$g(x)$ is an anti derivative of $f(x)=1+2^x \log 2$ and the graph of $y=g(x)$ passes through $\left(-1,…
$g(x)$ is an anti derivative of $f(x)=1+2^x \log 2$ and the graph of $y=g(x)$ passes through $\left(-1, \frac{1}{2}\right)$. Then the curve meets the $\mathrm{Y}$ - axis at
$(0,1)$
$(0,2)$
$(0,-2)$
$(1,1)$
Solution
$f(x)=1+2^x \log 2$
$\begin{aligned} & \Rightarrow \int f(x)=\int\left(1+2^x \log 2\right) d x \\ & \Rightarrow g(x)=x+\frac{2^x \log 2}{\log 2}+c\end{aligned}$
$\Rightarrow \mathrm{g}(x)=x+2^x+c \Rightarrow y=x+2^x+c$ ...(i)
Equation (i) passes through $\left(-1, \frac{1}{2}\right)$
$\therefore \frac{1}{2}=-1+2^{-1}+c \Rightarrow c=1$
Putting value of $c$ in $\mathrm{eq}^{\mathrm{n}}$ (i), we get
$y=x+2^x+1$ ...(ii)
When curve (ii) meets $y$-axis: $x=0$
$\therefore y=0+2^{\circ}+1=2$
$\therefore$ Curve meet $y$-axis at the point $(0,2)$