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]

Answer/Explanation

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.
Answer/Explanation

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 \(x = \pi \) .

Answer/Explanation

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.
Answer/Explanation

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.
Answer/Explanation

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

(ii) correct answer     A1     N1

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.
Answer/Explanation

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.
Answer/Explanation

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.
Answer/Explanation

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.
Answer/Explanation

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.
Answer/Explanation

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.
Answer/Explanation

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.
Answer/Explanation

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.

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