JEE Mains · Maths · STD 12 - 1. relation and function
For \(x \in R\), two real valued functions \(f(x)\) and \(g(x)\) are such that, \(g(x)=\sqrt{x}+1\) and \(f o g(x)=x+3-\sqrt{x}\). Then \(f(0)\) is equal to
- A \(1\)
- B \(-3\)
- C \(5\)
- D \(0\)
Answer & Solution
Correct Answer
(C) \(5\)
Step-by-step Solution
Detailed explanation
\(g(x)=\sqrt{x}+1\) \(f \circ g(x)=x+3-\sqrt{x}\) \(=(\sqrt{x}+1)^2-3(\sqrt{x}+1)+5\) \(=g^2(x)-3 g(x)+5\) \(\Rightarrow f(x)=x^2-3 x+5\) \(\therefore f(0)=5\)
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