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Maths Question 1 – JEE-MAIN 2025

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If the domain of the function f(x)=110+3x−x2+1|x|+x is (a,b), then (1+a)2+b2 is equal to :

For a function involving square roots in the denominator, the expression inside the square root must be strictly positive.

Step 1: Determine the domain for each term✦ Active

For the first term, 110+3x−x2, the expression under the square root must be strictly positive:

10+3x−x2>0⟹x2−3x−10<0⟹(x−5)(x+2)<0

This inequality holds for x∈(−2,5). For the second term, 1|x|+x, the expression under the square root must be strictly positive:

|x|+x>0

If x≥0, then |x|=x, so x+x>0⟹2x>0⟹x>0. This is consistent with x≥0. If x<0, then |x|=−x, so −x+x>0⟹0>0, which is false. Thus, no solutions for x<0. Therefore, for the second term, x∈(0,∞).

Step 2: Find the overall domain of the function○ Expand

The domain of f(x) is the intersection of the domains found for each term:

(−2,5)∩(0,∞)=(0,5)

Given that the domain is (a,b), we have a=0 and b=5.

💡 Teacher's Secret Hint

Remember to take the intersection of all valid intervals for the function to be defined.

Step 3: Calculate the required expression○ Expand

Substitute the values of a and b into the expression (1+a)2+b2:

(1+0)2+52=12+25=1+25=26
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