Home / IBDP Maths analysis and approaches Topic: SL 5.1 The derivative interpreted as a gradient function and as a rate of change HL Paper 1

# IBDP Maths analysis and approaches Topic: SL 5.1 The derivative interpreted as a gradient function and as a rate of change HL Paper 1

### Question

Consider the function f ( x ) = ax3 + bx2 + cx + d, where x ∈ R and a, b, c , d ∈ R.

1. (i) Write down an expression for f ( x ).

(ii) Hence, given that f 1 does not exist, show that b2 3ac > 0 . [4]

2. Consider the function g ( x ) $$\frac{1}{2}$$  x3 3x2 + 6x – 8 , where x ∈ R.

1. Show that g1 exists.

2. g ( x ) can be written in the form p ( x – 2 )3 + q , where p , q ∈ R. Find the value of p and the value of q .

3. Hence find g1 ( x ). [8]

The graph of y = g ( x ) may be obtained by transforming the graph of y = x3 using a sequence of three transformations.

3. State each of the transformations in the order in which they are applied. [3]

4. Sketch the graphs of y = g ( x ) and y = g1 ( x ) on the same set of axes, indicating the points where each graph crosses the coordinate axes. [5]

Ans:

1. (a) (i) f'(x)=$$3ax^{2}$$+2bx+c
2. (a) (ii)
1. $$since f^{-1}$$dose not exist , there must be two turning points
2. (f(x)= 0)has more than one solutin
3. using the disriminat Δ >0
4.  $$4b^{2}-12ac>0$$
5. $$b^{2}$$-3ac>0
3. (b) (i)
1. METHOD 1
2. $$b^{2}-3ac=(-3)^{2}-3\times \frac{1}{2}\times 6$$ = 9-9 = 0
3. hence g-1 exists
4. METHOD 2
5. g'(x)= $$\frac{3}{2}x^{2}-6x+6$$
6. $$\Delta = (-6)^{2}-4\times \frac{3}{2}\times 6$$
7. Δ = 36 -36 = 0 ⇒ there is (only) one point with gradient of 0 and this must be a point of inflexion (since g (x ) is a cubic.)

hence g-1 exists

1. (b) (ii)
1. p= $$\frac{1}{2}$$
2. $$(x-2)^{3}=x^{3}-6x^{2}$$+12x=8
3. $$\frac{1}{2}(x^{3}-6x^{3}+12x-8)$$= $$\frac{1}{2}x^{3}-3x^{2}+6x-4$$
4. g(x)=$$\frac{1}{2}(x-2)^{3}-4$$
2. (b) (iii)
1. x= $$\frac{1}{2}(y-2)^{3}-4$$
3. (c)
1. translation through $$\begin{pmatrix} 2\\ 0 \end{pmatrix}$$
2. EITHER
3. a stretch scale factor $$\frac{1}{2}$$ parallel to the y-axis then a translation through $$(_{-4}^{0})$$
4. OR
5. a translation through  $$(_{-8}^{0})$$ then a stretch scale factor $$\frac{1}{2}$$ parallel to the y-axis
4. (d)

### Question

The acceleration, a ms-2 , of a particle moving in a horizontal line at time t seconds, t ≥ 0 , is given by a = – (1+v) where v ms-1 is the particle’s velocity and v > -1.
At t = 0 , the particle is at a fixed origin O and has initial velocity v0 ms-1 .

(a) By solving an appropriate differential equation, show that the particle’s velocity at time t is given by v ( t ) = (1 + v0) e-t – 1 . [6]

(b) Initially at O, the particle moves in the positive direction until it reaches its maximum displacement from O. The particle then returns to O.

Let s metres represent the particle’s displacement from O and smax its maximum displacement from O.
(i) Show that the time T taken for the particle to reach smax satisfies the equation eT = 1 + v0 .
(ii) By solving an appropriate differential equation and using the result from part (b) (i), find an expression for smax in terms of v0 . [7]

Let v (T – k) represent the particle’s velocity k seconds before it reaches smax , where v (T – k) = (1 + v0) e-(T – k) – 1 .

(c) By using the result to part (b) (i), show that v (T – k) = ek – 1 . [2]

Similarly, let v (T + k) represent the particle’s velocity k seconds after it reaches smax .

