Let f(x) = frac{2^{x+2} + 16}{2^{2x+1} + 2^{x | vedclass

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(A) Given $f(x) = \frac{2^{x+2} + 16}{2^{2x+1} + 2^{x+4} + 32} = \frac{4 \cdot 2^x + 16}{2 \cdot (2^x)^2 + 16 \cdot 2^x + 32} = \frac{4(2^x + 4)}{2((2

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$118$

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(A) Given $f(x) = \frac{2^{x+2} + 16}{2^{2x+1} + 2^{x+4} + 32} = \frac{4 \cdot 2^x + 16}{2 \cdot (2^x)^2 + 16 \cdot 2^x + 32} = \frac{4(2^x + 4)}{2((2^x)^2 + 8 \cdot 2^x + 16)} = \frac{2(2^x + 4)}{(2^x + 4)^2} = \frac{2}{2^x + 4}$.Consider $f(4-x) = \frac{2}{2^{4-x} + 4} = \frac{2}{\frac{16}{2^x} + 4} = \frac{2 \cdot 2^x}{16 + 4 \cdot 2^x} = \frac{2^x}{8 + 2 \cdot 2^x} = \frac{2^x}{2(4 + 2^x)}$.Then $f(x) + f(4-x) = \frac{2}{2^x + 4} + \frac{2^x}{2(2^x + 4)} = \frac{4 + 2^x}{2(2^x + 4)} = \frac{1}{2}$.The sum is $S = \sum_{k=1}^{59} f \left( \frac{k}{15} \right)$. Note that $f(x) + f(4-x) = \frac{1}{2}$ implies $f \left( \frac{k}{15} \right) + f \left( 4 - \frac{k}{15} \right) = f \left( \frac{k}{15} \right) + f \left( \frac{60-k}{15} \right) = \frac{1}{2}$.There are $59$ terms. The middle term is $f \left( \frac{30}{15} \right) = f(2) = \frac{2}{2^2 + 4} = \frac{2}{8} = \frac{1}{4}$.There are $29$ pairs summing to $\frac{1}{2}$ and one middle term $\frac{1}{4}$.$S = 29 \cdot \frac{1}{2} + \frac{1}{4} = \frac{58 + 1}{4} = \frac{59}{4}$.The required value is $8 \cdot S = 8 \cdot \frac{59}{4} = 2 \cdot 59 = 118$.

List-$I$	List-$II$
$(A)$ $\sinh x$	$(I)$ Domain is $(-1, 1)$, even function
$(B)$ $\text{sech } x$	$(II)$ Domain is $[1, \infty)$, neither even nor odd function
$(C)$ $\tanh x$	$(III)$ Even function
$(D)$ $\text{cosech}^{-1} x$	$(IV)$ Range is $\mathbb{R}$, odd function
	$(V)$ Range is $(-1, 1)$, odd function

Let $f(x) = \frac{2^{x+2} + 16}{2^{2x+1} + 2^{x+4} + 32}$. Then the value of $8 \left( f \left( \frac{1}{15} \right) + f \left( \frac{2}{15} \right) + \dots + f \left( \frac{59}{15} \right) \right)$ is equal to

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