MCQ
Let $a_1, a_2, \ldots, a_{100}$ be non-zero real numbers such that $a_1+a_2+\ldots+a_{100}=0$ Then,
  • $\sum \limits_{i=1}^{100} a_i 2^{a_i} > 0$ and $\sum \limits_{i=1}^{100} a_i 2^{-a_i} < 0$
  • B
    $\sum \limits_{i=1}^{100} a_i 2^{a_i} \geq 0$ and $\sum \limits_{i=1}^{100} a_i 2^{-a_i} \geq 0$
  • C
    $\sum \limits_{i=1}^{100} a_i 2^{a_i} \leq 0$ and $\sum \limits_{i=1}^{100} a_i 2^{-a i} \leq 0$
  • D
    The sign of $\sum \limits_{i=1}^{100} a_i 2^{a_i}$ or $\sum \limits_{i=1}^{100} a_i 2^{-a_i}$ depends on the choice of $a_i^{\prime} s$

Answer

Correct option: A.
$\sum \limits_{i=1}^{100} a_i 2^{a_i} > 0$ and $\sum \limits_{i=1}^{100} a_i 2^{-a_i} < 0$
a
(a)

We have, $a_1, a_2, a_3, \ldots, a_{100}$ be non-zero real number and

$a_1+a_2+a_3+\ldots+a_{100}=0$

$a_i \cdot 2^{a_i} > a_i$ and $a_i \cdot 2^{-a_i} < a_i$

$\therefore \sum \limits_{i=1}^{100} a_1 \cdot 2^{a i} > \sum \limits_{i=1}^{100} a_i \text { and } \sum \limits_{i=1}^{100} a_1 \cdot 2^{-a_i} < \sum \limits_{i=1}^{100} a_i$

$\Rightarrow \sum \limits_{i=1}^{100} a_1 \cdot 2^{a_i} > 0 \text { and } \sum \limits_{i=1}^{100} a_1 \cdot 2^{-a_i} < 0$

Hence, option $(a)$ is correct.

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