(d) Deduce a similar expression for v (T + k) in terms of k . [2]

(e) Hence, show that v (T – k) + v (T + k) ≥ 0 . [3]

Ans:

(a) Since $a=\frac{\text{d}v}{\text{d}t}$, and $a=-\left(1+v\right)$, we have
$$\begin{eqnarray} \frac{\text{d}v}{\text{d}t} = -1-v \nonumber \\ \frac{\text{d}v}{\text{d}t}+v = -1. \end{eqnarray}$$
Let $\text{I}\left(x\right)=\text{e}^{\int 1 \text{d}t}=\text{e}^t$.<br>
Then, we have
$$\begin{eqnarray} \frac{\text{d}}{\text{d}t}\left(v\text{e}^t\right) = -\text{e}^t \nonumber \\ v\text{e}^t = -\int\text{e}^t \text{d}t \nonumber \\ v\text{e}^t = -\text{e}^t+c \nonumber \\ v = c\text{e}^{-t}-1. \end{eqnarray}$$
Since $v=v_0$ when $t=0$, we have $c=1+v_0$, i.e., $v=\left(1+v_0\right)\text{e}^{-t}-1$.<br>
(b)(i) At $s_{\text{max}}$, $t=T$, i.e., we have
$$\begin{eqnarray} v\left(T\right) = 0 \nonumber \\ \left(1+v_0\right)\text{e}^{-T}-1 = 0 \nonumber \\ \left(1+v_0\right)\text{e}^{-T} = 1 \nonumber \\ 1+v_0 = \text{e}^T. \end{eqnarray}$$
(b)(ii) Since $\frac{\text{d}s}{\text{d}t}=\left(1+v_0\right)\text{e}^{-t}-1$, integrating both sides with respect to $t$, we have
$$\begin{eqnarray} s &=& \int\left(1+v_0\right)\text{e}^{-t}-1\text{d}t \nonumber \\ &=& -\left(1+v_0\right)\text{e}^{-t}-t+c. \end{eqnarray}$$
When $t=0$, $s=0$, i.e.,
$$\begin{eqnarray} 0 = -\left(1+v_0\right)+c \nonumber \\ c = \left(1+v_0\right). \end{eqnarray}$$
Thus, $s=\left(1+v_0\right)-\left(1+v_0\right)\text{e}^{-t}-t$.<br>
Since from (b)(i) we have $1+v_0=\text{e}^T$, $T=\ln\left(1+v_0\right)$, i.e., $s_{\text{max}}=v_0-\ln \left(1+v_0\right)$.<br>
(c) Since $\text{e}^T=1+v_0$, we have $\left(1+v_0\right)\text{e}^{-T}=1$, i.e.,
$$\begin{eqnarray} v\left(T-k\right) &=& \left(1+v_0\right)\text{e}^{-\left(T-k\right)}-1 \nonumber \\ &=& \left(1+v_0\right)\text{e}^{-T}\text{e}^k-1 \nonumber \\ &=& \text{e}^k-1. \end{eqnarray}$$
(d) Similarly, as in (c),
$$\begin{eqnarray} v\left(T+k\right) &=& \left(1+v_0\right)\text{e}^{-\left(T+k\right)}-1 \nonumber \\ &=& \left(1+v_0\right)\text{e}^{-T}\text{e}^{-k}-1 \nonumber \\ &=& \text{e}^{-k}-1. \end{eqnarray}$$
(e)
$$\begin{eqnarray} v\left(T-k\right)+v\left(T+k\right) &=& \text{e}^k-1+\text{e}^{-k}-1 \nonumber \\ &=& \text{e}^k+\text{e}^{-k}-2 \nonumber \\ &=& \frac{\text{e}^{2k}-2\text{e}^k+1}{\text{e}^k} \nonumber \\ &=& \frac{\left(\text{e}^k-1\right)^2}{\text{e}^k} \geq 0. \end{eqnarray}$$

### Question

Consider the function f defined by f (x) = ln (x2 – 16) for x > 4 .
The following diagram shows part of the graph of f which crosses the x-axis at point A, with
coordinates ( a , 0 ). The line L is the tangent to the graph of f at the point B.

