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Question 17 of 37

Q.The slope of a tangent to the curve y=3x2−x+1y = 3x^2 - x + 1 at (1,3)(1, 3) is ______.

(a) 5
(b) −5-5
(c) −15\dfrac{-1}{5}
(d) 15\dfrac{1}{5}
Maharashtra MsbshseMaharashtra HSC (MSBSHSE) Board 2022MCQ· 1mImportance★★★★★
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The tangent's slope is dydx\dfrac{dy}{dx} evaluated at the point. Here dydx=6x−1\dfrac{dy}{dx} = 6x - 1, and at (1,3)(1, 3) this equals 6(1)−1=56(1) - 1 = 5.

The slope of the tangent to a curve y=f(x)y = f(x) at a point is given by the derivative dydx\dfrac{dy}{dx} at that point.

Differentiating y=3x2−x+1y = 3x^2 - x + 1 term by term:

dydx=6x−1.\frac{dy}{dx} = 6x - 1.

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