JEE Mains · Maths · STD 11- 2. Relation and Function
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 :
- A \(26\)
- B \(29\)
- C \(25\)
- D \(30\)
Answer & Solution
Correct Answer
(A) \(26\)
Step-by-step Solution
Detailed explanation
\(\mathrm{x}+|\mathrm{x}| \gt 0 \quad \Rightarrow \mathrm{x} \in(0, \infty)\) ...(1) \(\begin{aligned} & \& 10+3 x-x^2 \gt 0 \\ & \Rightarrow x^2-3 x-10 \lt 0\end{aligned}\) \(\Rightarrow x \in(-2,5)\) ...(2)…
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