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Let $a, b, c$ be non-zero real numbers such that $a+b+c=0$, let $q=a^2+b^2+c^2$ and $r=a^4+b^4+c^4$. Then,
$q^2 < 2 r$ always
$q^2=2 r$ always
$q^2 > 2 r$ always
$q^2-2 r$ can take both positive and negative values
Solution
(b)
Given, $a+b+c=0 \Rightarrow a, b, c \in R$
$\Rightarrow \quad a^2+b^2+c^2=q$
$\Rightarrow \quad a^4+b^4+c^4=r$
$\Rightarrow \quad\left(a^2+b^2+c^2\right)^2=a^4+b^4+c^4 +2\left(a^2 b^2+b^2 c^2+c^2 a^2\right)$
$\left(a^2+b^2+c^2\right)^2=a^4+b^4+c^4+2 \left[(a b+b c+c a)^2-2 a b c(a+b+c)\right]$
$q^2=r+2\left[(a b+b c+c a)^2-2(a b c)(0)\right]$
$q^2=r+2[a b+b c+c a]^2$
$q^2=r+2 \left[\begin{array}{c}(a+b+c)^2-\left(a^2+b^2+c^2\right) \\ 2\end{array}\right]^2$
$q^2=r+2\left(\frac{0-q}{2}\right)^2$
$q^2=r+\frac{2 q^2}{4}$
$q^2-\frac{1}{2} q^2=r$
$q^2=2 r$
