# IB Math AA: Question bank-Topic SL 5.1 Derivative interpreted as gradient function and as rate of change SL Paper 1

### Question

Consider the graph of the function f (x) = x2 − $$\frac{k}{x}$$ .

1. Write down f ′(x) . [3]

The equation of the tangent to the graph of y = f (x) at x = −2 is 2 y = 4 5x .

2. Write down the gradient of this tangent. [1]

3. Find the value of k . [2]

Ans:

(a) $$2x+ \frac{k}{x^{2}}$$

(b)-2.5 ($$\frac{-5}{2}$$)

(c)

$$-2.5=2\times (-2)+\frac{k}{-2^{2}}$$

k=6

## Question

Let $$f(x) = 1 + {{\text{e}}^{ – x}}$$ and $$g(x) = 2x + b$$, for $$x \in \mathbb{R}$$, where $$b$$ is a constant.

Find $$(g \circ f)(x)$$.

[2]
a.

Given that $$\mathop {\lim }\limits_{x \to + \infty } (g \circ f)(x) = – 3$$, find the value of $$b$$.

[4]
b.

## Markscheme

attempt to form composite     (M1)

eg$$\,\,\,\,\,$$$$g(1 + {{\text{e}}^{ – x}})$$

correct function     A1     N2

eg$$\,\,\,\,\,$$$$(g \circ f)(x) = 2 + b + 2{{\text{e}}^{ – x}},{\text{ }}2(1 + {{\text{e}}^{ – x}}) + b$$

[2 marks]

a.

evidence of $$\mathop {\lim }\limits_{x \to \infty } (2 + b + 2{{\text{e}}^{ – x}}) = 2 + b + \mathop {\lim }\limits_{x \to \infty } (2{{\text{e}}^{ – x}})$$     (M1)

eg$$\,\,\,\,\,$$$$2 + b + 2{{\text{e}}^{ – \infty }}$$, graph with horizontal asymptote when $$x \to \infty$$

Note:     Award M0 if candidate clearly has incorrect limit, such as $$x \to 0,{\text{ }}{{\text{e}}^\infty },{\text{ }}2{{\text{e}}^0}$$.

evidence that $${{\text{e}}^{ – x}} \to 0$$ (seen anywhere)     (A1)

eg$$\,\,\,\,\,$$$$\mathop {\lim }\limits_{x \to \infty } ({{\text{e}}^{ – x}}) = 0,{\text{ }}1 + {{\text{e}}^{ – x}} \to 1,{\text{ }}2(1) + b = – 3,{\text{ }}{{\text{e}}^{{\text{large negative number}}}} \to 0$$, graph of $$y = {{\text{e}}^{ – x}}$$ or

$$y = 2{{\text{e}}^{ – x}}$$ with asymptote $$y = 0$$, graph of composite function with asymptote $$y = – 3$$

correct working     (A1)

eg$$\,\,\,\,\,$$$$2 + b = – 3$$

$$b = – 5$$     A1     N2

[4 marks]

b.

## Question

Let $$f(x) = {{\rm{e}}^x}\cos x$$ . Find the gradient of the normal to the curve of f at $$x = \pi$$ .

## Markscheme

evidence of choosing the product rule     (M1)

$$f'(x) = {{\rm{e}}^x} \times ( – \sin x) + \cos x \times {{\rm{e}}^x}$$ $$( = {{\rm{e}}^x}\cos x – {{\rm{e}}^x}\sin x)$$     A1A1

substituting $$\pi$$     (M1)

e.g.  $$f'(\pi ) = {{\rm{e}}^\pi }\cos \pi – {{\rm{e}}^\pi }\sin \pi$$ , $${{\rm{e}}^\pi }( – 1 – 0)$$ , $$– {{\rm{e}}^\pi }$$

taking negative reciprocal      (M1)

e.g. $$– \frac{1}{{f'(\pi )}}$$

gradient is $$\frac{1}{{{{\rm{e}}^\pi }}}$$     A1     N3

[6 marks]

## Question

Let $$g(x) = 2x\sin x$$ .

Find $$g'(x)$$ .

[4]
a.

Find the gradient of the graph of g at $$x = \pi$$ .

[3]
b.

