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Beyond Horizon Set 4 Paper 1 · Q8

Paper 1Calculus

Given that

dydx=4x2−3−2xx3,x≠0\frac{dy}{dx}=4x^2-\frac{3-2x}{x^3},\quad x\neq0

and y=6y=6 when x=1x=1, find yy in terms of xx.

Ay=43x3+32x−2−23x−1+316y=\frac43x^3+\frac32x^{-2}-\frac23x^{-1}+\frac{31}{6}
By=2x3+x−2−x−1+5y=2x^3+x^{-2}-x^{-1}+5
Cy=43x3+32x−2−2x−3+236y=\frac43x^3+\frac32x^{-2}-2x^{-3}+\frac{23}{6}
Dy=43x3+32x−2−2x−1+316y=\frac43x^3+\frac32x^{-2}-2x^{-1}+\frac{31}{6}
Ey=43x3+32x−2−23x−3+4y=\frac43x^3+\frac32x^{-2}-\frac23x^{-3}+4
Fy=3x3+x−2−x−1+3y=3x^3+x^{-2}-x^{-1}+3

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