g(x) is an antiderivative of f(x)=1+2^x log 2 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given $f(x)=1+2^x \log 2$. Since $g(x)$ is an antiderivative of $f(x)$,we have $g(x) = \int f(x) dx$. $g(x) = \int (1+2^x \log 2) dx = x + \frac{2

Vedclass · Accepted Answer

$(0,2)$

Vedclass · Answer

(B) Given $f(x)=1+2^x \log 2$. Since $g(x)$ is an antiderivative of $f(x)$,we have $g(x) = \int f(x) dx$. $g(x) = \int (1+2^x \log 2) dx = x + \frac{2^x \log 2}{\log 2} + c = x + 2^x + c$. Thus,$y = x + 2^x + c$ ... $(i)$. The curve passes through $\left(-1, \frac{1}{2}\right)$,so $\frac{1}{2} = -1 + 2^{-1} + c$. $\frac{1}{2} = -1 + \frac{1}{2} + c \implies c = 1$. Substituting $c=1$ in $(i)$,we get $y = x + 2^x + 1$. To find the intersection with the $Y$-axis,set $x=0$: $y = 0 + 2^0 + 1 = 1 + 1 = 2$. Therefore,the curve meets the $Y$-axis at $(0,2)$.

$g(x)$ is an antiderivative 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 $Y$-axis at

Similar Questions

Find the following integral:
$\int \frac{1-\sin x}{\cos ^{2} x} d x$

The value of $\int \frac{1}{x^2 - 5x + 8} dx$ is equal to...

If $\int \frac{e^x-1}{e^x+1} dx = f(x) + c$, then $f(x)$ is equal to

$\int \frac{1 - \tan x}{1 + \tan x} \, dx = $

$\int \frac{1}{x^2}(2 x+1)^3 d x$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module