## Markscheme

evidence of choosing the product rule     (M1)

e.g. $$uv’ + vu’$$

correct derivatives $$\cos x$$ , 2     (A1)(A1)

$$g'(x) = 2x\cos x + 2\sin x$$     A1     N4

[4 marks]

a.

attempt to substitute into gradient function     (M1)

e.g. $$g'(\pi )$$

correct substitution     (A1)

e.g. $$2\pi \cos \pi + 2\sin \pi$$

$${\text{gradient}} = – 2\pi$$     A1     N2

[3 marks]

b.

## Question

Let $$f(x) = {x^3}$$. The following diagram shows part of the graph of f .

The point $${\rm{P}}(a,f(a))$$ , where $$a > 0$$ , lies on the graph of f . The tangent at P crosses the x-axis at the point $${\rm{Q}}\left( {\frac{2}{3},0} \right)$$ . This tangent intersects the graph of f at the point R(−2, −8) .

The equation of the tangent at P is $$y = 3x – 2$$ . Let T be the region enclosed by the graph of f , the tangent [PR] and the line $$x = k$$ , between $$x = – 2$$ and $$x = k$$ where $$– 2 < k < 1$$ . This is shown in the diagram below.

(i)     Show that the gradient of [PQ] is $$\frac{{{a^3}}}{{a – \frac{2}{3}}}$$ .

(ii)    Find $$f'(a)$$ .

(iii)   Hence show that $$a = 1$$ .

[7]
a(i), (ii) and (iii).

Given that the area of T is $$2k + 4$$ , show that k satisfies the equation $${k^4} – 6{k^2} + 8 = 0$$ .

[9]
b.

## Markscheme

(i) substitute into gradient $$= \frac{{{y_1} – {y_2}}}{{{x_1} – {x_2}}}$$     (M1)

e.g. $$\frac{{f(a) – 0}}{{a – \frac{2}{3}}}$$

substituting $$f(a) = {a^3}$$

e.g. $$\frac{{{a^3} – 0}}{{a – \frac{2}{3}}}$$     A1

gradient $$\frac{{{a^3}}}{{a – \frac{2}{3}}}$$     AG     N0

e.g. $$3{a^2}$$ , $$f'(a) = 3$$ , $$f'(a) = \frac{{{a^3}}}{{a – \frac{2}{3}}}$$

(iii) METHOD 1

evidence of approach     (M1)

e.g. $$f'(a) = {\rm{gradient}}$$ , $$3{a^2} = \frac{{{a^3}}}{{a – \frac{2}{3}}}$$

simplify     A1

e.g. $$3{a^2}\left( {a – \frac{2}{3}} \right) = {a^3}$$

rearrange     A1

e.g. $$3{a^3} – 2{a^2} = {a^3}$$

evidence of solving     A1

e.g. $$2{a^3} – 2{a^2} = 2{a^2}(a – 1) = 0$$

$$a = 1$$     AG     N0

METHOD 2

gradient RQ $$= \frac{{ – 8}}{{ – 2 – \frac{2}{3}}}$$     A1

simplify     A1

e.g. $$\frac{{ – 8}}{{ – \frac{8}{3}}},3$$

evidence of approach     (M1)

e.g. $$f'(a) = {\rm{gradient}}$$ , $$3{a^2} = \frac{{ – 8}}{{ – 2 – \frac{2}{3}}}$$ , $$\frac{{{a^3}}}{{a – \frac{2}{3}}} = 3$$

simplify     A1

e.g. $$3{a^2} = 3$$ , $${a^2} = 1$$

$$a = 1$$     AG     N0

[7 marks]

a(i), (ii) and (iii).

approach to find area of T involving subtraction and integrals    (M1)

e.g. $$\int {f – (3x – 2){\rm{d}}x}$$ , $$\int_{ – 2}^k {(3x – 2) – \int_{ – 2}^k {{x^3}} }$$ , $$\int {({x^3} – 3x + 2)}$$

correct integration with correct signs     A1A1A1

e.g. $$\frac{1}{4}{x^4} – \frac{3}{2}{x^2} + 2x$$ , $$\frac{3}{2}{x^2} – 2x – \frac{1}{4}{x^4}$$

correct limits $$– 2$$ and k (seen anywhere)     A1

e.g. $$\int_{ – 2}^k {({x^3} – 3x + 2){\rm{d}}x}$$ , $$\left[ {\frac{1}{4}{x^4} – \frac{3}{2}{x^2} + 2x} \right]_{ – 2}^k$$

attempt to substitute k and $$– 2$$     (M1)

correct substitution into their integral if 2 or more terms     A1

e.g. $$\left( {\frac{1}{4}{k^4} – \frac{3}{2}{k^2} + 2k} \right) – (4 – 6 – 4)$$

setting their integral expression equal to $$2k + 4$$ (seen anywhere)     (M1)

simplifying     A1

e.g. $$\frac{1}{4}{k^4} – \frac{3}{2}{k^2} + 2 = 0$$

$${k^4} – 6{k^2} + 8 = 0$$     AG     N0

[9 marks]

b.

