Continuity MCQ Questions & Answers in Calculus | Maths

$$\eqalign{ & {\text{RH limit}} = \mathop {\lim }\limits_{h \to 0} \frac{{\cos \left\{ {\sin \left( {0 + h} \right)} \right\} - \cos \left( {0 + h} \right)}}{{{{\left( {0 + h} \right)}^2}}} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{\cos \,\sin \,h - \cos \,h}}{{{h^2}}} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{ - \sin \left( {\sin \,h} \right) \times \cos \,h + \sin \,h}}{{2h}} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{ - \cos \left( {\sin \,h} \right) \times {{\cos }^2}h + \sin \left( {\sin \,h} \right) \times \sin \,h + \cos \,h}}{2} \cr & = 0 \cr} $$
Similarly, LH limit $$=0.$$
As $$f\left( x \right)$$  is continuous at $$x = 0,\,f\left( 0 \right) = 0$$

$$\eqalign{ & {\text{RH limit}} = \mathop {\lim }\limits_{h \to 0} {e^{ - \left| {0 + h} \right|}} = \mathop {\lim }\limits_{h \to 0} {e^{ - h}} = 1 \cr & {\text{LH limit}} = \mathop {\lim }\limits_{h \to 0} {e^{ - \left| {0 - h} \right|}} = \mathop {\lim }\limits_{h \to 0} {e^{ - h}} = 1 \cr} $$
So, the function is continuous at $$x=0$$
$$\eqalign{ & {\text{RH derivative}} = \mathop {\lim }\limits_{h \to 0} \frac{{{e^{ - \left| {0 + h} \right|}} - {e^{ - \left| 0 \right|}}}}{h} = \mathop {\lim }\limits_{h \to 0} \frac{{{e^{ - h}} - 1}}{h} = - 1 \cr & {\text{LH derivative}} = \mathop {\lim }\limits_{h \to 0} \frac{{{e^{ - \left| {0 - h} \right|}} - {e^{ - \left| 0 \right|}}}}{{ - h}} = \mathop {\lim }\limits_{h \to 0} \frac{{{e^{ - h}} - 1}}{{ - h}} = 1 \cr} $$
So, the function is not differentiable at $$x=0.$$

$$f\left( x \right) = x - \left| x \right|.\left| {1 - x} \right|.$$     We know that $$x,\,\left| x \right|,\,\left| {1 - x} \right|$$    are continuous everywhere. As the product and algebraic sum of continuous functions are continuous, $$f\left( x \right)$$  is continuous everywhere.

Given function is \[f\left( x \right) = \left\{ \begin{array}{l} \,\,\,mx + 1,\,\,\,\,\,\,\,x \le \frac{\pi }{2}\\ \sin \,x + n,\,\,\,\,\,x > \frac{\pi }{2} \end{array} \right.\]
As given this function is continuous at $$x = \frac{\pi }{2}$$
So, limit of function when $$x \to \frac{\pi }{2} = f\left( {\frac{\pi }{2}} \right)$$
$$\eqalign{ & \Rightarrow \mathop {\lim }\limits_{x \to \frac{\pi }{2} + } \left( {\sin \,x + n} \right) = f\left( {\frac{\pi }{2}} \right) \cr & \Rightarrow \mathop {\lim }\limits_{h \to 0} \left( {\sin \left( {\frac{\pi }{2} + h} \right) + n} \right) = \frac{{mx}}{2} + 1 \cr & \Rightarrow \sin \frac{\pi }{2} + n = \frac{{m\pi }}{2} + 1 \cr & \Rightarrow 1 + n = \frac{{m\pi }}{2} + 1 \cr & \Rightarrow n = \frac{{m\pi }}{2} \cr} $$

As we know,
A function $$f\left( x \right)$$  is said to be continuous at a point $$x = a$$  iff
$$\mathop {\lim }\limits_{x \to a} f\left( x \right) = f\left( a \right),$$    otherwise not continuous.
Thus $$f\left( x \right)$$ is continuous at $$x = a$$  iff
$$\mathop {\lim }\limits_{x \to {a^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {a^ + }} f\left( x \right) = f\left( a \right)$$
\[{\rm{Since, }}f\left( x \right) = \left\{ \begin{array}{l} \,\,\,{5^{\frac{1}{x}}},\,\,\,\,\,\,\,x < 0\\ \lambda \left[ x \right],\,\,\,\,x \ge 0 \end{array} \right.{\rm{ \,and\,\, }}\lambda \, \in \,R\]
$$\eqalign{ & {\text{R}}{\text{.H}}{\text{.L}}{\text{. at }}x = 0:\mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {0^ + }} \lambda \left[ x \right] = \mathop {\lim }\limits_{h \to 0} \lambda \left[ h \right] = 0 \cr & {\text{L}}{\text{.H}}{\text{.L}}{\text{. at }}x = 0:\mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {0^ - }} {5^{\frac{1}{x}}} = \mathop {\lim }\limits_{h \to 0} {5^{ - \frac{1}{h}}} = {5^\infty } = \infty \cr & {\text{and }}f\left( 0 \right) = \lambda \left[ 0 \right]0 \cr & {\text{Since, L}}{\text{.H}}{\text{.L}}{\text{.}} \ne {\text{R}}{\text{.H}}{\text{.L}}{\text{.}} \cr} $$
$$\therefore \,f\left( x \right)$$   is not continuous.

