If the domain of the function f(x)=1/√(10+3 x-x²)+1/√(x+|x|) is (a, b), then (1+a)²+b² is equal to :

Solution

Maths sir · 4 steps

Each square root sits in a denominator, so what is inside must be strictly positive. Intersect the two.

1. First root. Since √(10+3x-x²) is in a denominator, we need 10+3x-x²>0, that is x²-3x-10<0. This factors as (x-5)(x+2)<0, so -2<x<5.
2. Second root. We need x+|x|>0. For x≤ 0, |x|=-x and the sum is 0, which fails. For x>0 the sum is 2x>0, so this condition gives x>0.
3. Intersect. Both conditions must hold, so the domain is (-2,5)∩(0,∞)=(0,5). Hence a=0 and b=5.

   (1+a)²+b²=(1+0)²+5²=1+25=26

4. So. The value is 26, which is option (d).

The idea to keep. A root in the denominator needs a strictly positive inside, not just ≥ 0. Also remember that x+|x| is 0 for every x≤ 0, so it is positive only for x>0.

Answer from the NTA final answer key. Solution by PrepWiser.