Question

The value of${(\mathrm{a}+\mathrm{b})}^2+{(\mathrm{a}–\mathrm{b})}^2$ is

• A. $2\mathrm{a}+2\mathrm{b}$
• B. $2\mathrm{a}–2\mathrm{b}$
• C. $2{\mathrm{a}}^2+2{\mathrm{b}}^2$
• D. $2{\mathrm{a}}^2–2{\mathrm{b}}^2$

Answer 1

The correct option is C

$2{\mathrm{a}}^2+2{\mathrm{b}}^2$

Explanation for the correct option:

Find the value of${(\mathrm{a}+\mathrm{b})}^2+{(\mathrm{a}–\mathrm{b})}^2$.

$\begin{array}{rcl}{(\mathrm{a}+\mathrm{b})}^2+{(\mathrm{a}–\mathrm{b})}^2& =& {\mathrm{a}}^2+2\mathrm{ab}+{\mathrm{b}}^2+{\mathrm{a}}^2-2\mathrm{ab}+{\mathrm{b}}^2\left[\because {\left(\mathrm{a}+\mathrm{b}\right)}^2={\mathrm{a}}^2+2\mathrm{ab}+{\mathrm{b}}^2\mathrm{and}{\left(\mathrm{a}+\mathrm{b}\right)}^2={\mathrm{a}}^2-2\mathrm{ab}+{\mathrm{b}}^2\right]\\ & =& 2{\mathrm{a}}^2+2{\mathrm{b}}^2\\ \therefore {(\mathrm{a}+\mathrm{b})}^2+{(\mathrm{a}–\mathrm{b})}^2& =& 2{\mathrm{a}}^2+2{\mathrm{b}}^2\end{array}$

$\therefore$The correct option is (C).