If the domain of the function f(x) = frac{1}{ | vedclass

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Question

(A) For the function to be defined,the expressions inside the square roots in the denominator must be strictly positive.$1) \ x + |x| > 0$If $x \leq 0

Vedclass · Accepted Answer

$26$

Vedclass · Answer

(A) For the function to be defined,the expressions inside the square roots in the denominator must be strictly positive.$1) \ x + |x| > 0$If $x \leq 0$,then $x + |x| = 0$,which makes the denominator zero. Thus,we must have $x > 0$.So,$x \in (0, \infty)$.$2) \ 10 + 3x - x^2 > 0$Multiplying by $-1$,we get $x^2 - 3x - 10 $(x - 5)(x + 2) This inequality holds for $x \in (-2, 5)$.Taking the intersection of $x \in (0, \infty)$ and $x \in (-2, 5)$,we get the domain $(a, b) = (0, 5)$.Therefore,$a = 0$ and $b = 5$.Calculating the required value: $(1+a)^2 + b^2 = (1+0)^2 + 5^2 = 1 + 25 = 26$.

If the domain of the function $f(x) = \frac{1}{\sqrt{10+3x-x^2}} + \frac{1}{\sqrt{x+|x|}}$ is $(a, b)$,then $(1+a)^2 + b^2$ is equal to :

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