The range of the real valued function f(x) = | vedclass

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

Question

(C) Given $f(x) = \frac{x^2 + 2x - 15}{2x^2 + 13x + 15}$.Factorizing the numerator and denominator: $f(x) = \frac{(x+5)(x-3)}{(2x+3)(x+5)}$.For $x 
e

Vedclass · Accepted Answer

$R - \left\{\frac{1}{2}, \frac{8}{7}\right\}$

Vedclass · Answer

(C) Given $f(x) = \frac{x^2 + 2x - 15}{2x^2 + 13x + 15}$.Factorizing the numerator and denominator: $f(x) = \frac{(x+5)(x-3)}{(2x+3)(x+5)}$.For $x 
eq -5$,$f(x) = \frac{x-3}{2x+3}$.Let $y = \frac{x-3}{2x+3}$.$y(2x+3) = x-3$ $\Rightarrow 2xy + 3y = x - 3$ $\Rightarrow x(2y-1) = -3y - 3$.$x = \frac{-3y-3}{2y-1}$.Since $x$ is defined,$2y-1 
eq 0 \Rightarrow y 
eq \frac{1}{2}$.Also,$x 
eq -5 \Rightarrow \frac{-3y-3}{2y-1} 
eq -5$.$-3y-3 
eq -10y + 5$ $\Rightarrow 7y 
eq 8$ $\Rightarrow y 
eq \frac{8}{7}$.Thus,the range is $R - \left\{\frac{1}{2}, \frac{8}{7}ight\}$.

The range of the real valued function $f(x) = \frac{x^2 + 2x - 15}{2x^2 + 13x + 15}$ is

Similar Questions

The domain of the function $f(x) = \log_e(x - [x])$ is

If $D$ is the domain and $G$ is the range of the real-valued function $f(x)=\sqrt{\frac{1-x^2}{1+x^2}}$, then $D \cap G=$

The domain of $f(x) = \frac{\log_{(x+1)}(x-2)}{x^2 - (2x + 3)}$ for $x \in R$ is

If the domain of the function $f(x) = \log_7(1 - \log_4(x^2 - 9x + 18))$ is $(\alpha, \beta) \cup (\gamma, \delta)$,then $\alpha + \beta + \gamma + \delta$ is equal to

Find the domain and the range of the real function $f$ defined by $f(x) = \sqrt{x-1}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module