Let f : R to R be defined by f(x) = frac{x}{1 | vedclass

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

Question

(A) Let $y = \frac{x}{x^2 + 1}$.Rearranging the equation,we get $y(x^2 + 1) = x$,which implies $yx^2 - x + y = 0$.Since $x$ is a real number,the discr

Vedclass · Accepted Answer

$[ - \frac{1}{2}, \frac{1}{2} ]$

Vedclass · Answer

(A) Let $y = \frac{x}{x^2 + 1}$.Rearranging the equation,we get $y(x^2 + 1) = x$,which implies $yx^2 - x + y = 0$.Since $x$ is a real number,the discriminant $D$ of this quadratic equation in $x$ must be greater than or equal to $0$.$D = (-1)^2 - 4(y)(y) \ge 0$.$1 - 4y^2 \ge 0$.$4y^2 \le 1$,which means $y^2 \le \frac{1}{4}$.Taking the square root,we get $|y| \le \frac{1}{2}$,which implies $y \in [ - \frac{1}{2}, \frac{1}{2} ]$.Thus,the range of $f$ is $[ - \frac{1}{2}, \frac{1}{2} ]$.

Let $f : R \to R$ be defined by $f(x) = \frac{x}{1 + x^2}, x \in R$. Then the range of $f$ is

Similar Questions

If $R-(\alpha, \beta)$ is the range of $\frac{x+3}{(x-1)(x+2)}$,then the sum of the intercepts of the line $\alpha x+\beta y+1=0$ on the coordinate axes is

Let $f:[2, \infty) \rightarrow R$ be the function defined by $f(x)=x^{2}-4x+5$. Then the range of $f$ is:

The function $f: R \rightarrow R$ is defined by $f(x) = \cos^2 x + \sin^4 x$ for $x \in R$. Then $f(R)$ is equal to:

Domain of the function $\sqrt{\log \left\{ \frac{5x - x^2}{6} \right\}}$ is

If $f: R \rightarrow R$ is defined by $f(x)=[2x]-2[x]$ for $x \in R$, where $[x]$ is the greatest integer not exceeding $x$, then the range of $f$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module