## Question

The following diagram shows the graph of $$f(x) = a\sin (b(x – c)) + d$$ , for $$2 \le x \le 10$$ .

There is a maximum point at P(4, 12) and a minimum point at Q(8, −4) .

Use the graph to write down the value of

(i)     a ;

(ii)    c ;

(iii)   d .

[3]
a(i), (ii) and (iii).

Show that $$b = \frac{\pi }{4}$$ .

[2]
b.

Find $$f'(x)$$ .

[3]
c.

At a point R, the gradient is $$– 2\pi$$ . Find the x-coordinate of R.

[6]
d.

## Markscheme

(i) $$a = 8$$     A1     N1

(ii) $$c = 2$$     A1     N1

(iii) $$d = 4$$     A1     N1

[3 marks]

a(i), (ii) and (iii).

METHOD 1

recognizing that period $$= 8$$     (A1)

correct working     A1

e.g. $$8 = \frac{{2\pi }}{b}$$ , $$b = \frac{{2\pi }}{8}$$

$$b = \frac{\pi }{4}$$     AG     N0

METHOD 2

attempt to substitute     M1

e.g. $$12 = 8\sin (b(4 – 2)) + 4$$

correct working     A1

e.g. $$\sin 2b = 1$$

$$b = \frac{\pi }{4}$$     AG     N0

[2 marks]

b.

evidence of attempt to differentiate or choosing chain rule     (M1)

e.g. $$\cos \frac{\pi }{4}(x – 2)$$ , $$\frac{\pi }{4} \times 8$$

$$f'(x) = 2\pi \cos \left( {\frac{\pi }{4}(x – 2)} \right)$$ (accept $$2\pi \cos \frac{\pi }{4}(x – 2)$$ )     A2     N3

[3 marks]

c.

recognizing that gradient is $$f'(x)$$     (M1)

e.g. $$f'(x) = m$$

correct equation     A1

e.g. $$– 2\pi = 2\pi \cos \left( {\frac{\pi }{4}(x – 2)} \right)$$ , $$– 1 = \cos \left( {\frac{\pi }{4}(x – 2)} \right)$$

correct working     (A1)

e.g. $${\cos ^{ – 1}}( – 1) = \frac{\pi }{4}(x – 2)$$

using $${\cos ^{ – 1}}( – 1) = \pi$$ (seen anywhere)     (A1)

e.g. $$\pi = \frac{\pi }{4}(x – 2)$$

simplifying     (A1)

e.g. $$4 = (x – 2)$$

$$x = 6$$     A1     N4

[6 marks]

d.

## Question

Let $$f(x) = {{\rm{e}}^{6x}}$$ .

Write down $$f'(x)$$ .

[1]
a.

The tangent to the graph of f at the point $${\text{P}}(0{\text{, }}b)$$ has gradient m .

(i)     Show that $$m = 6$$ .

(ii)    Find b .

[4]
b(i) and (ii).

Hence, write down the equation of this tangent.

[1]
c.

## Markscheme

$$f'(x) = 6{{\rm{e}}^{6x}}$$     A1     N1

[1 mark]

a.

(i) evidence of valid approach     (M1)

e.g. $$f'(0)$$ ,  $$6{{\rm{e}}^{6 \times 0}}$$

correct manipulation     A1

e.g. $$6{{\rm{e}}^0}$$ , $$6 \times 1$$

$$m = 6$$    AG     N0

(ii) evidence of finding $$f(0)$$     (M1)

e.g. $$y = {{\rm{e}}^{6(0)}}$$

$$b = 1$$     A1     N2

[4 marks]

b(i) and (ii).

$$y = 6x + 1$$     A1     N1

[1 mark]

c.

## Question

Part of the graph of $$f(x) = a{x^3} – 6{x^2}$$ is shown below.

The point P lies on the graph of $$f$$ . At P,  x = 1.

Find $$f'(x)$$ .

[2]
a.

The graph of $$f$$ has a gradient of $$3$$ at the point P. Find the value of $$a$$ .

[4]
b.

