Gx is an antiderivative of fx-1-2xlog2 whose graph passes through -1- 1-2 - The curve y-gx meets y -axis at

The correct option is A (0,2) f(x)=1+2^x log #160;2 g(x) is antiderivative of f (x) ∫ f(x) dx = g(x) =∫ (1+2^xlog #160;2)· ...

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Standard XII

Physics

Ideal Gas Equation

gx is an anti...

Question

$g (x)$ is an antiderivative of $f (x) = 1 + 2^{x} log 2$ whose graph passes through $(- 1, \frac{1}{2})$ . The curve $y = g (x)$ meets $y$ -axis at

(0, 1)

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(0, 2)

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(0, - 2)

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(1, 1)

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Solution

The correct option is A $(0, 2)$
$f (x) = 1 + 2^{x} l o g 2$
$g (x)$ is antiderivative of $f (x)$
$\int f (x) d x = g (x)$
$= \int (1 + 2^{x} l o g 2) \cdot d x$
$= \int d x + \int 2^{x} l o g 2 \cdot d x$
$= x + \frac{2^{x}}{l o g 2} \cdot l o g 2 + c \cdot$
$g (x) = x + 2^{x} + c$
$g (- 1) = 1 / 2 = - 1 + 1 / 2 + c$
$= 1 /_{2} + c = 1 / 2$
$\Rightarrow C = 1$
$g (x) = 1 + x + 2^{x}$
On y-axis $x = 0$
$g (0) = 1 + 1 = 2$
$(0, 2) \cdot$

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g(x) is an antiderivative of f(x)=1+2xlog2 whose graph passes through (−1,12) . The curve y=g(x) meets y -axis at

The correct option is A (0,2)f(x)=1+2xlog2g(x) is antiderivative of f(x)∫f(x)dx=g(x)=∫(1+2xlog2)⋅dx=∫dx+∫2xlog2⋅dx=x+2xlog2⋅log2+c⋅g(x)=x+2x+cg(−1)=1/2=−1+1/2+c=1/2+c=1/2⇒C=1g(x)=1+x+2xOn y-axis x=0g(0)=1+1=2(0,2)⋅

$g (x)$ is an antiderivative of $f (x) = 1 + 2^{x} log 2$ whose graph passes through $(- 1, \frac{1}{2})$ . The curve $y = g (x)$ meets $y$ -axis at