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Exercise 11.13 · Q4

Q.The gradient (slope) of a curve at any point (x,y)(x,y) is x2−4x2\dfrac{x^2-4}{x^2}. If the curve passes through the point (2,7)(2,7), then the equation of the curve is

(1) y=x+4x+3y=x+\dfrac4x+3
(2) y=x+4x+4y=x+\dfrac4x+4
(3) y=x2+3x+4y=x^2+3x+4
(4) y=x2−3x+6y=x^2-3x+6
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Integrate the slope, then fix the constant with the given point.

Step 1. dydx=x2−4x2=1−4x2\dfrac{dy}{dx}=\dfrac{x^2-4}{x^2}=1-\dfrac4{x^2}.

Step 2. Integrate: y=∫(1−4x2)dx=x+4x+Cy=\displaystyle\int\left(1-\dfrac4{x^2}\right)dx=x+\dfrac4x+C. …

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