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JEE Main 20252 AprilEvening Shift
Exam
JEE Main
Year
2025
Date
2 April
Shift
Evening Shift
Subject
Mathematics
Chapter
Functions

JEE Main 2025 · 2 April · Evening Shift · Mathematics · Functions

If the domain of the function f(x)= 1 10+3 x-x^2 + 1 x+|x| is (a,...

If the domain of the function
f(x)=110+3x−x2+1x+∣x∣f(x)=\frac{1}{\sqrt{10+3 x-x^2}}+\frac{1}{\sqrt{x+|x|}} is (a,b)(a, b), then (1+a)2+b2(1+a)^2+b^2 is equal to :

Options

  1. A2626
  2. B2929
  3. C2525
  4. D3030

Correct answer

A. 2626

Step-by-step solution

x+∣x∣>0⇒x∈(0,∞)\mathrm{x}+|\mathrm{x}| > 0 \quad \Rightarrow \mathrm{x} \in(0, \infty) ...(1)10+3x−x2>0⇒x2−3x−10<0\begin{aligned} & & 10+3 x-x^2 > 0 \\ & \Rightarrow x^2-3 x-10 < 0\end{aligned}\Rightarrow x \in(-2,5)...(2)...(2) from  (1)(2)x∈(0,5)∴a=0b=5∴(1+a2)+b2=1+25=26\begin{aligned} & \text{\text{ from } }(1) &(2) \quad \mathrm{x} \in(0,5) \\ & \therefore \mathrm{a}=0 & \mathrm{ b}=5 \\ & \therefore\left(1+\mathrm{a}^2\right)+\mathrm{b}^2=1+25=26\end{aligned}$
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