Question

f(x) = 5ax +3b if>1 differentiable
[x²+5 if x<1
at x =1 find the values ?
a and b​

Answer 1

Appropriate Question

If

$\begin{gathered}\begin{gathered}\bf\: \rm :\longmapsto\:f(x) = \begin{cases} &\sf{5ax + 3b, \: when \: x \geqslant 1} \\ &\sf{ {x}^{2} + 5, \: when \: x < 1} \end{cases}\end{gathered}\end{gathered}$

is differentiable at x = 1, find the values of a and b.

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

Given that,

$\begin{gathered}\begin{gathered}\bf\: \rm :\longmapsto\:f(x) = \begin{cases} &\sf{5ax + 3b, \: when \: x \geqslant 1} \\ &\sf{ {x}^{2} + 5, \: when \: x < 1} \end{cases}\end{gathered}\end{gathered}$

Now, It is given that f(x) is differentiable at x = 1. It implies, f(x) is continuous at x = 1 also.

[ Because every differentiable function is always continuous. ]

We know,

A function f (x) is said to be continuous at x = a, iff

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

Since, it is given that f(x) is continuous at x = 1

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

$\red{\rm :\longmapsto\:\displaystyle\lim_{x \to 1^-}( {x}^{2} + 5) = 5a + 3b}$

$\boxed{ \rm{ Put \: x = 1-h, \: \: \: as \: \: x \to 1, \: \: so \: \: h \to 0}}$

$\red{\rm :\longmapsto\:\displaystyle\lim_{h \to \: 0}( {(1 - h)}^{2} + 5) = 5a + 3b}$

$\red{\rm :\longmapsto \: 1 + 5 = 5a + 3b}$

$\red{\rm :\longmapsto \:5a + 3b = 6 - - - - (1)}$

Now, It is given that f(x) is differentiable at x = 1.

We know,

A function f(x) is said to be differentiable at x = a, iff

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

So, as it is given that f(x) is differentiable at x = 1, so

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

${\rm :\longmapsto\:\displaystyle\lim_{x \to 1^-}\dfrac{ {x}^{2} + 5 - 5a - 3b}{x - 1} = \:\displaystyle\lim_{x \to 1^ + }\dfrac{5ax + 3b - 5a - 3b}{x - 1}}$

${\rm :\longmapsto\:\displaystyle\lim_{x \to 1^-}\dfrac{ {x}^{2} + 5 - 6}{x - 1} = \:\displaystyle\lim_{x \to 1^ + }\dfrac{5ax - 5a}{x - 1}}$

$\: \: \: \: \: \: \: \: \: \: \red{\rm :\longmapsto \: \: (\because \: 5a + 3b = 6 )}$

${\rm :\longmapsto\:\displaystyle\lim_{x \to 1^-}\dfrac{ {x}^{2} - 1}{x - 1} = \:\displaystyle\lim_{x \to 1^ + }\dfrac{5a(x - 1)}{x - 1}}$

${\rm :\longmapsto\:\displaystyle\lim_{x \to 1^-}\dfrac{({x} - 1)(x + 1)}{x - 1} = \:\displaystyle\lim_{x \to 1^ + }5a}$

${\rm :\longmapsto\:\displaystyle\lim_{x \to 1^-}(x + 1) = 5a}$

$\boxed{ \rm{ Put \: x = 1-h, \: \: \: as \: \: x \to 1, \: \: so \: \: h \to 0}}$

${\rm :\longmapsto\:\displaystyle\lim_{h \to 0}(1 - h+ 1) = 5a}$

${\rm :\longmapsto\:\displaystyle\lim_{h \to 0}(2 - h) = 5a}$

$\rm :\longmapsto\:2 = 5a$

$\bf\implies \:a = \dfrac{2}{5}$

On substituting the value of a in equation (1), we get

$\rm :\longmapsto\:2 + 3b = 6$

$\rm :\longmapsto\:3b = 6 - 2$

$\rm :\longmapsto\:3b = 4$

$\bf\implies \:b = \dfrac{4}{3}$