(a) Find the exact value of a . [3]

(b) Given that the gradient of L is $$\frac{1}{3}$$ , find the x-coordinate of B. [6]

Ans:

(a) When f(x)=0, we have
ln(x2−16)=0

x2−16=1

x=±√17
However, since x>4, we have x=√17
(b) Differentiating f(x) with respect to x, we have f′(x)= $$\frac{2x}{x^2-16}$$
At B, f′(x)=$$\frac{1}{3}$$ , i.e., we have
x2−16=6x

x2−6x−16=0

(x+2)(x−8)=0.
Thus, x=−2 (reject) or x=8.

### Question

a.Using the definition of a derivative as $$f'(x) = \mathop {\lim }\limits_{h \to 0} \left( {\frac{{f(x + h) – f(x)}}{h}} \right)$$ , show that the derivative of $$\frac{1}{{2x + 1}}{\text{ is }}\frac{{ – 2}}{{{{(2x + 1)}^2}}}$$.[4]

b.Prove by induction that the $${n^{{\text{th}}}}$$ derivative of $${(2x + 1)^{ – 1}}$$ is $${( – 1)^n}\frac{{{2^n}n!}}{{{{(2x + 1)}^{n + 1}}}}$$.[9]

### Markscheme

let $$f(x) = \frac{1}{{2x + 1}}$$ and using the result $$f'(x) = \mathop {\lim }\limits_{h \to 0} \left( {\frac{{f(x + h) – f(x)}}{h}} \right)$$

$$f'(x) = \mathop {\lim }\limits_{h \to 0} \left( {\frac{{\frac{1}{{2(x + h) + 1}} – \frac{1}{{2x + 1}}}}{h}} \right)$$     M1A1

$$\Rightarrow f'(x) = \mathop {\lim }\limits_{h \to 0} \left( {\frac{{[2x + 1] – [2(x + h) + 1]}}{{h[2(x + h) + 1][2x + 1]}}} \right)$$     A1

$$\Rightarrow f'(x) = \mathop {\lim }\limits_{h \to 0} \left( {\frac{{ – 2}}{{[2(x + h) + 1][2x + 1]}}} \right)$$     A1

$$\Rightarrow f'(x) = \frac{{ – 2}}{{{{(2x + 1)}^2}}}$$     AG

[4 marks]

a.

let $$y = \frac{1}{{2x + 1}}$$

we want to prove that $$\frac{{{{\text{d}}^n}y}}{{{\text{d}}{x^n}}} = {( – 1)^n}\frac{{{2^n}n!}}{{{{(2x + 1)}^{n + 1}}}}$$

let $$n = 1 \Rightarrow \frac{{{\text{d}}y}}{{{\text{d}}x}} = {( – 1)^1}\frac{{{2^1}1!}}{{{{(2x + 1)}^{1 + 1}}}}$$     M1

$$\Rightarrow \frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{ – 2}}{{{{(2x + 1)}^2}}}$$ which is the same result as part (a)

hence the result is true for $$n = 1$$     R1

assume the result is true for $$n = k{\text{ : }}\frac{{{{\text{d}}^k}y}}{{{\text{d}}{x^k}}} = {( – 1)^k}\frac{{{2^k}k!}}{{{{(2x + 1)}^{k + 1}}}}$$     M1

$$\frac{{{{\text{d}}^{k + 1}}y}}{{{\text{d}}{x^{k + 1}}}} = \frac{{\text{d}}}{{{\text{d}}x}}\left[ {{{( – 1)}^k}\frac{{{2^k}k!}}{{{{(2x + 1)}^{k + 1}}}}} \right]$$     M1

$$\Rightarrow \frac{{{{\text{d}}^{k + 1}}y}}{{{\text{d}}{x^{k + 1}}}} = \frac{{\text{d}}}{{{\text{d}}x}}\left[ {{{( – 1)}^k}{2^k}k!{{(2x + 1)}^{ – k – 1}}} \right]$$     (A1)

$$\Rightarrow \frac{{{{\text{d}}^{k + 1}}y}}{{{\text{d}}{x^{k + 1}}}} = {( – 1)^k}{2^k}k!( – k – 1){(2x + 1)^{ – k – 2}} \times 2$$     A1

$$\Rightarrow \frac{{{{\text{d}}^{k + 1}}y}}{{{\text{d}}{x^{k + 1}}}} = {( – 1)^{k + 1}}{2^{k + 1}}(k + 1)!{(2x + 1)^{ – k – 2}}$$     (A1)

$$\Rightarrow \frac{{{{\text{d}}^{k + 1}}y}}{{{\text{d}}{x^{k + 1}}}} = {( – 1)^{k + 1}}\frac{{{2^{k + 1}}(k + 1)!}}{{{{(2x + 1)}^{k + 2}}}}$$     A1

hence if the result is true for $$n = k$$ , it is true for $$n = k + 1$$

since the result is true for $$n = 1$$ , the result is proved by mathematical induction     R1

Note: Only award final R1 if all the M marks have been gained.

