Let f: R rightarrow R be a function defined b | vedclass

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

Question

(A) Given that: $f(x) = \frac{2x+1}{3}$ ...$(i)$Also,$f(\alpha) = \frac{1}{\alpha}$.Substituting $\alpha$ in the function definition:$\frac{2\alpha+1}

Vedclass · Accepted Answer

$\frac{-1}{2}$

Vedclass · Answer

(A) Given that: $f(x) = \frac{2x+1}{3}$ ...$(i)$Also,$f(\alpha) = \frac{1}{\alpha}$.Substituting $\alpha$ in the function definition:$\frac{2\alpha+1}{3} = \frac{1}{\alpha}$$\Rightarrow 2\alpha^2 + \alpha = 3$$\Rightarrow 2\alpha^2 + \alpha - 3 = 0$Factoring the quadratic equation:$2\alpha^2 + 3\alpha - 2\alpha - 3 = 0$$\Rightarrow \alpha(2\alpha + 3) - 1(2\alpha + 3) = 0$$\Rightarrow (\alpha - 1)(2\alpha + 3) = 0$Thus,the possible values for $\alpha$ are $\alpha = 1$ and $\alpha = -\frac{3}{2}$.The sum of all possible values of $\alpha$ is $1 + (-\frac{3}{2}) = 1 - \frac{3}{2} = -\frac{1}{2}$.

Let $f: R \rightarrow R$ be a function defined by $f(x) = \frac{2x+1}{3}$. If $\alpha$ is an element in the domain of $f$ whose image is $\frac{1}{\alpha}$,then the sum of all possible values of such $\alpha$ is

Similar Questions

$f: R \rightarrow R$ is defined by $f(x+y)=f(x)+12y, \forall x, y \in R$. If $f(1)=6$,then $\sum_{r=1}^n f(r)=$

$f$ is a real-valued function satisfying the relation $f\left(3x + \frac{1}{2x}\right) = 9x^2 + \frac{1}{4x^2}$. If $f\left(x + \frac{1}{x}\right) = 1$, then $x =$

If $f(x)$ is a polynomial function satisfying $f(x) \cdot f(\frac{1}{x}) = f(x) + f(\frac{1}{x})$ and $f(4) = 65$,then the value of $f(6)$ is:

Let $f(x)$ be defined for all $x > 0$ and be continuous. Let $f(x)$ satisfy $f\left( \frac{x}{y} \right) = f(x) - f(y)$ for all $x, y > 0$ and $f(e) = 1$. Then:

Let $f(x)$ be a polynomial of degree $3$ such that $f(k) = -\frac{2}{k}$ for $k = 2, 3, 4, 5$. Then the value of $52 - 10 f(10)$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module