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Provide solution for RD Sharma Class 12 Chapter Functions Exercise 2.3 Question 1 Sub question (ix) maths textbook solution.

Answers (1)

Answer :

\begin{aligned} &f \circ g(x)=\frac{3 x^{2}-4 x+2}{(1-x)^{2}} \\ &g \circ f(x)=\frac{\left(x^{2}+2\right)}{\left(x^{2}+1\right)} \end{aligned}

Hint : f: R\rightarrow [2,\infty ]

Given : Here given that f(x)=x^{2}+1 \text { and } g(x)=1-\frac{1}{1-x}

For domain of g=1-x\neq 0

\Rightarrow \; \; \; \; \; \; \; x\neq 1.

Domain of  g=R-\{1\}

                  \\g(x)=1-\frac{1}{x} \\ =\frac{1-x-1}{1-x} \\ =\frac{-x}{1-x}

For range of g

\begin{array}{ll} & y=\frac{-x}{1-x} \\ \Rightarrow & y-x y=-x \\ \Rightarrow & y=x y-x \\ \Rightarrow & y=x(y-1) \\ \Rightarrow & x=\frac{y}{y-1} \end{array}

Range of g=R-\{1\}

               \therefore g: R-\{1\} \rightarrow R-\{1\}

Solution :

First, we compute f\; o\; g

Here clearly the range of g is subset of domain of f.

\begin{aligned} &\Rightarrow \quad f \circ g: R-\{1\} \rightarrow R \\ &\Rightarrow \quad f \circ g(x)=f(g(x)) \\ &\qquad \begin{aligned} \Rightarrow & f\left(\frac{-x}{1-x}\right) \\ &=\left(\frac{x}{(1-x)}\right)^{2}+2 \\ &=\frac{x^{2}+2 x^{2}+2-4 x}{(1-x)^{2}} \\ & f \circ g(x)=\frac{3 x^{2}-4 x+2}{(1-x)^{2}} \end{aligned} \end{aligned}

Now, we will compute g\; o\; f

Clearly the range of f is a subset of the domain of g

\begin{aligned} \Rightarrow \quad g \circ f: R \rightarrow R \\ g \circ f(x) &=g(f(x)) \\ &=g\left(x^{2}+2\right) \\ &=1-\frac{1}{1-\left(x^{2}+2\right)} \\ &=\frac{-1}{1-\left(x^{2}+2\right)} \\ &=\frac{x^{2}+2}{x^{2}+1} \\ g \circ f(x) &=\frac{x^{2}+2}{x^{2}+1} \end{aligned}

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