KCET · Maths · Functions
Let \(\boldsymbol{f}: \boldsymbol{R} \rightarrow \boldsymbol{R}\) be defined by \(f(x)=3 x^2-5\) and \(g: R \rightarrow R\) by \(g(x)=\frac{x}{x^2+1}\), then \(g \circ f\) is
- A \(\frac{3 x^2-5}{9 x^4-6 x^2+26}\)
- B \(\frac{3 x^2}{x^4+2 x^2-4}\)
- C \(\frac{3 x^2}{9 x^4+30 x^2-2}\)
- D \(\frac{3 x^2-5}{9 x^4-30 x^2+26}\)
Answer & Solution
Correct Answer
(D) \(\frac{3 x^2-5}{9 x^4-30 x^2+26}\)
Step-by-step Solution
Detailed explanation
Given that \(f(x)=3 x^2-5\)
and \(g(x)=\frac{x}{x^2+1}\)
\(g o t=g\{f(x)\}=g\left(3 x^2-5\right)\)
\(\begin{aligned} & =\frac{3 x^2-5}{\left(3 x^2-5\right)^2+1}=\frac{3 x^2-5}{9 x^4-30 x^2+25+1} \\ & =\frac{3 x^2-5}{9 x^4-30 x^2+26}\end{aligned}\)
and \(g(x)=\frac{x}{x^2+1}\)
\(g o t=g\{f(x)\}=g\left(3 x^2-5\right)\)
\(\begin{aligned} & =\frac{3 x^2-5}{\left(3 x^2-5\right)^2+1}=\frac{3 x^2-5}{9 x^4-30 x^2+25+1} \\ & =\frac{3 x^2-5}{9 x^4-30 x^2+26}\end{aligned}\)
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