[9 marks]

b.

### Question

A curve is defined by $$xy = {y^2} + 4$$.

a.Show that there is no point where the tangent to the curve is horizontal.[4]

b.Find the coordinates of the points where the tangent to the curve is vertical.[4]

### Markscheme

$$x\frac{{{\text{d}}y}}{{{\text{d}}x}} + y = 2y\frac{{{\text{d}}y}}{{{\text{d}}x}}$$     M1A1

a horizontal tangent occurs if $$\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0$$ so $$y = 0$$     M1

we can see from the equation of the curve that this solution is not possible $$(0 = 4)$$ and so there is not a horizontal tangent     R1

[4 marks]

a.

$$\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{y}{{2y – x}}$$ or equivalent with $$\frac{{{\text{d}}x}}{{{\text{d}}y}}$$

the tangent is vertical when $$2y = x$$     M1

substitute into the equation to give $$2{y^2} = {y^2} + 4$$     M1

$$y = \pm 2$$     A1

coordinates are $$(4,{\text{ }}2),{\text{ }}( – 4,{\text{ }} – 2)$$     A1

[4 marks]

Total [8 marks]

b.

### Question

A curve is defined by $$xy = {y^2} + 4$$.

a.Show that there is no point where the tangent to the curve is horizontal.[4]

b.Find the coordinates of the points where the tangent to the curve is vertical.[4]

### Markscheme

$$x\frac{{{\text{d}}y}}{{{\text{d}}x}} + y = 2y\frac{{{\text{d}}y}}{{{\text{d}}x}}$$     M1A1

a horizontal tangent occurs if $$\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0$$ so $$y = 0$$     M1

we can see from the equation of the curve that this solution is not possible $$(0 = 4)$$ and so there is not a horizontal tangent     R1

[4 marks]

a.

$$\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{y}{{2y – x}}$$ or equivalent with $$\frac{{{\text{d}}x}}{{{\text{d}}y}}$$

the tangent is vertical when $$2y = x$$     M1

substitute into the equation to give $$2{y^2} = {y^2} + 4$$     M1

$$y = \pm 2$$     A1

coordinates are $$(4,{\text{ }}2),{\text{ }}( – 4,{\text{ }} – 2)$$     A1

[4 marks]

Total [8 marks]

b.

### Question

Find the $$x$$-coordinates of all the points on the curve $$y = 2{x^4} + 6{x^3} + \frac{7}{2}{x^2} – 5x + \frac{3}{2}$$ at which

the tangent to the curve is parallel to the tangent at $$( – 1,{\text{ }}6)$$.

## Markscheme

$$\frac{{{\text{d}}y}}{{{\text{d}}x}} = 8{x^3} + 18{x^2} + 7x – 5$$    A1

when $$x = – 1,{\text{ }}\frac{{{\text{d}}y}}{{{\text{d}}x}} = – 2$$     A1

$$8{x^3} + 18{x^2} + 7x – 5 = – 2$$    M1

$$8{x^3} + 18{x^2} + 7x – 3 = 0$$

$$(x + 1)$$ is a factor     A1

$$8{x^3} + 18{x^2} + 7x – 3 = (x + 1)(8{x^2} + 10x – 3)$$    (M1)

Note:     M1 is for attempting to find the quadratic factor.

$$(x + 1)(4x – 1)(2x + 3) = 0$$

$$(x = – 1),{\text{ }}x = 0.25,{\text{ }}x = – 1.5$$    (M1)A1

Note:     M1 is for an attempt to solve their quadratic factor.

[7 marks]

### Question

The function $$f$$ is defined as $$f(x) = a{x^2} + bx + c$$ where $$a,{\text{ }}b,{\text{ }}c \in \mathbb{R}$$.

Hayley conjectures that $$\frac{{f({x_2}) – f({x_1})}}{{{x_2} – {x_1}}} = \frac{{f'({x_2}) + f'({x_1})}}{2},{\text{ }}x1 \ne x2$$.

