(iv)\quad a=1$

.

$1<x<3$ .

There is one value for $x$ , which is $2$ .

$(v)\quad 1<a<\sqrt{2}$

$a^2\leq x<2a+1$ .

Since $1<a^2<2$ and $3<2a+1<1+2\sqrt{2}<4$ ,

There are two values for $x$ , $2$ and $3$ , which doesn't satisfy the problem.

From $(i)$ through $(v)$ , we can say that

All possible values for $a$ are $\dfrac{1}{2}$ and $1$ .

.

$S=\dfrac{3}{2}$ ,

$6000S=6000\times\dfrac{3}{2}=\boxed{9000}$ .