Pearson Edexcel IAL Mathematics Unit FP3: Further Pure Mathematics 3 Question Bank

Practise FP3 techniques across hyperbolic functions, coordinate systems, calculus, vectors and matrices using exact algebra and proof.

Covered syllabus points

FP3.1 - Hyperbolic functions FP3.2 - Further coordinate systems FP3.3 - Differentiation FP3.4 - Integration FP3.5 - Vectors FP3.6 - Further matrix algebra

[Maximum number: 9]

(a)

Use the definitions of hyperbolic functions in terms of exponentials to show that

\sinh(A+B) \equiv \sinh A \cosh B + \cosh A \sinh B

[ 3 ]

\sinh A\cosh B+\cosh A\sinh B =\frac{e^A-e^{-A}}{2}\frac{e^B+e^{-B}}{2} +\frac{e^A+e^{-A}}{2}\frac{e^B-e^{-B}}{2}

=\frac{2e^{A+B}-2e^{-(A+B)}}{4}=\sinh(A+B)

M1: replaces two of the four hyperbolic functions with correct exponential expressions.
M1: expands the numerator with at least four terms.
A1*: reaches sinh(A+B) with no errors.
There is no credit for proofs that do not use exponential definitions.

(b)

Hence express $10\sinh x+8\cosh x$ in the form $R\sinh(x+\alpha)$ where R>0,
giving $\alpha$ in the form $\ln p$ where p is an integer.

[ 4 ]

R\sinh(x+\alpha)=R\cosh\alpha\sinh x+R\sinh\alpha\cosh x

R\cosh\alpha=10,\quad R\sinh\alpha=8

R^2=10^2-8^2=36\Rightarrow R=6

\tanh\alpha=\frac{8}{10}=\frac45

\alpha=\operatorname{artanh}\left(\frac45\right)=\frac12\ln\left(\frac{1+4/5}{1-4/5}\right)=\ln 3

10\sinh x+8\cosh x=6\sinh(x+\ln 3)

B1/M1: equates coefficients correctly.
M1: complete attempt to find positive R.
M1: complete attempt to find alpha = ln p.
A1: R = 6, alpha = ln 3, so the expression is 6sinh(x + ln 3).

(c)

Hence solve the equation

10\sinh x+8\cosh x=18\sqrt7

giving your answer in the form $\ln(\sqrt7+q)$ where q is a rational number to be determined.

[ 2 ]

Using part (b),

6\sinh(x+\ln 3)=18\sqrt{7}

\sinh(x+\ln 3)=3\sqrt{7}

x+\ln 3=\operatorname{arsinh}(3\sqrt{7})

x=\ln\left(3\sqrt{7}+\sqrt{(3\sqrt{7})^2+1}\right)-\ln 3

x=\ln\left(3\sqrt{7}+8\right)-\ln 3=\ln\left(\sqrt{7}+\frac{8}{3}\right)

M1: uses part (b) to form the equation in hyperbolic sine.
M1: applies the inverse hyperbolic sine form.
A1: $x=\ln\left(\sqrt{7}+\frac{8}{3}\right)$ , so $q=\frac{8}{3}$ .

[Maximum number: 10]

The ellipse E has equation

\frac{x^2}{9}+\frac{y^2}{4}=1

The line l has equation y=kx-3, where k is a constant.
Given that E and l meet at 2 distinct points P and Q,

(a)

show that the x coordinates of P and Q are solutions of the equation

(9k^2+4)x^2-54kx+45=0

The point M is the midpoint of PQ.

[ 2 ]

\frac{x^2}{9}+\frac{(kx-3)^2}{4}=1

4x^2+9(k^2x^2-6kx+9)=36

(9k^2+4)x^2-54kx+45=0

M1 A1*

(b)

Determine, in simplest form in terms of k, the coordinates of M.

[ 3 ]

Using the sum of roots,

x_M=\frac12\left(\frac{54k}{9k^2+4}\right)=\frac{27k}{9k^2+4}

Then

y_M=k\left(\frac{27k}{9k^2+4}\right)-3=\frac{27k^2-27k^2-12}{9k^2+4}=-\frac{12}{9k^2+4}

M\left(\frac{27k}{9k^2+4},-\frac{12}{9k^2+4}\right)

M1 dM1 A1

(c)

Hence show that, as k varies, M lies on the curve with equation

x^2+py^2=qy

where p and q are constants to be determined.

[ 5 ]

From part (b),

x^2=\frac{729k^2}{(9k^2+4)^2},\qquad y^2=\frac{144}{(9k^2+4)^2}

x^2+py^2=\frac{729k^2+144p}{(9k^2+4)^2}

Since

y=-\frac{12}{9k^2+4}

match

\frac{729k^2+144p}{(9k^2+4)^2}=-\frac{12q}{9k^2+4}

This gives

p=\frac94,\qquad q=-\frac{27}{4}

Hence

x^2+\frac94y^2=-\frac{27}{4}y

M1 dM1 A1 ddM1 A1

[Maximum number: 5]

\mathrm{g}(x)=\arctan \left(\mathrm{e}^{2 x}\right)

Show that

\mathrm{g}^{\prime \prime}(x)=k \operatorname{sech}(2 x) \tanh (2 x)

where k is a constant to be found.

$g(x)=\arctan(e^{2x})$ $g'(x)=\frac{2e^{2x}}{e^{4x}+1}=\operatorname{sech}(2x)$ $g''(x)=-2\operatorname{sech}(2x)\tanh(2x)$
So k=-2.

M1A1
M1
M1
A1

[Maximum number: 8]

In this question you must show all stages of your working. Solutions relying on calculator technology are not acceptable.

Figure 1

Figure 1 shows the curve with equation

y=\ln\left(\tanh\frac{x}{2}\right) \quad 1\leqslant x\leqslant 2

(a)

Show that the length, s, of the curve is given by

s=\int_1^2 \coth x\,\mathrm{d}x

[ 4 ]

\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{1}{\tanh(x/2)}\cdot\frac12\operatorname{sech}^2(x/2)

=\frac{1}{2\sinh(x/2)\cosh(x/2)}=\operatorname{cosech}x

\sqrt{1+\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2}=\sqrt{1+\operatorname{cosech}^2x}

=\sqrt{\coth^2x}=\coth x\quad (1\le x\le2)

s=\int_1^2\coth x\,\mathrm{d}x

M1: obtains an appropriate expression for dy/dx.
M1: applies the arc length formula.
ddM1: uses identities to obtain 1+(dy/dx)^2 in terms of x.
A1*: reaches the printed integral with no errors.

(b)

Hence show that

s=\ln\left(e+\frac1e\right)

[ 4 ]

s=\int_1^2\coth x\,\mathrm{d}x

=\left[\ln(\sinh x)\right]_1^2=\ln(\sinh2)-\ln(\sinh1)

=\ln\left(\frac{\sinh2}{\sinh1}\right)

\sinh2=2\sinh1\cosh1\Rightarrow \frac{\sinh2}{\sinh1}=2\cosh1=e+e^{-1}

s=\ln\left(e+\frac1e\right)

B1: integrates coth x to ln(sinh x).
M1/dM1: substitutes limits and uses hyperbolic/exponential identities or log laws.
A1*: obtains the required result by algebraic integration.

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Covered syllabus points

Unit FP3: Further Pure Mathematics 3 question 1

Question (a)

Question (b)

Question (c)

Unit FP3: Further Pure Mathematics 3 question 2

Question (a)

Question (b)

Question (c)

Unit FP3: Further Pure Mathematics 3 question 3

Unit FP3: Further Pure Mathematics 3 question 4

Question (a)

Question (b)