# C3 2017 5 Edexcel

(a)

Note that if x = -2, then y = 2 \ln(5 - 4) - \dfrac 3 2 (-2) = 3.

Differentiating we have:

\begin{align*}\frac {\mathrm dy} {\mathrm dx} & = \frac {2 \times 2 } {2x + 5} - \frac 3 2 \\ & = \frac 4 {2x + 5} - \frac 3 2\end{align*}

using the chain rule. When x = -2 we have:

\displaystyle \frac {\mathrm dy} {\mathrm dx} = \frac 4 {5 - 4} - \frac 3 2 = \frac 5 2

So the normal to C at P has gradient:

-\dfrac 2 5

So the normal has equation:

\displaystyle y - 3 = -\dfrac 2 5 (x + 2)

So:

5y - 15 = -2(x + 2) = -2x - 4

Giving:

5y + 2x = 11

(b)

Say Q = (x, y), we would have:

5y + 2x - 11 = 0

and:

y = 2 \ln(2x + 5) - \dfrac {3x} 2

so:

\displaystyle 5 \left(2 \ln(2x + 5) - \frac {3x} 2\right) + 2x - 11 = 0

So:

\displaystyle 10 \ln (2x + 5) - \frac {15 x} 2 + 2x - 11 = 0

We can write this:

\displaystyle 10 \ln (2x + 5) - 11 = \frac {11} 2 x

Multiplying by \dfrac 2 {11} we have:

\displaystyle x = \frac {20} {11} \ln (2x + 5) - 2

as required.

We have:

x_2 = \dfrac {20} {11} \ln (4 + 5) - 2 = 1.9950 \text { (4dp)}

and:

x_3 = \dfrac {20} {11} \ln (2x_2 + 5) - 2 = 1.9929 \text{ (4dp)}

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When x=-2, y= 2 ln(1) +3 = 3 so P has coordinates (-2,3)
Differentiating:

dy/dx = 4/(2x+5) -3/2

At P, dy/dx = 4-3/2 = 5/2

The normal to C at P has dy/dx= -2/5

Hence equation of normal to C at P is y-3= -2/5 (x+2) which is 5y+2x-11 =0

(b) Let the coordinates of Q be (x,y)

Q lies on C and hence has y-coordinates 2ln(2x+5) -3/2
Q also satisfies 5y+2x-11 =0 as P and Q are collinear

Hence,
5(2ln(2x+5)-3x/2) + 2x-11 = 0
10 ln (2x+5) -15x/2 + 2x +11 =0
10 ln (2x+5) -11 = 11/2 x
x = 20/11 ln(2x+5) - 2

(Part C)
X2 = 20/11 ln (9) -2 =1.9950
X3 = 20/11 ln (23) -2 =1.9929

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Nice  