## Markscheme

$$f'(x) = 3a{x^2} – 12x$$     A1A1     N2

Note: Award A1 for each correct term.

[2 marks]

a.

setting their derivative equal to 3 (seen anywhere)     A1

e.g. $$f'(x) = 3$$

attempt to substitute $$x = 1$$ into $$f'(x)$$     (M1)

e.g. $$3a{(1)^2} – 12(1)$$

correct substitution into $$f'(x)$$     (A1)

e.g. $$3a – 12$$ , $$3a = 15$$

$$a = 5$$    A1     N2

[4 marks]

b.

## Question

Consider $$f(x) = {x^2}\sin x$$ .

Find $$f'(x)$$ .

[4]
a.

Find the gradient of the curve of $$f$$ at $$x = \frac{\pi }{2}$$ .

[3]
b.

## Markscheme

evidence of choosing product rule     (M1)

eg   $$uv’ + vu’$$

correct derivatives (must be seen in the product rule) $$\cos x$$ , $$2x$$     (A1)(A1)

$$f'(x) = {x^2}\cos x + 2x\sin x$$     A1 N4

[4 marks]

a.

substituting $$\frac{\pi }{2}$$ into their $$f'(x)$$     (M1)

eg   $$f’\left( {\frac{\pi }{2}} \right)$$ , $${\left( {\frac{\pi }{2}} \right)^2}\cos \left( {\frac{\pi }{2}} \right) + 2\left( {\frac{\pi }{2}} \right)\sin \left( {\frac{\pi }{2}} \right)$$

correct values for both $$\sin \frac{\pi }{2}$$ and $$\cos \frac{\pi }{2}$$ seen in $$f'(x)$$     (A1)

eg   $$0 + 2\left( {\frac{\pi }{2}} \right) \times 1$$

$$f’\left( {\frac{\pi }{2}} \right) = \pi$$     A1 N2

[3 marks]

b.

## Question

Consider $$f(x) = \ln ({x^4} + 1)$$ .

The second derivative is given by $$f”(x) = \frac{{4{x^2}(3 – {x^4})}}{{{{({x^4} + 1)}^2}}}$$ .

The equation $$f”(x) = 0$$ has only three solutions, when $$x = 0$$ , $$\pm \sqrt[4]{3}$$ $$( \pm 1.316 \ldots )$$ .

Find the value of $$f(0)$$ .

[2]
a.

Find the set of values of $$x$$ for which $$f$$ is increasing.

[5]
b.

(i)     Find $$f”(1)$$ .

(ii)     Hence, show that there is no point of inflexion on the graph of $$f$$ at $$x = 0$$ .

[5]
c.

There is a point of inflexion on the graph of $$f$$ at $$x = \sqrt[4]{3}$$ $$(x = 1.316 \ldots )$$ .

Sketch the graph of $$f$$ , for $$x \ge 0$$ .

[3]
d.

## Markscheme

substitute $$0$$ into $$f$$     (M1)

eg   $$\ln (0 + 1)$$ , $$\ln 1$$

$$f(0) = 0$$     A1 N2

[2 marks]

a.

$$f'(x) = \frac{1}{{{x^4} + 1}} \times 4{x^3}$$ (seen anywhere)     A1A1

Note: Award A1 for $$\frac{1}{{{x^4} + 1}}$$ and A1 for $$4{x^3}$$ .

recognizing $$f$$ increasing where $$f'(x) > 0$$ (seen anywhere)     R1

eg   $$f'(x) > 0$$ , diagram of signs

attempt to solve $$f'(x) > 0$$     (M1)

eg   $$4{x^3} = 0$$ , $${x^3} > 0$$

$$f$$ increasing for $$x > 0$$ (accept $$x \ge 0$$ )     A1     N1

[5 marks]

b.

(i)     substituting $$x = 1$$ into $$f”$$     (A1)

eg   $$\frac{{4(3 – 1)}}{{{{(1 + 1)}^2}}}$$ , $$\frac{{4 \times 2}}{4}$$

$$f”(1) = 2$$     A1     N2

(ii)     valid interpretation of point of inflexion (seen anywhere)     R1

eg   no change of sign in $$f”(x)$$ , no change in concavity,

$$f’$$ increasing both sides of zero

attempt to find $$f”(x)$$ for $$x < 0$$     (M1)

eg   $$f”( – 1)$$ , $$\frac{{4{{( – 1)}^2}(3 – {{( – 1)}^4})}}{{{{({{( – 1)}^4} + 1)}^2}}}$$ , diagram of signs

correct working leading to positive value     A1

eg   $$f”( – 1) = 2$$ , discussing signs of numerator and denominator

there is no point of inflexion at $$x = 0$$     AG     N0

[5 marks]

c.

