The value of -22-p log1-x-1-x-1-q log1-x-1-x2-r-dx depends on

The correct option is D The value of r In our problem g(x)=log (1 x/1+x) is an odd function nbsp; ∫_ 2^2[ p g(x)+2q g(x)+r]dx=0+0+∫_ 2^2r dx ...

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Standard XII

Mathematics

Integration of Piecewise Continuous Functions

The value of ...

Question

The value of $\int_{- 2}^{2} [p l o g {(\frac{1 - x}{1 + x})}^{- 1} + q l o g {(\frac{1 - x}{1 + x})}^{2} + r] d x$ depends on

The value of

p

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The value of

q

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The value of

p

and

q

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The value of

r

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Solution

The correct option is D The value of $r$
In our problem $g (x) = l o g (\frac{1 - x}{1 + x})$ is an odd function
$\int_{- 2}^{2} [- p g (x) + 2 q g (x) + r] d x = 0 + 0 + \int_{- 2}^{2} r d x$
$= \int_{- 2}^{2} d x = r \int_{- 2}^{2} 1 d x = 4 r$
$\Rightarrow$ Value of given integral depends on r

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The value of ∫2−2[plog(1−x1+x)−1+qlog(1−x1+x)2+r]dx depends on

The correct option is D The value of rIn our problem g(x)=log(1−x1+x) is an odd function ∫2−2[−pg(x)+2qg(x)+r]dx=0+0+∫2−2rdx=∫2−2dx=r∫2−21dx=4r⇒ Value of given integral depends on r

The value of $\int_{- 2}^{2} [p l o g {(\frac{1 - x}{1 + x})}^{- 1} + q l o g {(\frac{1 - x}{1 + x})}^{2} + r] d x$ depends on

The correct option is D The value of $r$
In our problem $g (x) = l o g (\frac{1 - x}{1 + x})$ is an odd function
$\int_{- 2}^{2} [- p g (x) + 2 q g (x) + r] d x = 0 + 0 + \int_{- 2}^{2} r d x$
$= \int_{- 2}^{2} d x = r \int_{- 2}^{2} 1 d x = 4 r$
$\Rightarrow$ Value of given integral depends on r