Q51 / 90 · KVPY 2017 (19 Nov SB/SX) · Mathematics
Let
$$
\mathrm{S}=\left\{\frac{\mathrm{a}^{2}+\mathrm{b}^{2}+\mathrm{c}^{2}}{\mathrm{ab}+\mathrm{bc}+\mathrm{ca}}: \mathrm{a}, \mathrm{b}, \mathrm{c} \in \mathrm{R}, \mathrm{ab}+\mathrm{bc}+\mathrm{ca} \neq 0\right\}
$$
where $R$ is the set of real numbers. Then $S$ equals
[2017]
- A$(-\infty,-1] \cup[1, \infty)$
- B$(-\infty, 0) \cup(0, \infty)$
- C$(-\infty,-1] \cup[2, \infty)$
- Correct answerD$(-\infty,-2] \cup[1, \infty)$
Solution
Case-I
$\begin{array}{l}
(a-b)^{2}+(b-c)^{2}+(c-a)^{2} \geq 0 \\
a^{2}+b^{2}+c^{2}-a b-b c-c a \geq 0 \\
a^{2}+b^{2}+c^{2} \geq a b+b c+c a \\
\text { If } a b+b c+c a>0
\end{array}$
Then $\frac{a^{2}+b^{2}+c^{2}}{a b+b c+c a} \geq 1$
Case-II
$\text { Let } a b+b c+c a < 0$
$\begin{array}{l}
\frac{(a+b+c)^{2}}{a b+b c+c a} \leq 0 \\
\frac{a^{2}+b^{2}+c^{2}+2(a b+b c+c a)}{a b+b c+c a} \leq 0 \\
\frac{a^{2}+b^{2}+c^{2}}{a b+b c+c a}+2 \leq 0 \\
\frac{a^{2}+b^{2}+c^{2}}{a b+b c+c a} \leq-2
\end{array}$
So, Range is $(-\infty,-2] \cup[1, \infty)$
Hence, the correct answer is option: 'D'.