Points at which the tangents to the curve y=x−1x+1 are parallel to the line y=2x+1 are

dydx=ddx(x−1x+1)=(x+1).1−(x−1).1(x+1)2=2(x+1)2 Slope of y=2x+1 is 2∴ 2/(x+1)2=2⇒(x+1)2=1⇒x+1=±1⇒x=0, −2 and =−1,3. Points =(0,−1),(−2,3)

Points at which the tangents to the curve y=x−1x+1 are parallel to the line y=2x+1 are

dydx=ddx(x−1x+1)=(x+1).1−(x−1).1(x+1)2=2(x+1)2 Slope of y=2x+1 is 2∴ 2/(x+1)2=2⇒(x+1)2=1⇒x+1=±1⇒x=0, −2 and =−1,3. Points =(0,−1),(−2,3)

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Points at which the tangents to the curve $y = \frac{x - 1}{x + 1}$ are parallel to the line $y = 2 x + 1$ are

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$(0, 2), (3, - 5)$

$(0, - 1), (- 2, 3)$

$(0, - 1), (2, - 3)$

$(0, 1), (3, 0)$

answer is B.

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Detailed Solution

$\frac{d y}{d x} = \frac{d}{d x} (\frac{x - 1}{x + 1}) = \frac{(x + 1) .1 - (x - 1) .1}{{(x + 1)}^{2}} = \frac{2}{{(x + 1)}^{2}}$ Slope of $y = 2 x + 1$ is 2
$∴ 2 / {(x + 1)}^{2} = 2 \Rightarrow {(x + 1)}^{2} = 1 \Rightarrow x + 1 = \pm 1 \Rightarrow x = 0$ , $- 2$ and $= - 1, 3$ . Points $= (0, - 1), (- 2, 3)$