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  • A function  is defined as <img alt="\mathrm{f(x)}=\begin{Bmatrix} \mathrm{e^{

A function \mathrm{f(x)} is defined as \mathrm{f(x)}=\begin{Bmatrix} \mathrm{e^{x}}, &\mathrm{x<1} \\ \mathrm{\ln x+a x^2+b x} & \mathrm{x \geq 1} \end{Bmatrix}, where \mathrm{x \in \mathbb{R}}. Which one of the following statement is TRUE ?

 

Option: 1

\mathrm{f(x)} is NOT differentiable at x = 1 for any values of a and b 


Option: 2

\mathrm{f(x)} is differentiable at x = 1 for the unique value of a and b


Option: 3

\mathrm{f(x)} is differentiable at x=1 for all values of a and b such that a+b=e.


Option: 4

\mathrm{f(x)} is differentiable at x = 1 for all values of a and b 


Answers (1)

best_answer

\text{f(x)}=\left\{\begin{array}{cc} \mathrm{e^x,} & \mathrm{x<1} \\ \mathrm{\ln x+a x^2+b x,} & \mathrm{x \geq 1} \end{array}\right.

Left hand derivative \mathrm{=\lim _{x \rightarrow 1} \frac{f(x)-f(1)}{x-1}}

Where \mathrm{f(1)=\ln (1)+a(1)^2+b(1)=a+b}

So, Left hand derivatives (LHD) 

                      \mathrm{=\lim _{x \rightarrow 1} \frac{e^x-(a+b)}{x-1}}

this limit exists if \mathrm{e=a+b}

because with this condition expression comes in \frac{0}{0} from and we can use L'Hospital Rule i.e.

                     \mathrm{\mathrm{LHD}=\lim _{x \rightarrow 1} \frac{e^x}{1}=e}

Right hand derivative (RHD)

                               \begin{aligned} & \mathrm{=\lim _{x \rightarrow 1} \frac{f(x)-f(1)}{x-1}} \\ \\& \mathrm{=\lim _{x \rightarrow 1} \frac{\ln x+a x^2+b x-(a+b)}{x-1}} \end{aligned}

this limit exist if \mathrm{1+2 a+b=a+b}

Or                     \mathrm{a=-1}                    ......(2)

Hence by using L'Hospital Rule

                 \mathrm{R H D=\lim _{x \rightarrow 1} \frac{\frac{1}{x}+2 a x+b}{1}=1+2 a+b}

LHD will be equal to RHD if a=-1 and b=e+1 Hence, f(x) is differentiable at x=1 for unique values of a and b.

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