Let a_1, a_2, ldots, a_{100} be non-zero real | vedclass

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(A) Consider the function $f(x) = x 2^x$. Note that $f'(x) = 2^x + x 2^x \ln 2 = 2^x(1 + x \ln 2)$.For $x > 0$,$f(x) > x$ because $2^x > 1$.For $x < 0

Vedclass · Accepted Answer

$\sum_{i=1}^{100} a_i 2^{a_i} > 0$ and $\sum_{i=1}^{100} a_i 2^{-a_i} < 0$

Vedclass · Answer

(A) Consider the function $f(x) = x 2^x$. Note that $f'(x) = 2^x + x 2^x \ln 2 = 2^x(1 + x \ln 2)$.For $x > 0$,$f(x) > x$ because $2^x > 1$.For $x  0$. Then $x 2^x = -y 2^{-y}$. Since $2^y > 1$,$2^{-y}  -y$,which means $x 2^x > x$.Thus,for all $x 
eq 0$,$x 2^x > x$.Summing over $i=1$ to $100$: $\sum_{i=1}^{100} a_i 2^{a_i} > \sum_{i=1}^{100} a_i = 0$.Similarly,consider $g(x) = x 2^{-x}$. For $x > 0$,$2^{-x}  0)$,then $x 2^{-x} = -y 2^y Thus,for all $x 
eq 0$,$x 2^{-x} Summing over $i=1$ to $100$: $\sum_{i=1}^{100} a_i 2^{-a_i} Therefore,option $A$ is correct.

$(a)$ $x\|x\|$	$(i)$ continuous in $(-1, 1)$
$(b)$ $\sqrt{\|x\|}$	$(ii)$ differentiable in $(-1, 1)$
$(c)$ $x+[x]$	$(iii)$ strictly increasing in $(-1, 1)$
$(d)$ $\|x-1\|+\|x+1\|$	$(iv)$ not differentiable at,at least one point in $(-1, 1)$

Let $a_1, a_2, \ldots, a_{100}$ be non-zero real numbers such that $a_1+a_2+\ldots+a_{100}=0$. Then,

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