Range of the function f(x) = frac{x^2 + x + 2 | vedclass

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

Question

(C) Let $y = \frac{x^2 + x + 2}{x^2 + x + 1}$.We can rewrite the function as $y = \frac{(x^2 + x + 1) + 1}{x^2 + x + 1} = 1 + \frac{1}{x^2 + x + 1}$.T

Vedclass · Accepted Answer

$(1, 7/3]$

Vedclass · Answer

(C) Let $y = \frac{x^2 + x + 2}{x^2 + x + 1}$.We can rewrite the function as $y = \frac{(x^2 + x + 1) + 1}{x^2 + x + 1} = 1 + \frac{1}{x^2 + x + 1}$.To find the range,we need to find the range of the expression $g(x) = x^2 + x + 1$.Completing the square for $g(x)$,we get $g(x) = (x + 1/2)^2 + 3/4$.The minimum value of $g(x)$ is $3/4$ (at $x = -1/2$) and the maximum value is $\infty$.Thus,the range of $g(x)$ is $[3/4, \infty)$.Now,the range of $\frac{1}{g(x)}$ is $(0, 1/(3/4)] = (0, 4/3]$.Adding $1$ to this range,the range of $f(x)$ is $(1, 1 + 4/3] = (1, 7/3]$.

Range of the function $f(x) = \frac{x^2 + x + 2}{x^2 + x + 1}; x \in R$ is

Similar Questions

In a survey of a town,it was found that $65 \%$ of the people surveyed watch the news on $T.V.$,$40 \%$ read a newspaper,and $25 \%$ read a newspaper and watch the news on $T.V.$ What percent of the people surveyed neither watch the news on $T.V.$ nor read a newspaper?

Domain and range of $f(x) = \frac{|x - 3|}{x - 3}$ are respectively

In a certain class,$72 \%$ of the students took Biology and $44 \%$ took Mathematics. If each student took at least one of Biology or Mathematics and $40$ students took both of these subjects,the total number of students in the class is:

Find the cardinal number of the following set ${x: x = 2n, n \in N, 4 \leq x \leq 11}$.

Let $A, B, C$ be three sets such that $n(A \cap B) = n(B \cap C) = n(C \cap A) = n(A \cap B \cap C) = 2$. Then $n((A \times B) \cap (B \times C))$ is equal to -

Vedclass Test Series

Exam Paper Generator

Online Exam Module