Let f: R rightarrow R be defined as f(x) = 2x | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given $f(x) = 2x - 1$ and $g(x) = \frac{x - 1/2}{x - 1} = \frac{2x - 1}{2(x - 1)}$.Calculate the composition $f(g(x))$:$f(g(x)) = 2(g(x)) - 1 = 2

Vedclass · Accepted Answer

one-one but not onto

Vedclass · Answer

(C) Given $f(x) = 2x - 1$ and $g(x) = \frac{x - 1/2}{x - 1} = \frac{2x - 1}{2(x - 1)}$.Calculate the composition $f(g(x))$:$f(g(x)) = 2(g(x)) - 1 = 2 \left( \frac{2x - 1}{2(x - 1)} \right) - 1$$= \frac{2x - 1}{x - 1} - 1 = \frac{2x - 1 - (x - 1)}{x - 1} = \frac{x}{x - 1} = 1 + \frac{1}{x - 1}$.For one-one check:Let $f(g(x_1)) = f(g(x_2))$.$1 + \frac{1}{x_1 - 1} = 1 + \frac{1}{x_2 - 1} \implies x_1 - 1 = x_2 - 1 \implies x_1 = x_2$.Thus,$f(g(x))$ is one-one.For onto check:The range of $f(g(x)) = 1 + \frac{1}{x - 1}$ is $R - \{1\}$.Since the codomain is $R$,the range is not equal to the codomain.Therefore,$f(g(x))$ is not onto.Conclusion: $f(g(x))$ is one-one but not onto.

Let $f: R \rightarrow R$ be defined as $f(x) = 2x - 1$ and $g: R - \{1\} \rightarrow R$ be defined as $g(x) = \frac{x - 1/2}{x - 1}$. Then the composition function $f(g(x))$ is:

Similar Questions

If $f(x) = \frac{4x+3}{6x-4}$,$x \neq \frac{2}{3}$ and $(f \circ f)(x) = g(x)$,where $g: R - \{\frac{2}{3}\} \rightarrow R - \{\frac{2}{3}\}$,then $(g \circ g \circ g)(4)$ is equal to

If $g(x) = x^2 + x - 2$ and $\frac{1}{2}g(f(x)) = 2x^2 - 5x + 2$,then $f(x)$ is

If $f(x)=e^x$ and $h(x)=(f \circ f)(x)$, then $\frac{h^{\prime}(x)}{h(x)}=$

Let $f: R - \{-\frac{1}{2}\} \rightarrow R$ and $g: R - \{-\frac{5}{2}\} \rightarrow R$ be defined as $f(x) = \frac{2x+3}{2x+1}$ and $g(x) = \frac{|x|+1}{2x+5}$. Then the domain of the function $f \circ g$ is:

Let $f(x) = e^x$ and $g(x) = x^2$. Then,the number of solutions of $f(g(x)) = g(f(x))$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module