Let f, g :(1, infty) rightarrow mathbb{R} be | vedclass

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

Question

(D) Given $f(x) = \frac{2x+3}{5x+2}$ and $g(x) = \frac{2-3x}{1-x}$.We need to find $(f \circ g)(x) = f(g(x))$.$(f \circ g)(x) = f\left(\frac{2-3x}{1-x

Vedclass · Accepted Answer

$56$

Vedclass · Answer

(D) Given $f(x) = \frac{2x+3}{5x+2}$ and $g(x) = \frac{2-3x}{1-x}$.We need to find $(f \circ g)(x) = f(g(x))$.$(f \circ g)(x) = f\left(\frac{2-3x}{1-x}\right) = \frac{2\left(\frac{2-3x}{1-x}\right)+3}{5\left(\frac{2-3x}{1-x}\right)+2}$Multiplying numerator and denominator by $(1-x)$:$(f \circ g)(x) = \frac{2(2-3x) + 3(1-x)}{5(2-3x) + 2(1-x)} = \frac{4-6x+3-3x}{10-15x+2-2x} = \frac{7-9x}{12-17x}$.Now,we evaluate the function at the endpoints of the interval $[2, 4]$:$(f \circ g)(2) = \frac{7-9(2)}{12-17(2)} = \frac{7-18}{12-34} = \frac{-11}{-22} = \frac{1}{2}$.$(f \circ g)(4) = \frac{7-9(4)}{12-17(4)} = \frac{7-36}{12-68} = \frac{-29}{-56} = \frac{29}{56}$.Since the function is monotonic in the interval $[2, 4]$,the range is $[\alpha, \beta] = [\frac{1}{2}, \frac{29}{56}]$.Here,$\alpha = \frac{1}{2}$ and $\beta = \frac{29}{56}$.Then $\beta - \alpha = \frac{29}{56} - \frac{1}{2} = \frac{29-28}{56} = \frac{1}{56}$.Therefore,$\frac{1}{\beta-\alpha} = 56$.

Let $f, g :(1, \infty) \rightarrow \mathbb{R}$ be defined as $f(x) = \frac{2x+3}{5x+2}$ and $g(x) = \frac{2-3x}{1-x}$. If the range of the function $f \circ g : [2, 4] \rightarrow \mathbb{R}$ is $[\alpha, \beta]$,then $\frac{1}{\beta-\alpha}$ is equal to

Similar Questions

Let $R = \{(1, 3), (2, 2), (3, 2)\}$ and $S = \{(2, 1), (3, 2), (2, 3)\}$ be two relations on the set $A = \{1, 2, 3\}$. Then $R \circ S = $

Let $g(x) = 1 + x - [x]$ and $f(x) = \begin{cases} -1, & \text{if } x < 0 \\ 0, & \text{if } x = 0 \\ 1, & \text{if } x > 0 \end{cases}$. Then for all values of $x$,the value of $f(g(x))$ is:

If $f(x)=e^x$ and $g(x)=\ln(x)$ for all $x \in [1, \infty)$,then $f \circ g$ is . . . . . .

If $f(x) = \frac{\alpha x}{x + 1}$,$x \ne -1$,for what value of $\alpha$ is $f(f(x)) = x$?

If $f(x) = \begin{cases} \sin x, & x \neq n\pi, n \in \mathbb{Z} \\ 0, & \text{otherwise} \end{cases}$ and $g(x) = \begin{cases} x^2 + 1, & x \neq 0, 2 \\ 4, & x = 0 \\ 5, & x = 2 \end{cases}$,then $\lim_{x \to 0} g(f(x)) = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module