Let f be the function defined by f(x)=ax^2-√2 for some positive a. If f(f(√2))=-√2, then a=

1990 AMC 12 Problem 12

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12.

Let

f

be the function defined by

f(x)=ax^2-\sqrt2

for some positive

a.

f(f(\sqrt2))=-\sqrt2,

then

a=

\frac{2-\sqrt2}{2}

\frac12

2-\sqrt2

\frac{\sqrt2}{2}

\frac{2+\sqrt2}{2}

For positive

a,

when can

f(y)

equal

-\sqrt2?

Set the inner value

f(\sqrt2)

equal to zero

Because

a\gt0,

the equation

f(y)=ay^2-\sqrt2=-\sqrt2

forces

y=0.

Hence

f(\sqrt2)=2a-\sqrt2=0,

a=\frac{\sqrt2}{2}.

Thus the correct answer is D.