Show that Hayley’s conjecture is correct.

## Markscheme

$$\frac{{f({x_2}) – f({x_1})}}{{{x_2} – {x_1}}} = \frac{{ax_2^2 + b{x_2} + c – (ax_1^2 + b{x_1} + c)}}{{{x_2} – {x_1}}}$$     (M1)

$$= \frac{{a(x_2^2 – x_1^2) + b({x_2} – {x_1})}}{{{x_2} – {x_1}}}$$     A1

$$= \frac{{a({x_2} – {x_1})({x_2} + {x_1}) + b({x_2} – {x_1})}}{{{x_2} – {x_1}}}$$     (A1)

$$= a({x_2} + {x_1}) + b\,\,\,\,\,({x_1} \ne {x_2})$$     A1

$$\frac{{f'({x_2}) + f'({x_1})}}{2} = \frac{{(2a{x_2} + b) + (2a{x_1} + b)}}{2}$$     M1

$$= \frac{{2a({x_2} + {x_1}) + 2b}}{2}$$

$$= a({x_2} + {x_1}) + b$$     A1

so Hayley’s conjecture is correct     AG

[6 marks]

### Question

A curve has equation $$3x – 2{y^2}{{\text{e}}^{x – 1}} = 2$$.

a.Find an expression for $$\frac{{{\text{d}}y}}{{{\text{d}}x}}$$ in terms of $$x$$ and $$y$$.[5]

b.Find the equations of the tangents to this curve at the points where the curve intersects the line $$x = 1$$.[4]

### Markscheme

attempt to differentiate implicitly     M1

$$3 – \left( {4y\frac{{{\text{d}}y}}{{{\text{d}}x}} + 2{y^2}} \right){{\text{e}}^{x – 1}} = 0$$    A1A1A1

Note: Award A1 for correctly differentiating each term.

$$\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{3 \bullet {{\text{e}}^{1 – x}} – 2{y^2}}}{{4y}}$$    A1

Note: This final answer may be expressed in a number of different ways.

[5 marks]

a.

$$3 – 2{y^2} = 2 \Rightarrow {y^2} = \frac{1}{2} \Rightarrow y = \pm \sqrt {\frac{1}{2}}$$    A1

$$\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{3 – 2 \bullet \frac{1}{2}}}{{ \pm 4\sqrt {\frac{1}{2}} }} = \pm \frac{{\sqrt 2 }}{2}$$    M1

at $$\left( {1,{\text{ }}\sqrt {\frac{1}{2}} } \right)$$ the tangent is $$y – \sqrt {\frac{1}{2}} = \frac{{\sqrt 2 }}{2}(x – 1)$$ and     A1

at $$\left( {1,{\text{ }} – \sqrt {\frac{1}{2}} } \right)$$ the tangent is $$y + \sqrt {\frac{1}{2}} = – \frac{{\sqrt 2 }}{2}(x – 1)$$     A1

Note: These equations simplify to $$y = \pm \frac{{\sqrt 2 }}{2}x$$.

Note: Award A0M1A1A0 if just the positive value of $$y$$ is considered and just one tangent is found.

[4 marks]

b.

### Question

Let $$y = {{\text{e}}^x}\sin x$$.

Consider the function $$f$$  defined by $$f(x) = {{\text{e}}^x}\sin x,{\text{ }}0 \leqslant x \leqslant \pi$$.

The curvature at any point $$(x,{\text{ }}y)$$ on a graph is defined as $$\kappa = \frac{{\left| {\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}}} \right|}}{{{{\left( {1 + {{\left( {\frac{{{\text{d}}y}}{{{\text{d}}x}}} \right)}^2}} \right)}^{\frac{3}{2}}}}}$$.

a.Find an expression for $$\frac{{{\text{d}}y}}{{{\text{d}}x}}$$.[2]

b.Show that $$\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} = 2{{\text{e}}^x}\cos x$$.[2]

c.Show that the function $$f$$ has a local maximum value when $$x = \frac{{3\pi }}{4}$$.[2]

d.Find the $$x$$-coordinate of the point of inflexion of the graph of $$f$$.[2]

e.Sketch the graph of $$f$$, clearly indicating the position of the local maximum point, the point of inflexion and the axes intercepts.[3]

f.Find the area of the region enclosed by the graph of $$f$$ and the $$x$$-axis.

