It will save a lot of time for many problems if you keep in mind that (m+n)^2 = m^2+n^2+2mn and (m-n)^2 = m^2+n^2-2mn. Here it obviously gives you the answer:

f(m+n)+f(m-n)= m^2 + n^2 + 2mn + m^2 + n^2 - 2mn = 2m^2 + 2n^2

So C

But this also comes in handy when it comes to remembering the inequality:

m^2 + n^2 <= (m+n)^2

Explanation

The problem provides the function \(f(x) = x^2\) and asks for the quantity \(f(m + n) + f(m - n)\).

Plug into this function twice—first, to insert m + n in place of x, and then to insert \(m - n\) in place of x:

\(f(m + n) = (m + n)^2 = m^2 + 2mn + n^2\)

\(f(m - n) = (m - n)2 = m^2 - 2mn + n^2\)

Now add the two:

\((m^2 + 2mn + n^2) + (m^2 - 2mn + n^2) = 2m^2 + 2n^2\)
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Is it m^2 + n^2 <= (m+n)^2
or
m^2 + n^2 < (m+n)^2 (Without = sign).

If m and n are positive or both strictly negative (i.e non 0). Then:

\(m^2 + n^2 < (m+n)^2\) because \((m+n)^2= m^2 + n^2 + 2mn\) now \(2mn\) is positive for m and n being positive.

For \(m=n=0\) they are equal.

If m and n have different signs then the inequality flips and \(m^2+n^2 > (m+n)^2\).
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