The points on the curve $y = x^{3} + x - 2$ at which its tangents are parallel to the straight line $y = 4 x - 1$ are :

(1, 0), (- 1, - 4)

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(2, 7), (- 2, - 11)

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(0, - 2), (2^{1 / 3}, 2^{1 / 3})

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(- 2^{1 / 3}, - 2^{1 / 3}), (0, - 4)

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Solution

The correct option is A $(1, 0), (- 1, - 4)$
The tangent is parallel to the line $y = 4 x - 1$
So, slope of the tangent is $4$
$\frac{d y}{d x} = 3 x^{2} + 1 = 4 \Rightarrow x = \pm 1 \Rightarrow y {∣ ∣}_{x = 1} = 0, y {∣ ∣}_{x = - 1} = - 4$

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