The curvature at any point $$(x,{\text{ }}y)$$ on a graph is defined as $$\kappa = \frac{{\left| {\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}}} \right|}}{{{{\left( {1 + {{\left( {\frac{{{\text{d}}y}}{{{\text{d}}x}}} \right)}^2}} \right)}^{\frac{3}{2}}}}}$$.[6]

g.Find the value of the curvature of the graph of $$f$$ at the local maximum point.[3]
h.Find the value $$\kappa$$ for $$x = \frac{\pi }{2}$$ and comment on its meaning with respect to the shape of the graph.[2]

### Markscheme

$$\frac{{{\text{d}}y}}{{{\text{d}}x}} = {{\text{e}}^x}\sin x + {{\text{e}}^x}\cos x{\text{ }}\left( { = {{\text{e}}^x}(\sin x + \cos x)} \right)$$    M1A1

[2 marks]

a.

$$\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} = {{\text{e}}^x}(\sin x + \cos x) + {{\text{e}}^x}(\cos x – \sin x)$$    M1A1

$$= 2{{\text{e}}^x}\cos x$$    AG

[2 marks]

b.

$$\frac{{{\text{d}}y}}{{{\text{d}}x}} = {{\text{e}}^{\frac{{3\pi }}{4}}}\left( {\sin \frac{{3\pi }}{4} + \cos \frac{{3\pi }}{4}} \right) = 0$$    R1

$$\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} = 2{{\text{e}}^{\frac{{3\pi }}{4}}}\cos \frac{{3\pi }}{4} < 0$$    R1

hence maximum at $$x = \frac{{3\pi }}{4}$$     AG

[2 marks]

c.

$$\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} = 0 \Rightarrow 2{{\text{e}}^x}\cos x = 0$$    M1

$$\Rightarrow x = \frac{\pi }{2}$$    A1

Note: Award M1A0 if extra zeros are seen.

[2 marks]

d.

correct shape and correct domain     A1

max at $$x = \frac{{3\pi }}{4}$$, point of inflexion at $$x = \frac{\pi }{2}$$     A1

zeros at $$x = 0$$ and $$x = \pi$$     A1

Note: Penalize incorrect domain with first A mark; allow FT from (d) on extra points of inflexion.

[3 marks]

e.

EITHER

$$\int_0^x {{{\text{e}}^x}\sin x{\text{d}}x = [{{\text{e}}^x}\sin x]_0^\pi – \int_0^\pi {{{\text{e}}^x}\cos x{\text{d}}x} }$$    M1A1

$$\int_0^\pi {{{\text{e}}^x}\sin x{\text{d}}x = [{{\text{e}}^x}\sin x]_0^\pi – \left( {[{{\text{e}}^x}\cos x]_0^x + \int_0^\pi {{{\text{e}}^x}\sin x{\text{d}}x} } \right)}$$    A1

OR

$$\int_0^\pi {{{\text{e}}^x}\sin x{\text{d}}x = [ – {{\text{e}}^x}\cos x]_0^\pi + \int_0^\pi {{{\text{e}}^x}\cos x{\text{d}}x} }$$    M1A1

$$\int_0^\pi {{{\text{e}}^x}\sin x{\text{d}}x = [ – {{\text{e}}^x}\cos x]} _0^\pi + \left( {[{{\text{e}}^x}\sin x]_0^\pi – \int_0^\pi {{{\text{e}}^x}\sin x{\text{d}}x} } \right)$$    A1

THEN

$$\int_0^\pi {{{\text{e}}^x}\sin x{\text{d}}x = \frac{1}{2}\left( {[{{\text{e}}^x}\sin x]_0^x – [{{\text{e}}^x}\cos x]_0^x} \right)}$$    M1A1

$$\int_0^\pi {{{\text{e}}^x}\sin x{\text{d}}x = \frac{1}{2}({{\text{e}}^x} + 1)}$$    A1

[6 marks]

f.

$$\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0$$    (A1)

$$\frac{{{d^2}y}}{{d{x^2}}} = 2{e^{\frac{{3\pi }}{4}}}\cos \frac{{3\pi }}{4} = – \sqrt 2 {e^{\frac{{3\pi }}{4}}}$$ (A1)

$$\kappa = \frac{{\left| { – \sqrt 2 {{\text{e}}^{\frac{{3\pi }}{4}}}} \right|}}{1} = \sqrt 2 {{\text{e}}^{\frac{{3\pi }}{4}}}$$    A1

[3 marks]

g.

$$\kappa = 0$$    A1

the graph is approximated by a straight line     R1

[2 marks]

h.