$$\eqalign{ & f\left( a \right) = 0 \cr & \mathop {\lim }\limits_{x \to a - } f\left( x \right) = \mathop {\lim }\limits_{x \to a - } \left( {\frac{{{x^2}}}{a} - a} \right) = \mathop {\lim }\limits_{h \to 0} \left\{ {\frac{{{{\left( {a - h} \right)}^2}}}{a} - a} \right\} = 0 \cr & {\text{and }}\mathop {\lim }\limits_{x \to a + } f\left( x \right) = \mathop {\lim }\limits_{h \to 0} \left\{ {a - \frac{{{{\left( {a + h} \right)}^2}}}{a}} \right\} = 0 \cr} $$
Hence it is continuous at $$x = a.$$

For a function to be continuous at a point the limit should be exist and should be equal to the value of the function at the point.
Here point is $$x = 0$$
$$\eqalign{ & {\text{and }}\mathop {\lim }\limits_{x \to 0} f\left( x \right) \cr & = \mathop {\lim }\limits_{x \to 0} {\left( {x + 1} \right)^{\cot \,x}} \cr & = \mathop {\lim }\limits_{x \to 0} {\left( {1 + x} \right)^{\cot \,x}} \cr & = \mathop {\lim }\limits_{x \to 0} {\left( {1 + x} \right)^{\frac{1}{x}.x\,\cot \,x}} \cr & = \mathop {\lim }\limits_{x \to 0} {\left( {x + 1} \right)^{\frac{1}{x}.\mathop {\lim }\limits_{x \to 0} \frac{x}{{\tan \,x}}}} \cr & = {e^1} \cr & = e \cr} $$
Since, limiting value of $$f\left( x \right) = e,$$   when $$x \to 0,\,f\left( 0 \right)$$   should also be equal to $$e.$$

Clearly, $$0 \leqslant f\left( 0 \right) \leqslant 1$$    and $$0 \leqslant f\left( 1 \right) \leqslant 1.$$    As $$f\left( x \right)$$  is continuous, $$f\left( x \right)$$  attains all values between $$f\left( 0 \right)$$  to $$f\left( 1 \right),$$  and the graph will have no breaks. So, the graph will cut the line $$y=x$$  at one point $$x$$ at least where $$0 \leqslant x \leqslant 1.$$   So, $$f\left( x \right) = x$$   at that point.

For $$f\left( x \right)$$ to be continuous at $$x =0$$
$$\eqalign{ & f\left( 0 \right) = \mathop {\lim }\limits_{x \to 0} f\left( x \right) \cr & = \mathop {\lim }\limits_{x \to 0} \frac{{\ln \left( {1 + ax} \right) - \ln \left( {1 - bx} \right)}}{x}\,\,\left[ {{\text{using}}\,\,\mathop {\lim }\limits_{x \to 0} \frac{{\ln \left( {1 + x} \right)}}{x} = 1} \right] \cr & = a + b \cr} $$

$$\eqalign{ & f\left( { - 1} \right) = b\left( {1 - 1} \right) + 1 = 1; \cr & \mathop {\lim }\limits_{h \to 0} f\left( { - 1 + h} \right) = 1 \cr & \mathop {\lim }\limits_{h \to 0} f\left( { - 1 - h} \right) = \mathop {\lim }\limits_{h \to 0} \sin \left( {\pi \left( { - 1 - h} \right) + \pi a} \right) \cr & = \sin \left( { - \pi + \pi a} \right) \cr & = - \sin \,\pi a \cr & {\text{For continuous }}\sin \,\pi a = - 1 = \sin \left( {2n\pi + \frac{{3\pi }}{2}} \right) \cr & \Rightarrow \pi a = 2n\pi + \frac{{3\pi }}{2} \cr & \Rightarrow a = 2n + \frac{3}{2} \cr & {\text{Hence, }}a = 2n + \frac{3}{2},\,n\, \in \,I{\text{ and }}b\, \in \,R \cr} $$