Question

Consider the following in respect of the function \[f\left( x \right) = \left\{ \begin{array}{l} 2 + x,\,\,\,\,\,x \ge 0\\ 2 - x,\,\,\,\,\,x < 0 \end{array} \right.\]
$$\eqalign{ & 1.\,\,\mathop {\lim }\limits_{x \to 1} f\left( x \right)\,{\text{does not exist}} \cr & 2.{\text{ }}\,f\left( x \right)\,{\text{is differentiable at }}x = 0 \cr & 3.{\text{ }}\,f\left( x \right)\,{\text{is continuous at }}x = 0 \cr} $$
Which of the above statements is/are correct ?

A. 1 only
B. 3 only
C. 2 and 3 only
D. 1 and 3 only  
Answer :   1 and 3 only
Solution :
$$\eqalign{ & {\text{For}}\,x \geqslant 0 \cr & \mathop {\lim }\limits_{x \to 1} f\left( x \right) = \mathop {\lim }\limits_{x \to 1} 2 + x = 2 + 1 = 3 \cr & {\text{For}}\,x < 0 \cr & \mathop {\lim }\limits_{x \to 1} f\left( x \right) = \mathop {\lim }\limits_{x \to 1} 2 - x = 2 - 1 = 1 \cr & {\text{So, }}\mathop {\lim }\limits_{x \to 1} f\left( x \right){\text{ does not exist}} \cr & {\text{At }}x = 0 \cr & {\text{R}}{\text{.H}}{\text{.L}}{\text{. :}}\mathop {\lim }\limits_{h \to {0^ + }} f\left( {0 + h} \right) = \mathop {\lim }\limits_{h \to 0} 2 + h = 2 \cr & {\text{L}}{\text{.H}}{\text{.L}}{\text{. :}}\mathop {\lim }\limits_{h \to {0^ - }} f\left( {0 - h} \right) = \mathop {\lim }\limits_{h \to 0} 2 - h = 2 \cr & f\left( 0 \right) = 2 + 0 = 2 \cr & {\text{So, R}}{\text{.H}}{\text{.L}}{\text{.}} = {\text{L}}{\text{.H}}{\text{.L}}{\text{.}} = f\left( 0 \right) \cr & \Rightarrow \,f\left( x \right){\text{ is continuous at}}\,x = 0 \cr & {\text{Differentiability at }}x = 0 \cr & {\text{L}}{\text{.H}}{\text{.D}}{\text{. : }}\mathop {\lim }\limits_{h \to {0^ - }} \frac{{f\left( {0 - h} \right) - f\left( 0 \right)}}{{ - h}} = \mathop {\lim }\limits_{h \to {0^ - }} \frac{{2 + h - 2}}{{ - h}} = \frac{{ - h}}{h} = - 1 \cr & {\text{R}}{\text{.H}}{\text{.D}}{\text{. : }}\mathop {\lim }\limits_{h \to {0^ + }} \frac{{f\left( {0 + h} \right) - f\left( 0 \right)}}{h} = \mathop {\lim }\limits_{h \to {0^ + }} \frac{{2 + h - 2}}{h} = 1 \cr & {\text{Since, L}}{\text{.H}}{\text{.D}}{\text{.}} \ne {\text{R}}{\text{.H}}{\text{.D}}{\text{.}} \cr & {\text{So, }}f\left( x \right){\text{ is not differentiable at }}x = 0. \cr} $$

Releted MCQ Question on
Calculus >> Differentiability and Differentiation

Releted Question 1

There exist a function $$f\left( x \right),$$  satisfying $$f\left( 0 \right) = 1,\,f'\left( 0 \right) = - 1,\,f\left( x \right) > 0$$       for all $$x,$$ and-

A. $$f''\left( x \right) > 0$$   for all $$x$$
B. $$ - 1 < f''\left( x \right) < 0$$    for all $$x$$
C. $$ - 2 \leqslant f''\left( x \right) \leqslant - 1$$    for all $$x$$
D. $$f''\left( x \right) < - 2$$   for all $$x$$
Releted Question 2

If $$f\left( a \right) = 2,\,f'\left( a \right) = 1,\,g\left( a \right) = - 1,\,g'\left( a \right) = 2,$$         then the value of $$\mathop {\lim }\limits_{x \to a} \frac{{g\left( x \right)f\left( a \right) - g\left( a \right)f\left( x \right)}}{{x - a}}$$      is-

A. $$-5$$
B. $$\frac{1}{5}$$
C. $$5$$
D. none of these
Releted Question 3

Let $$f:R \to R$$   be a differentiable function and $$f\left( 1 \right) = 4.$$   Then the value of $$\mathop {\lim }\limits_{x \to 1} \int\limits_4^{f\left( x \right)} {\frac{{2t}}{{x - 1}}} dt$$     is-

A. $$8f'\left( 1 \right)$$
B. $$4f'\left( 1 \right)$$
C. $$2f'\left( 1 \right)$$
D. $$f'\left( 1 \right)$$
Releted Question 4

Let [.] denote the greatest integer function and $$f\left( x \right) = \left[ {{{\tan }^2}x} \right],$$    then:

A. $$\mathop {\lim }\limits_{x \to 0} f\left( x \right)$$     does not exist
B. $$f\left( x \right)$$  is continuous at $$x = 0$$
C. $$f\left( x \right)$$  is not differentiable at $$x =0$$
D. $$f'\left( 0 \right) = 1$$

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Differentiability and Differentiation


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