### Question

Consider the function $$f$$ defined by $$f(x) = {x^2} – {a^2},{\text{ }}x \in \mathbb{R}$$ where $$a$$ is a positive constant.

The function $$g$$ is defined by $$g(x) = x\sqrt {f(x)}$$ for $$\left| x \right| > a$$.

a.i.Showing any $$x$$ and $$y$$ intercepts, any maximum or minimum points and any asymptotes, sketch the following curves on separate axes.

$$y = f(x)$$;[2]

a.ii.

Showing any $$x$$ and $$y$$ intercepts, any maximum or minimum points and any asymptotes, sketch the following curves on separate axes.

$$y = \frac{1}{{f(x)}}$$;[4]

a.iii.

Showing any $$x$$ and $$y$$ intercepts, any maximum or minimum points and any asymptotes, sketch the following curves on separate axes.

$$y = \left| {\frac{1}{{f(x)}}} \right|$$.[2]

b.

Find $$\int {f(x)\cos x{\text{d}}x}$$.[5]

c.

By finding $$g'(x)$$ explain why $$g$$ is an increasing function.[4]

### Markscheme

A1 for correct shape

A1 for correct $$x$$ and $$y$$ intercepts and minimum point

[2 marks]

a.i.

A1 for correct shape

A1 for correct vertical asymptotes

A1 for correct implied horizontal asymptote

A1 for correct maximum point

[??? marks]

a.ii.

A1 for reflecting negative branch from (ii) in the $$x$$-axis

A1 for correctly labelled minimum point

[2 marks]

a.iii.

EITHER

attempt at integration by parts     (M1)

$$\int {({x^2} – {a^2})\cos x{\text{d}}x = ({x^2} – {a^2})\sin x – \int {2x\sin x{\text{d}}x} }$$     A1A1

$$= ({x^2} – {a^2})\sin x – 2\left[ { – x\cos x + \int {\cos x{\text{d}}x} } \right]$$     A1

$$= ({x^2} – {a^2})\sin x + 2x\cos – 2\sin x + c$$     A1

OR

$$\int {({x^2} – {a^2})\cos x{\text{d}}x = \int {{x^2}\cos x{\text{d}}x – \int {{a^2}\cos x{\text{d}}x} } }$$

attempt at integration by parts     (M1)

$$\int {{x^2}\cos x{\text{d}}x = {x^2}\sin x – \int {2x\sin x{\text{d}}x} }$$     A1A1

$$= {x^2}\sin x – 2\left[ { – x\cos x + \int {\cos x{\text{d}}x} } \right]$$     A1

$$= {x^2}\sin x + 2x\cos x – 2\sin x$$

$$– \int {{a^2}\cos x{\text{d}}x = – {a^2}\sin x}$$

$$\int {({x^2} – {a^2})\cos x{\text{d}}x = ({x^2} – {a^2})\sin x + 2x\cos x – 2\sin x + c}$$     A1

[5 marks]

b.

$$g(x) = x{({x^2} – {a^2})^{\frac{1}{2}}}$$

$$g'(x) = {({x^2} – {a^2})^{\frac{1}{2}}} + \frac{1}{2}x{({x^2} – {a^2})^{ – \frac{1}{2}}}(2x)$$     M1A1A1

Note:     Method mark is for differentiating the product. Award A1 for each correct term.

$$g'(x) = {({x^2} – {a^2})^{\frac{1}{2}}} + {x^2}{({x^2} – {a^2})^{ – \frac{1}{2}}}$$

both parts of the expression are positive hence $$g'(x)$$ is positive     R1

and therefore $$g$$ is an increasing function (for $$\left| x \right| > a$$)     AG

[4 marks]

c.

