The real-valued function f: R rightarrow [ fr | vedclass

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(B) Given $f(x) = | 2x + 1 | + | x - 2 |$. We can write the function as: $f(x) = \begin{cases} -(2x+1) - (x-2) = -3x + 1, & x < -\frac{1}{2} \ (2x+1)

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Onto function but not one-one

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(B) Given $f(x) = | 2x + 1 | + | x - 2 |$. We can write the function as: $f(x) = \begin{cases} -(2x+1) - (x-2) = -3x + 1, & x At $x = -\frac{1}{2}$,$f(-\frac{1}{2}) = 0 + |-\frac{1}{2} - 2| = \frac{5}{2}$. At $x = 2$,$f(2) = |4+1| + 0 = 5$. The minimum value of the function is $\frac{5}{2}$ at $x = -\frac{1}{2}$. Since the function is not strictly increasing or decreasing,it is not one-one (e.g.,$f(0) = 3$ and $f(1) = 4$,but there exist values where $f(x_1) = f(x_2)$ for $x_1 
eq x_2$). Since the codomain is $[ \frac{5}{2}, \infty )$ and the range of the function is also $[ \frac{5}{2}, \infty )$,the function is onto. Therefore,the function is onto but not one-one.

The real-valued function $f: R \rightarrow [ \frac{5}{2}, \infty )$ defined by $f(x) = | 2x + 1 | + | x - 2 |$ is

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