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JEE Main 2017 · Mathematics · Differentiation and Application of Derivatives · easy

The equation of the tangent to the curve y=x^2+3x at the point where x=1 is

Single correct option

The equation of the tangent to the curve y=x2+3xy=x^2+3x at the point where x=1x=1 is
  1. A. y=5x−1y=5x-1
  2. B. y=5x+1y=5x+1
  3. C. y=2x+2y=2x+2
  4. D. y=4xy=4x
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Answer: (A) y=5x−1y=5x-1

At x=1x=1, y=1+3=4y=1+3=4, so the point is (1,4)(1,4). The slope is dydx=2x+3=5\dfrac{dy}{dx}=2x+3=5. The tangent is y−4=5(x−1)y-4=5(x-1), i.e. y=5x−1y=5x-1. Check: the point (1,4)(1,4) satisfies 4=5−14=5-1. Option y=4xy=4x passes through the point but has the wrong slope.
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Original question written in the JEE Main 2017 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake