CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

At x = 0 

$^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{0}^{-}}\text{f(x)} = ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{0}}\text{(x)}\ [ \text{Put}\ \text{x} = 0 - \text{h}, \text{h}>0\ \text{so that}\ \text{h}\rightarrow 0\ \text{as}\ \text{x}\rightarrow 0^-]$

$=\ ^{\ \ \text{Lt}}_{\text{h}\rightarrow\text{0}}(0-\text{h})$

$=\ ^{\ \ \text{Lt}}_{\text{h}\rightarrow\text{0}}\text{(-h)} = (0) = 0$

$^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{0}^{+}}\text{f(x)} = ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{0}^{+}}\text{(x)}\ [ \text{Put}\ \text{x} = 0 + \text{h}, \text{h}>0\ \text{so that}\ \text{h}\rightarrow 0\ \text{as}\ \text{x}\rightarrow 0^+]$

$=\ ^{\ \ \text{Lt}}_{\text{h}\rightarrow\text{0}}(0 + \text{h})$

= 0 + 0 = 0

$\therefore\ ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{0}^{-}}\text{f(x)} = ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{0}^{+}}\text{f(x)} = 0$

$\therefore \ ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{0}}\text{f(x)} = 0$

Also f(0) = value of x at x = 0

= 0

$\therefore$ f is continous at x = 0.

At x = 1

$^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{1}^{-}}\text{f(x)} = ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{1}^{-}}\text{x}\ [ \text{Put}\ \text{x} = 1 - \text{h}, \text{h}>0,\ \text{so that}\ \text{h}\rightarrow 0\ \text{on}\ \text{x}\rightarrow 1^-]$

$^{\ \ \text{Lt}}_{\text{h}\rightarrow\text{0}}(1 - \text{h}) = 1 - 0 = 1$

$^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{1}^{+}}\text{f(x)} = ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{1}^{+}}(5) = 5$

$\therefore\ ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{1}^{-}}\text{f(x)}\neq\ ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{1}^{+}}\text{f(x)}$

$\therefore \ ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{1}^{-}}\text{f(x)}$ does not exist

$\therefore$ f is discontinuous at x =1.

At x = 2

$^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{2}}\text{f(x)} = ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{2}}(5) = 5$

Also f(2) = 5

$\therefore\ ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{2}}\text{f(x)} = \text{f(2)} = 5$

$\therefore$ f is continous at x = 2.