If `f(x) = {{:(kx , "for ". x le 2),(3, "for " x gt 2):}` is continuous at x=2 then the value of k is ………………………. .

Explanation

`f(x) = {{:(x^(2)+kx+3",",,"for "xge0),(2kx+3",",,"for "x lt0):}`
Graph of `y=x^(2)+kx+3` is a variable parabola intersecting the y-axis at (0, 3).
Graph of `y=2kx+3` is a variable straight line passing through (0, 3).
Also the function is continuous, so the possible graph for different values of `k` is as shown in the following figures.
Graph for `k lt 0`

Graph for `k=0`

Graph for `k gt 0`

From the above graphs, the possible value of `k` for which the function is injective is `k gt 0`.