Question

If $f\left(x\right)=\{\begin{array}{cc}\frac{2-(128-6x{)}^{\frac{1}{7}}}{(4x+16{)}^{\frac{1}{4}}-2},& x\ne 0\\ k,& x=0\end{array}$ is continuous at $x=0$ and $g$ is a function with the property that $g(x+y)=g\left(x\right)+g\left(y\right)+xy$ and $\underset{h\to 0}{lim}\frac{g\left(h\right)}{h}=28f\left(0\right)$

• A. $g$ is a differentiable function
• B. $g\left(x\right)=3x+\frac{x^2}{2}+1$
• C. $g\left(x\right)=3x+\frac{x^2}{2}$
• D. $g\left(x\right)=3+x$

Answer 1

Answer: (A), (C)

$g\left(x\right)=3x+\frac{x^2}{2}$

$\underset{x\to 0}{lim}f\left(x\right)=\underset{x\to 0}{lim}\frac{2-(128-6x{)}^{\frac{1}{7}}}{(4x+16{)}^{\frac{1}{4}}-2}\left(\frac{0}{0}\right)$
Applying L'Hospital Rule,
$=\underset{x\to 0}{lim}\frac{0-\frac{1}{7}(-6)(128-6x{)}^{-6/7}}{\frac{1}{4}\times 4(4x+16{)}^{-3/4}-0}$
$=\frac{6\times (2^7{)}^{-6/7}}{7\times (2^4{)}^{-3/4}}$
$=\frac{3}{28}$

$\underset{h\to 0}{lim}\frac{g\left(h\right)}{h}=28f\left(0\right)=3$
Now we know that,
$g^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{g(x+h)-g\left(x\right)}{h}$
$\Rightarrow g^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{g\left(x\right)+g\left(h\right)+xh-g\left(x\right)}{h}$
$\Rightarrow g^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{g\left(h\right)}{h}+x\Rightarrow g^{\prime }\left(x\right)=3+x$
Integrating both the sides, we get
$g\left(x\right)=3x+\frac{x^2}{2}+c$
Putting $x=x,y=0$ in $g(x+y)$ we get $g\left(0\right)=0$
$g\left(x\right)=3x+\frac{x^2}{2}$