Question

solve it :

Question from Application of derivative​

Answer 1

$\large\underline{\sf{Solution-}}$

➢ Given function is

$\rm :\longmapsto\:f(x) = |x - 1| + {x}^{2}$

We know,

$\begin{gathered}\begin{gathered}\bf\: \rm :\longmapsto\: |x| = \begin{cases} &\sf{ - x \: \: if \: x < 0} \\ &\sf{x \: \: \: if \: x \geqslant 0} \end{cases}\end{gathered}\end{gathered}$

Let first define the function as

$\begin{gathered}\begin{gathered}\bf\: \rm :\longmapsto\:f(x) = \begin{cases} &\sf{ - (x - 1) + {x}^{2} \: \: if \: x < 1 } \\ &\sf{x - 1 + {x}^{2} \: \: if \: x \: \geqslant 1} \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\bf\: \rm :\longmapsto\:f(x) = \begin{cases} &\sf{ - x + 1+ {x}^{2} \: \: if \: x < 1 } \\ &\sf{{x}^{2} + x - 1 \: \: if \: x \: \geqslant 1} \end{cases}\end{gathered}\end{gathered}$

$\red{\begin{gathered}\begin{gathered}\bf\: \bf :\longmapsto\:f(x) = \begin{cases} &\sf{{x}^{2} - x + 1\: \: if \: x < 1 } \\ &\sf{{x}^{2} + x - 1 \: \: if \: x \: \geqslant 1} \end{cases}\end{gathered}\end{gathered}}$

Thus, Breaking point is x = 1, hence we have to check continuity and differentiability at x = 1.

Now, To check continuity at x = 1.

$\rm :\longmapsto\:f(1) = 1 + 1 - 1 = 0$

LHL

$\rm :\longmapsto\:\displaystyle\lim_{x \to 1^-} \: f(x)$

$\rm \: = \: \: \displaystyle\lim_{x \to 1^-}( {x}^{2} - x + 1)$

$\purple{ \sf{ \: Put \: x = 1 - h, \: As \: x \to \: 1, \: so \: h \: \to \: 0}}$

$\rm \: = \: \: \displaystyle\lim_{h \to 0}( {(1 - h)}^{2} - (1 - h) + 1)$

$\rm \: = \: \: 1 - 1 + 1$

$\rm \: = \: \: 1$

So,

$\bf :\longmapsto\:\displaystyle\lim_{x \to 1^-} \: f(x) = 1$

Now,

Consider,

RHL

$\rm :\longmapsto\:\displaystyle\lim_{x \to 1^ + } \: f(x)$

$\rm \: = \: \: \displaystyle\lim_{x \to 1^ + }( {x}^{2} + x - 1)$

$\purple{ \sf{ \: Put \: x = 1 + h, \: As \: x \to \: 1, \: so \: h \: \to \: 0}}$

$\rm \: = \: \: \displaystyle\lim_{h \to 0}( {(1 + h)}^{2} + (1 + h) - 1)$

$\rm \: = \: \: 1 + 1 - 1$

$\rm \: = \: \: 1$

So,

$\bf :\longmapsto\:\displaystyle\lim_{x \to 1^ + } \: f(x) = 1$

So, we concluded that

$\rm :\longmapsto\:\displaystyle\lim_{x \to 1^-}f(x) = \displaystyle\lim_{x \to 1^ + }f(x) = f(1) = 1$

$\pink{\bf\implies \:f(x) \: is \: continuous \: at \: x = 1}$

Now, To check differentiability at x = 1.

Consider, RHD

$\rm :\longmapsto\:\displaystyle\lim_{x \to 1^-}\dfrac{f(x) - f(1)}{x - 1}$

$\rm \: = \: \: \displaystyle\lim_{x \to 1^-}\dfrac{ {x}^{2} - x + 1 - 1 }{x - 1}$

$\rm \: = \: \: \displaystyle\lim_{x \to 1^-}\dfrac{ {x}^{2} - x }{x - 1}$

$\rm \: = \: \: \displaystyle\lim_{x \to 1^-}\dfrac{ x(x - 1) }{x - 1}$

$\rm \: = \: \: \displaystyle\lim_{x \to 1^-}x$

$\purple{ \sf{ \: Put \: x = 1 - h, \: As \: x \to \: 1, \: so \: h \: \to \: 0}}$

$\rm \: = \: \: \displaystyle\lim_{h \to 0}(1 - h)$

$\rm \: = \: \: 1$

So,

$\rm :\longmapsto\:\displaystyle\lim_{x \to 1^-}\dfrac{f(x) - f(1)}{x - 1} = 1$

Now,

Consider, RHD

$\rm :\longmapsto\:\displaystyle\lim_{x \to 1^ + }\dfrac{f(x) - f(1)}{x - 1}$

$\rm \: = \: \: \displaystyle\lim_{x \to 1^ + }\dfrac{ {x}^{2} + x - 1 - 1 }{x - 1}$

$\rm \: = \: \: \displaystyle\lim_{x \to 1^ + }\dfrac{ {x}^{2} + x - 2 }{x - 1}$

$\rm \: = \: \: \displaystyle\lim_{x \to 1^ + }\dfrac{ {x}^{2} + 2x - x - 2 }{x - 1}$

$\rm \: = \: \: \displaystyle\lim_{x \to 1^ + }\dfrac{ {x}(x + 2) - 1(x + 2) }{x - 1}$

$\rm \: = \: \: \displaystyle\lim_{x \to 1^ + }\dfrac{(x + 2)(x - 1) }{x - 1}$

$\rm \: = \: \: \displaystyle\lim_{x \to 1^ + }(x + 2)$

$\purple{ \sf{ \: Put \: x = 1 + h, \: As \: x \to \: 1, \: so \: h \: \to \: 0}}$

$\rm \: = \: \: \displaystyle\lim_{h \to 0 }(1 + h + 2)$

$\rm \: = \: \: 3$

So,

$\rm :\longmapsto\:\displaystyle\lim_{x \to 1^ + }\dfrac{f(x) - f(1)}{x - 1} = 3$

So, we concluded that

$\rm\implies\:LHD \: \ne \: RHD$

$\pink{\bf\implies \:f(x) \: is \:not \: differentiable \: at \: x = 1}$

Hence, Option (b) is correct.