If A subseteq Z and the function f: A rightar | vedclass

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(C) Given $f(x) = \frac{1}{\sqrt{64 - (0.5)^{24 + x - x^2}}}$.For $f(x)$ to be defined,the expression inside the square root must be strictly positive

Vedclass · Accepted Answer

$25$

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(C) Given $f(x) = \frac{1}{\sqrt{64 - (0.5)^{24 + x - x^2}}}$.For $f(x)$ to be defined,the expression inside the square root must be strictly positive:$64 - (0.5)^{24 + x - x^2} > 0$$64 > (0.5)^{24 + x - x^2}$$2^6 > (2^{-1})^{24 + x - x^2}$$2^6 > 2^{-(24 + x - x^2)}$Since the base $2 > 1$,we have:$6 > -24 - x + x^2$$x^2 - x - 30 $(x - 6)(x + 5) This inequality holds for $x \in (-5, 6)$.Since $A \subseteq Z$,the set $A$ consists of integers between $-5$ and $6$,which are $A = \{-4, -3, -2, -1, 0, 1, 2, 3, 4, 5\}$.The sum of the absolute values of the elements of $A$ is:$|-4| + |-3| + |-2| + |-1| + |0| + |1| + |2| + |3| + |4| + |5|$$= 4 + 3 + 2 + 1 + 0 + 1 + 2 + 3 + 4 + 5 = 25$.

If $A \subseteq Z$ and the function $f: A \rightarrow R$ is defined by $f(x) = \frac{1}{\sqrt{64 - (0.5)^{24 + x - x^2}}}$,then the sum of all absolute values of elements of $A$ is

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