KCET · Maths · Functions
\(\boldsymbol{f}: \boldsymbol{R} \rightarrow \boldsymbol{R}\) and \(g:[0, \infty) \rightarrow R\) defined by \(f(x)=x^2\) and \(g(x)=\sqrt{x}\). Which one of the following is not true?
- A \((f \circ g)(-4)=4\)
- B \((f \circ g)(2)=2\)
- C \((g \circ f)(-2)=2\)
- D \((g \circ f)(4)=4\)
Answer & Solution
Correct Answer
(A) \((f \circ g)(-4)=4\)
Step-by-step Solution
Detailed explanation
\(f o g(x)=f(g(x))\)
As domain of \(g(x) \rightarrow[0, \infty)\), so domain of \(f \circ g(\lambda)\) also \([0, \infty)\)
\(\therefore \quad f(g(x))=(\sqrt{x})^2\)
\(\begin{aligned} & g \circ f(x)=g(f(x)) \\ & \qquad g \circ f(x)=\sqrt{x^2}=|x|\end{aligned}\)
Hence, \(f o g(-4)=(\sqrt{-4})^2\) is not define.
As domain of \(g(x) \rightarrow[0, \infty)\), so domain of \(f \circ g(\lambda)\) also \([0, \infty)\)
\(\therefore \quad f(g(x))=(\sqrt{x})^2\)
\(\begin{aligned} & g \circ f(x)=g(f(x)) \\ & \qquad g \circ f(x)=\sqrt{x^2}=|x|\end{aligned}\)
Hence, \(f o g(-4)=(\sqrt{-4})^2\) is not define.
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