Let A subseteq R, B subseteq R and f: A right | vedclass

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(C) The function $f: A \rightarrow B$ is defined by $f(x)=x^2-3x+2$. To be a bijection,the function must be both one-one and onto. The derivative is $

Vedclass · Accepted Answer

$A=\left[\frac{3}{2}, \infty\right), B=\left[\frac{-1}{4}, \infty\right)$

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(C) The function $f: A \rightarrow B$ is defined by $f(x)=x^2-3x+2$. To be a bijection,the function must be both one-one and onto. The derivative is $f'(x) = 2x - 3$. Setting $f'(x) = 0$ gives the vertex at $x = \frac{3}{2}$. The minimum value of the function is $f\left(\frac{3}{2}\right) = \left(\frac{3}{2}\right)^2 - 3\left(\frac{3}{2}\right) + 2 = \frac{9}{4} - \frac{9}{2} + 2 = \frac{9-18+8}{4} = -\frac{1}{4}$. For the function to be one-one,we must restrict the domain to either $A = \left(-\infty, \frac{3}{2}\right]$ or $A = \left[\frac{3}{2}, \infty\right)$. For the function to be onto,the codomain $B$ must be equal to the range,which is $\left[-\frac{1}{4}, \infty\right)$. Comparing this with the given options,option $B$ and $C$ are candidates. However,standard convention for such problems usually selects the interval starting from the vertex. Thus,$A = \left[\frac{3}{2}, \infty\right)$ and $B = \left[-\frac{1}{4}, \infty\right)$ is a valid bijection.

Let $A \subseteq R, B \subseteq R$ and $f: A \rightarrow B$ be defined by $f(x)=x^2-3x+2$. If $f$ is a bijection,then

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