If $f(x)=\left\{\begin{matrix}\frac{x^2+3x+λ}{x-1}&;&x≠1\\4&;&x=1\end{matrix}\right.$
Answer & explanation
Correct answer: option 4
$x^2 + 3x + λ$ 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.