\[{\rm{If\, }}f\left( x \right) = \left\{ \begin{array}{l} p{x^2} - q,\,x \in \left[ {0,\,1} \right)\\ x + 1,\,x\, \in \left( {1,\,2} \right] \end{array} \right.\]
and $$f\left( 1 \right) = 2$$   then the value of the pair $$\left( {p,\,q} \right)$$  for which $$f\left( x \right)$$  cannot be continuous at $$x=1$$  is :

Solution :
$$f\left( x \right)$$  is continuous at $$x=1$$  if
$$\eqalign{ & \mathop {\lim }\limits_{h \to 0} f\left( {1 + h} \right) = \mathop {\lim }\limits_{h \to 0} f\left( {1 - h} \right) = f\left( 1 \right) = 2 \cr & \mathop {\lim }\limits_{h \to 0} f\left( {1 + h} \right) = \mathop {\lim }\limits_{h \to 0} \left\{ {1 + h + 1} \right\} = 2 \cr & \mathop {\lim }\limits_{h \to 0} f\left( {1 - h} \right) = \mathop {\lim }\limits_{h \to 0} \left\{ {p{{\left( {1 - h} \right)}^2} - q} \right\} = p - q \cr} $$
$$\therefore \,f\left( x \right)$$   is not continuous at $$x=1$$  if $$p - q \ne 2$$