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Q.The equation of the tangent to the curve y=x2+4x+1y = x^2 + 4x + 1 at the point x=3x = 3 is

(a) x+10y=8x + 10y = 8
(b) 10x+y=810x + y = 8
(c) 10x−y=810x - y = 8
(d) x−10y=8x - 10y = 8
Bihar BsebBihar Board Intermediate 2026MCQ· 1mImportance★★★★★
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The tangent at x=3x=3 is 10x−y=810x - y = 8.

Given y=x2+4x+1y = x^2+4x+1. The slope is

dydx=2x+4.\frac{dy}{dx} = 2x+4.

At x=3x=3: slope =2(3)+4=10= 2(3)+4 = 10, and y=9+12+1=22y = 9+12+1 = 22, so the point is (3,22)(3,22).

Equation of the tangent: …

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