If the domain of the function $f(x)=\frac{1}{\sqrt{10+3 x-x^2}}+\frac{1}{\sqrt{x+|x|}}$ is $(a, b)$, then $(1+a)^2+b^2$ is equal to :
Solution
<p>$$x+|x|= \begin{cases}2 x, & x \geq 0 \\ 0, & x<0\end{cases}$$</p>
<p>$\Rightarrow \frac{1}{\sqrt{x+|x|}}$, domain is $x>0$, as $2 x \neq 0$</p>
<p>Similarly,</p>
<p>$$\begin{aligned}
&\frac{1}{\sqrt{3 x+10-x^2}} \text { is defined when } 3 x+10-x^2>0\\
&\begin{aligned}
\Rightarrow & x^2-3 x-10<0 \\
& (x-5)(x+2)<0 \\
\Rightarrow & x \in(-2,5) \\
\Rightarrow & \text { Domain will be }(0, \infty) \cap(-2,5)=(0,5) \\
\Rightarrow & (1+a)^2+b^2=1+25=26
\end{aligned}
\end{aligned}$$</p>
About this question
Subject: Mathematics · Chapter: Sets, Relations and Functions · Topic: Sets and Operations
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