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 \(m^2+n^2 > (m+n)^2\).

Given that \(f(x) = x^2\) and we need to find the value of \(f(m + n) + f(m - n)\)

To find f(m + n) we need to compare what is inside the bracket in f(m + n) and f(x)

=> We get to substitute x with m + n in f(x) to get the value of f(m+n)

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

Similarly, \(f(m-n) = (m-n)^2\)

=> \(f(m + n) + f(m - n)\) = \((m+n)^2 + (m-n)^2\) = \(m^2 + 2mn + n^2 + m^2 - 2mn + n^2\) (Using \((a+b)^2 = a^2 + 2ab + b^2\) and \((a-b)^2 = a^2 - 2ab + b^2\) )
= \(2m^2 + 2n^2\)

So, Answer will be C
Hope it helps!

Watch the following video to learn the Basics of Functions and Custom Characters

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

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

The above two are special products whose expansions are

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

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

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

Hence answer is Choice C.


Note: The three special products which you should understand and memorize are:

\((a+b)^2 = a^2 + b^2 + 2ab\)

\((a-b)^2 = a^2 + b^2 - 2ab\)

\((a+b)(a-b) = a^2 - b^2\)