C1 2015 6 Edexcel

6

(a)

We have:

\begin{align*}y & = \frac {(x^2 + 4)(x - 3)} {2x} \\ & = \frac {x^3 - 3x^2 + 4x - 12} {2x} \\ & = \frac 1 2 \left(x^2 - 3x + 4 - 12x^{-1}\right)\end{align*}

Differentiating, we have:

\begin{align*}\frac {\mathrm dy} {\mathrm dx} & = \frac 1 2 \left(2x - 3 + 12x^{-2}\right) \\ & = x - \frac 3 2 + \frac 6 {x^2}\end{align*}

(b)

Substituting x = -1, the gradient of the tangent is:

\displaystyle -1 - \frac 3 2 + \frac 6 {(-1)^2} = 5 - \frac 3 2 = \frac 7 2

Noting that at x = -1, we have:

\begin{align*}y & = \frac {(-1)^2 + 4)(-1 - 3)} {-2} \\ & = \frac {5 \times (-4)} {-2} \\ & = 10\end{align*}

So the equation of the tangent is:

\displaystyle y - 10 = \frac 7 2 \left(x - (-1)\right)

So that:

\displaystyle 2y - 20 = 7x + 7

Then giving the equation of the tangent as:

\displaystyle 2y - 7x - 27 = 0

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