A1A1A1     N3

Notes: Award A1 for shape concave up left of POI and concave down right of POI.

Only if this A1 is awarded, then award the following:

A1 for curve through ($$0$$, $$0$$) , A1 for increasing throughout.

Sketch need not be drawn to scale. Only essential features need to be clear.

[3 marks]

d.

## Question

Let $$y = f(x)$$, for $$– 0.5 \le$$ x $$\le$$ $$6.5$$. The following diagram shows the graph of $$f’$$, the derivative of $$f$$.

The graph of $$f’$$ has a local maximum when $$x = 2$$, a local minimum when $$x = 4$$, and it crosses the $$x$$-axis at the point $$(5,{\text{ }}0)$$.

Explain why the graph of $$f$$ has a local minimum when $$x = 5$$.

[2]
a.

Find the set of values of $$x$$ for which the graph of $$f$$ is concave down.

[2]
b.

The following diagram shows the shaded regions $$A$$, $$B$$ and $$C$$.

The regions are enclosed by the graph of $$f’$$, the $$x$$-axis, the $$y$$-axis, and the line $$x = 6$$.

The area of region $$A$$ is 12, the area of region $$B$$ is 6.75 and the area of region $$C$$ is 6.75.

Given that $$f(0) = 14$$, find $$f(6)$$.

[5]
c.

The following diagram shows the shaded regions $$A$$, $$B$$ and $$C$$.

The regions are enclosed by the graph of $$f’$$, the x-axis, the y-axis, and the line $$x = 6$$.

The area of region $$A$$ is 12, the area of region $$B$$ is 6.75 and the area of region $$C$$ is 6.75.

Let $$g(x) = {\left( {f(x)} \right)^2}$$. Given that $$f'(6) = 16$$, find the equation of the tangent to the graph of $$g$$ at the point where $$x = 6$$.

[6]
d.

## Markscheme

METHOD 1

$$f'(5) = 0$$     (A1)

valid reasoning including reference to the graph of $$f’$$     R1

eg$$\;\;\;f’$$ changes sign from negative to positive at $$x = 5$$, labelled sign chart for $$f’$$

so $$f$$ has a local minimum at $$x = 5$$     AG     N0

Note:     It must be clear that any description is referring to the graph of $$f’$$, simply giving the conditions for a minimum without relating them to $$f’$$ does not gain the R1.

METHOD 2

$$f'(5) = 0$$     A1

valid reasoning referring to second derivative     R1

eg$$\;\;\;f”(5) > 0$$

so $$f$$ has a local minimum at $$x = 5$$     AG     N0

[2 marks]

a.

attempt to find relevant interval     (M1)

eg$$\;\;\;f’$$ is decreasing, gradient of $$f’$$ is negative, $$f” < 0$$

$$2 < x < 4\;\;\;$$(accept “between 2 and 4”)     A1     N2

Notes:     If no other working shown, award M1A0 for incorrect inequalities such as $$2 \le$$ $$x$$ $$\le$$ 4, or “from 2 to 4”

[2 marks]

b.

METHOD 1 (one integral)

correct application of Fundamental Theorem of Calculus     (A1)

eg$$\;\;\;\int_0^6 {f'(x){\text{d}}x = } f(6) – f(0),{\text{ }}f(6) = 14 + \int_0^6 {f'(x){\text{d}}x}$$

attempt to link definite integral with areas     (M1)

eg$$\;\;\;\int_0^6 {f'(x){\text{d}}x = – 12 – 6.75 + 6.75,{\text{ }}\int_0^6 {f'(x){\text{d}}x = {\text{Area }}A + {\text{Area }}B + {\text{ Area }}C} }$$

correct value for $$\int_0^6 {f'(x){\text{d}}x}$$     (A1)

eg$$\;\;\;\int_0^6 {f'(x){\text{d}}x} = – 12$$

correct working     A1

eg$$\;\;\;f(6) – 14 = – 12,{\text{ }}f(6) = – 12 + f(0)$$

$$f(6) = 2$$     A1     N3

METHOD 2 (more than one integral)

correct application of Fundamental Theorem of Calculus     (A1)

