Question
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 :
The correct answer is
26
Domain Calculation for Function $f(x)$
The function is given by $f(x) = \frac{1}{\sqrt{10+3x-x^2}} + \frac{1}{\sqrt{x+|x|}}$. To find the domain $(a, b)$, we need to consider the conditions under which each term is defined.
First Term Domain: $\frac{1}{\sqrt{10+3x-x^2}}$
- For the square root in the denominator to be defined and non-zero, the expression inside must be strictly positive: $10+3x-x^2 > 0$.
- Rewrite as: $x^2 - 3x - 10 < 0$.
- Factor the quadratic: $(x-5)(x+2) < 0$.
- This inequality holds true when $x$ is between the roots -2 and 5.
- So, the domain for the first term is $-2 < x < 5$.
Second Term Domain: $\frac{1}{\sqrt{x+|x|}}$
- The expression inside the square root must be strictly positive: $x+|x| > 0$.
- Consider cases for $|x|$:
- If $x > 0$, then $|x| = x$. The inequality becomes $x+x > 0 \implies 2x > 0 \implies x > 0$.
- If $x \leq 0$, then $|x| = -x$. The inequality becomes $x+(-x) > 0 \implies 0 > 0$, which is false.
- Therefore, the domain for the second term is $x > 0$.
Combined Domain $(a, b)$
The domain of the function $f(x)$ is the intersection of the domains of both terms.
- Condition 1: $-2 < x < 5$
- Condition 2: $x > 0$
- The intersection is $0 < x < 5$.
Thus, the domain is $(a, b) = (0, 5)$. This implies $a=0$ and $b=5$.
Calculation of $(1+a)^2+b^2$
Substitute the values of $a$ and $b$ into the expression:
- $(1+a)^2+b^2 = (1+0)^2 + 5^2$
- $= 1^2 + 25$
- $= 1 + 25$
- $= 26$
The value of the expression is 26.
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