JEE Main 2023 · 13 April · morning · Single correct · +4 / −1
For x ∈ ℝ, two real valued functions f(x) and g(x) are such that, g(x)=√x+1 and f ° g(x)=x+3-√x. Then f(0) is equal to
- (a) 5
- (b) 0
- (c) -3
- (d) 1
Answer and solution
Answer (a) 5
Solution
Maths sir · 4 steps
Call √x+1 a new letter t, rewrite everything in t, and you have f itself.
- Substitute. Let t = √x + 1. Then √x = t - 1 and x = (t-1)². Since f ° g(x) = f(g(x)), the left side becomes f(t).
- Rewrite in t. Replace x and √x on the right side by their values in t, then expand.
f(t) = (t-1)² + 3 - (t-1) = t² - 3t + 5
- Put in zero. With f(x) = x² - 3x + 5 as the rule for f, we get f(0) = 0 - 0 + 5 = 5.
- So. f(0) = 5, which is option (a).
The idea to keep. To recover f from f(g(x)), set t = g(x), solve for x in terms of t, and rewrite the whole right side in t. Never put the number straight into f ° g.
Answer from the NTA final answer key. Solution by PrepWiser.