Search any question & find its solution
Question:
Answered & Verified by Expert
Let $\mathrm{a}_{1}, \mathrm{a}_{2}, \ldots \ldots . \mathrm{a}_{100}$ be non-zero real numbers such that $\mathrm{a}_{1}+\mathrm{a}_{2}+\ldots \ldots .+\mathrm{a}_{100}=0$, Then
Options:
Solution:
2944 Upvotes
Verified Answer
The correct answer is:
$\sum_{j=1}^{100} a_{i} 2^{a_{i}}>0$ and $\sum_{i=1}^{100} a_{i} 2^{-a_{i}} < 0$
Note that for every real number $\mathrm{a}_{\mathrm{i}}$
$\begin{array}{ll}
& a_{i} \cdot 2^{a_{i}}>a_{i} \quad \text { and } \quad a_{i} \cdot 2^{-a_{i}} < a_{i} \\
\text { therefore, } & \sum_{i=1}^{100} a_{i} \cdot 2^{a_{i}}>\sum_{i=1}^{100} a_{i} \text { and } \sum_{i=1}^{100} a_{i} \cdot 2^{-a_{i}}>\sum_{i=1}^{100} a_{i}
\end{array}$
$\begin{array}{ll}
& a_{i} \cdot 2^{a_{i}}>a_{i} \quad \text { and } \quad a_{i} \cdot 2^{-a_{i}} < a_{i} \\
\text { therefore, } & \sum_{i=1}^{100} a_{i} \cdot 2^{a_{i}}>\sum_{i=1}^{100} a_{i} \text { and } \sum_{i=1}^{100} a_{i} \cdot 2^{-a_{i}}>\sum_{i=1}^{100} a_{i}
\end{array}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.