The equation of tangent to the curve $y = 3 x^{2} - x + 1$ at the point $(1, 3)$ is

y = 5 x + 2

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y = 5 x - 2

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y = \frac{1}{5} x + 2

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y = \frac{1}{5} x - 2

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Solution

The correct option is B $y = 5 x - 2$
Given, $y = 3 x^{2} - x + 1$ and the given point is $(1, 3)$
Therefore, $\frac{d y}{d x} = 6 x - 1$
$∴ {(\frac{d y}{d x})}_{(1, 3)} = 6 - 1 = 5$
$∴$ $m = 5$
Equation of tangent is
$y = m x + c$
$y = 5 x + c$ ...(i)
Put $(1, 3)$ in equation $(i)$
$3 = 5 \times 1 + c$
$∴$ $c = - 2$
Equation of tangent $y = 5 x - 2$ .

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