### Question

Consider the function $${f_n}(x) = (\cos 2x)(\cos 4x) \ldots (\cos {2^n}x),{\text{ }}n \in {\mathbb{Z}^ + }$$.

a.Determine whether $${f_n}$$ is an odd or even function, justifying your answer.[2]

b.By using mathematical induction, prove that

$${f_n}(x) = \frac{{\sin {2^{n + 1}}x}}{{{2^n}\sin 2x}},{\text{ }}x \ne \frac{{m\pi }}{2}$$ where $$m \in \mathbb{Z}$$.[8]

c.Hence or otherwise, find an expression for the derivative of $${f_n}(x)$$ with respect to $$x$$.[3]

d.Show that, for $$n > 1$$, the equation of the tangent to the curve $$y = {f_n}(x)$$ at $$x = \frac{\pi }{4}$$ is $$4x – 2y – \pi = 0$$.[8]

### Markscheme

even function     A1

since $$\cos kx = \cos ( – kx)$$ and $${f_n}(x)$$ is a product of even functions     R1

OR

even function     A1

since $$(\cos 2x)(\cos 4x) \ldots = \left( {\cos ( – 2x)} \right)\left( {\cos ( – 4x)} \right) \ldots$$     R1

Note:     Do not award A0R1.

[2 marks]

a.

consider the case $$n = 1$$

$$\frac{{\sin 4x}}{{2\sin 2x}} = \frac{{2\sin 2x\cos 2x}}{{2\sin 2x}} = \cos 2x$$     M1

hence true for $$n = 1$$     R1

assume true for $$n = k$$, ie, $$(\cos 2x)(\cos 4x) \ldots (\cos {2^k}x) = \frac{{\sin {2^{k + 1}}x}}{{{2^k}\sin 2x}}$$     M1

Note:     Do not award M1 for “let $$n = k$$” or “assume $$n = k$$” or equivalent.

consider $$n = k + 1$$:

$${f_{k + 1}}(x) = {f_k}(x)(\cos {2^{k + 1}}x)$$     (M1)

$$= \frac{{\sin {2^{k + 1}}x}}{{{2^k}\sin 2x}}\cos {2^{k + 1}}x$$     A1

$$= \frac{{2\sin {2^{k + 1}}x\cos {2^{k + 1}}x}}{{{2^{k + 1}}\sin 2x}}$$     A1

$$= \frac{{\sin {2^{k + 2}}x}}{{{2^{k + 1}}\sin 2x}}$$     A1

so $$n = 1$$ true and $$n = k$$ true $$\Rightarrow n = k + 1$$ true. Hence true for all $$n \in {\mathbb{Z}^ + }$$     R1

Note:     To obtain the final R1, all the previous M marks must have been awarded.

[8 marks]

b.

attempt to use $$f’ = \frac{{vu’ – uv’}}{{{v^2}}}$$ (or correct product rule)     M1

$${f’_n}(x) = \frac{{({2^n}\sin 2x)({2^{n + 1}}\cos {2^{n + 1}}x) – (\sin {2^{n + 1}}x)({2^{n + 1}}\cos 2x)}}{{{{({2^n}\sin 2x)}^2}}}$$     A1A1

Note:     Award A1 for correct numerator and A1 for correct denominator.

[3 marks]

c.

$${f’_n}\left( {\frac{\pi }{4}} \right) = \frac{{\left( {{2^n}\sin \frac{\pi }{2}} \right)\left( {{2^{n + 1}}\cos {2^{n + 1}}\frac{\pi }{4}} \right) – \left( {\sin {2^{n + 1}}\frac{\pi }{4}} \right)\left( {{2^{n + 1}}\cos \frac{\pi }{2}} \right)}}{{{{\left( {{2^n}\sin \frac{\pi }{2}} \right)}^2}}}$$     (M1)(A1)

$${f’_n}\left( {\frac{\pi }{4}} \right) = \frac{{({2^n})\left( {{2^{n + 1}}\cos {2^{n + 1}}\frac{\pi }{4}} \right)}}{{{{({2^n})}^2}}}$$     (A1)

$$= 2\cos {2^{n + 1}}\frac{\pi }{4}{\text{ }}( = 2\cos {2^{n – 1}}\pi )$$     A1

$${f’_n}\left( {\frac{\pi }{4}} \right) = 2$$     A1

$${f_n}\left( {\frac{\pi }{4}} \right) = 0$$     A1

Note:     This A mark is independent from the previous marks.

$$y = 2\left( {x – \frac{\pi }{4}} \right)$$     M1A1

$$4x – 2y – \pi = 0$$     AG

[8 marks]

d.
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