Find the values of $p$ and $q$, so that $f(x) = \left\{ {\begin{array}{llllllllllllllllllll}{{x^2} + 3x + p,{\rm{ if }}x \le 1}\\{qx + 2,{\rm{ }}\,\,\,\,\,\,\,\,\,\,\,{\rm{if }}x > 1}\end{array}} \right.$ is differentiable at $x = 1$}

We have, $f(x) = \left\{ {\begin{array}{llllllllllllllllllll}{{x^2} + 3x + p,{\rm{ if }}x \le 1}\\{qx + 2,{\rm{ }}\,\,\,\,\,\,\,\,\,\,{\rm{if }}x > 1}\end{array}} \right.$ is differentiable at $x = 1$

therefore,$L{f^\prime }(1) = \mathop {\lim }\limits_{x \to {1^ - }} \frac{{f(x) - f(1)}}{{x - 1}}$

$= \mathop {\lim }\limits_{x \to {1^ - }} \frac{{\left( {{x^2} + 3x + p} \right) - (1 + 3 + p)}}{{x - 1}}$

$= \mathop {\lim }\limits_{h \to 0} \frac{{\left[ {{{(1 - h)}^2} + 3(1 - h) + p} \right] - [1 + 3 + p]}}{{(1 - h) - 1}}$

$= \mathop {\lim }\limits_{h \to 0} \frac{{\left[ {1 + {h^2} - 2h + 3 - 3h + p} \right] - [4 + p]}}{{ - h}}$

$= \mathop {\lim }\limits_{h \to 0} \frac{{\left[ {{h^2} - 5h + p + 4 - 4 - p} \right]}}{{ - h}} = \mathop {\lim }\limits_{h \to 0} \frac{{h[h - 5]}}{{ - h}}$

$= \mathop {\lim }\limits_{h \to 0} - [h - 5] = 5$
$R{f^\prime }(1) = \mathop {\lim }\limits_{x \to {1^ + }} \frac{{f(x) - f(1)}}{{x - 1}} = \mathop {\lim }\limits_{x \to {1^ + }} \frac{{(qx + 2) - (1 + 3 + p)}}{{x - 1}}$

$= \mathop {\lim }\limits_{h \to 0} \frac{{[q(1 + h) + 2] - (4 + p)}}{{1 + h - 1}}$

$= \mathop {\lim }\limits_{h \to 0} \frac{{[q + qh + 2 - 4 - p]}}{h} = \mathop {\lim }\limits_{h \to 0} \frac{{qh + (q - 2 - p)}}{h}$

$\Rightarrow$ $q - 2 - p = 0 \Rightarrow p - q = - 2$ ………(i)

$\Rightarrow$ $\mathop {\lim }\limits_{h \to 0} \frac{{qh + 0}}{h} = q\quad$ [for limit to exist]

If $L{f^\prime }(1) = R{f^\prime }(1),$ then $5 = q$
$\Rightarrow$ $p - 5 = - 2 \Rightarrow p = 3$

Therefore the value of $p = 3$ and $q = 5$