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

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

Question

(A) Let $y = \frac{x^2-x+4}{x^2+x+4}$.Multiplying both sides by $(x^2+x+4)$, we get $y(x^2+x+4) = x^2-x+4$.Rearranging the terms, we get $(y-1)x^2 + (

Vedclass · Accepted Answer

$[\frac{3}{5}, \frac{5}{3}]$

Vedclass · Answer

(A) Let $y = \frac{x^2-x+4}{x^2+x+4}$.Multiplying both sides by $(x^2+x+4)$, we get $y(x^2+x+4) = x^2-x+4$.Rearranging the terms, we get $(y-1)x^2 + (y+1)x + (4y-4) = 0$.For $x$ to be a real number, the discriminant $D$ must be greater than or equal to $0$.$D = (y+1)^2 - 4(y-1)(4y-4) \geq 0$.$(y+1)^2 - 16(y-1)^2 \geq 0$.Using the identity $a^2 - b^2 = (a-b)(a+b)$, we get $((y+1) - 4(y-1))((y+1) + 4(y-1)) \geq 0$.$(y+1-4y+4)(y+1+4y-4) \geq 0$.$(5-3y)(5y-3) \geq 0$.Multiplying by $-1$ reverses the inequality: $(3y-5)(5y-3) \leq 0$.The roots are $y = \frac{5}{3}$ and $y = \frac{3}{5}$.Thus, the range is $y \in [\frac{3}{5}, \frac{5}{3}]$.

Let $f: R \rightarrow R$ be defined as $f(x) = \frac{x^2-x+4}{x^2+x+4}$. Then, the range of the function $f(x)$ is

Similar Questions

Let the domain of the function $f(x) = \cos^{-1}\left(\frac{4x+5}{3x-7}\right)$ be $[\alpha, \beta]$ and the domain of $g(x) = \log_2\left(2-6\log_{27}(2x+5)\right)$ be $(\gamma, \delta)$. Then $|7(\alpha+\beta)+4(\gamma+\delta)|$ is equal to . . . . . . .

If $f:[-3,2] \rightarrow [0, \sqrt[3]{x}]$ is an onto function defined by $f(n) = \begin{cases} 2+\sqrt[3]{n}, & -3 \leq n \leq -1 \\ n^{2/3}, & -1 < n \leq 2 \end{cases}$, then $x=$

If $R = \{ (x, y) | x, y \in Z, x^2 + y^2 \le 4 \}$ is a relation in $Z$,then the domain of $R$ is:

$A$ real valued function $f(x) = |x^2 - 3x + 2| + 2x - 3$ is defined on $[-2, 1]$. If $m$ and $M$ are absolute minimum and absolute maximum values of $f$ respectively, then $M - 4m =$

The domain of the real valued function $f(x) = \frac{\sqrt{2-x} + \sqrt{1+x}}{\sqrt{x+3}}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module