eg$$\;\;\;\int_0^2 {f'(x){\text{d}}x} = f(2) – f(0),{\text{ }}f(2) = 14 + \int_0^2 {f'(x)}$$

attempt to link definite integrals with areas     (M1)

eg$$\;\;\;\int_0^2 {f'(x){\text{d}}x} = 12,{\text{ }}\int_2^5 {f'(x){\text{d}}x = – 6.75} ,{\text{ }}\int_0^6 {f'(x)} = 0$$

correct values for integrals     (A1)

eg$$\;\;\;\int_0^2 {f'(x){\text{d}}x} = – 12,{\text{ }}\int_5^2 {f'(x)} {\text{d}}x = 6.75,{\text{ }}f(6) – f(2) = 0$$

one correct intermediate value     A1

eg$$\;\;\;f(2) = 2,{\text{ }}f(5) = – 4.75$$

$$f(6) = 2$$     A1     N3

[5 marks]

c.

correct calculation of $$g(6)$$ (seen anywhere)     A1

eg$$\;\;\;{2^2},{\text{ }}g(6) = 4$$

choosing chain rule or product rule     (M1)

eg$$\;\;\;g’\left( {f(x)} \right)f'(x),{\text{ }}\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{{\text{d}}y}}{{{\text{d}}u}} \times \frac{{{\text{d}}u}}{{{\text{d}}x}},{\text{ }}f(x)f'(x) + f'(x)f(x)$$

correct derivative     (A1)

eg$$\;\;\;g'(x) = 2f(x)f'(x),{\text{ }}f(x)f'(x) + f'(x)f(x)$$

correct calculation of $$g'(6)$$ (seen anywhere)     A1

eg$$\;\;\;2(2)(16),{\text{ }}g'(6) = 64$$

attempt to substitute their values of $$g'(6)$$ and $$g(6)$$ (in any order) into equation of a line     (M1)

eg$$\;\;\;{2^2} = (2 \times 2 \times 16)6 + b,{\text{ }}y – 6 = 64(x – 4)$$

correct equation in any form     A1     N2

eg$$\;\;\;y – 4 = 64(x – 6),{\text{ }}y = 64x – 380$$

[6 marks]

[Total 15 marks]

d.

## Question

Let $$f(x) = 1 + {{\text{e}}^{ – x}}$$ and $$g(x) = 2x + b$$, for $$x \in \mathbb{R}$$, where $$b$$ is a constant.

Find $$(g \circ f)(x)$$.

[2]
a.

Given that $$\mathop {\lim }\limits_{x \to + \infty } (g \circ f)(x) = – 3$$, find the value of $$b$$.

[4]
b.

## Markscheme

attempt to form composite     (M1)

eg$$\,\,\,\,\,$$$$g(1 + {{\text{e}}^{ – x}})$$

correct function     A1     N2

eg$$\,\,\,\,\,$$$$(g \circ f)(x) = 2 + b + 2{{\text{e}}^{ – x}},{\text{ }}2(1 + {{\text{e}}^{ – x}}) + b$$

[2 marks]

a.

evidence of $$\mathop {\lim }\limits_{x \to \infty } (2 + b + 2{{\text{e}}^{ – x}}) = 2 + b + \mathop {\lim }\limits_{x \to \infty } (2{{\text{e}}^{ – x}})$$     (M1)

eg$$\,\,\,\,\,$$$$2 + b + 2{{\text{e}}^{ – \infty }}$$, graph with horizontal asymptote when $$x \to \infty$$

Note:     Award M0 if candidate clearly has incorrect limit, such as $$x \to 0,{\text{ }}{{\text{e}}^\infty },{\text{ }}2{{\text{e}}^0}$$.

evidence that $${{\text{e}}^{ – x}} \to 0$$ (seen anywhere)     (A1)

eg$$\,\,\,\,\,$$$$\mathop {\lim }\limits_{x \to \infty } ({{\text{e}}^{ – x}}) = 0,{\text{ }}1 + {{\text{e}}^{ – x}} \to 1,{\text{ }}2(1) + b = – 3,{\text{ }}{{\text{e}}^{{\text{large negative number}}}} \to 0$$, graph of $$y = {{\text{e}}^{ – x}}$$ or

$$y = 2{{\text{e}}^{ – x}}$$ with asymptote $$y = 0$$, graph of composite function with asymptote $$y = – 3$$

correct working     (A1)

eg$$\,\,\,\,\,$$$$2 + b = – 3$$

$$b = – 5$$     A1     N2

[4 marks]

b.