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प्रश्न
Find the equation of tangent to the curve y = 3x2 − x + 1 at P(1, 3).
उत्तर
Equation of the curve is y = 3x2 − x + 1
Differentiating w.r.t. x, we get
`dy/dx = 6x - 1`
∴ Slope of tangent at P(1, 3) is
`(dy/dx)_((1","3))` = 6(1) − 1 = 5
∴ the equation of tangent at P(1, 3) is
y − 3 = 5(x − 1)
∴ y − 3 = 5x − 5
∴ 5x − y − 2 = 0
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