If $f(x)=\left\{\begin{matrix}\frac{x^2+3x+λ}{x-1}&;&x≠1\\4&;&x=1\end{matrix}\right.$ |
$λ = –4$ $λ = 4$ $λ = –5$ none of these |
none of these |
$x^2 + 3x + λ$ must be zero at x = 1 So, $λ = –4$ But $\underset{x→1}{\lim}\frac{x^2+3x-4}{(x-1)}=5$. So no value of λ exist. |