Let f: R-{-1/2} → R and g: R-{-5/2} → R be defined as f(x)=(2 x+3)/(2 x+1) and g(x)=(|x|+1)/(2 x+5). Then, the domain of the function fog is :

Solution

Maths sir · 4 steps

For fog, x must be allowed in g, and g(x) must be allowed in f. Check both, and respect the modulus.

1. Two conditions. f(g(x)) exists when x is in the domain of g, so x ≠ -5/2. It also needs g(x) in the domain of f, so g(x) ≠ -1/2.

   (|x|+1)/(2x+5) = -1/2 ⇒ 2|x| + 2x = -7

2. Split the modulus. For x ≥ 0: 2x + 2x = -7 gives x = -7/4, which is not ≥ 0, so it is rejected. For x < 0: -2x + 2x = 0, and 0 = -7 is impossible.
3. Second condition. So g(x) never equals -1/2, and the second condition removes no point. As a check, g(-7/4) = (11/4)/(3/2) = 11/6, not -1/2.
4. So. Only x = -5/2 is excluded, so the domain of fog is R - {-5/2}, option (d).

The idea to keep. Domain of fog: start with the domain of g, then remove the x where g(x) falls outside the domain of f. With a modulus, solve case by case and keep only roots that fit their case.

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