If the domain of the function f(x)=frac 1 sqrt 10+3 x-x 2 +frac 1 √(x+ x ) is (a, b) , then (1+a) 2 +b 2 is equal to :

Mathematics · Functions

JEE Main 2025 — 2 April, Evening Shift — Question 28

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 :

$x+|x|=\left\{\begin{array}{l}2 x, x \geq 0\\ 0, x<0\end{array}\right.$

$\Rightarrow \frac{1}{\sqrt{x+|x|}}$ , domain is $x>0$ , as $2 x \neq 0$

Similarly, $\frac{1}{\sqrt{3 x+10-x^{2}}}$ is defined when $3 x+10-x^{2}>0$

$\Rightarrow x^{2}-3 x-10<0$ $(x-5)(x+2)<0$

$\Rightarrow x \in(-2,5)$

$\Rightarrow$ Domain will be $(0, \infty) \cap(-2,5)=(0,5)$

$\Rightarrow(1+a)^{2}+b^{2}=1+25=26$ .

Answer key and solution verified before publishing.

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