IB Maths AA SL 5.1 Calculus Sl Content Questions

Practise IB Mathematics AA SL 5.1 by interpreting limits, derivatives, integrals, rates of change and optimisation in mathematical and contextual problems.

Exam points

SL syllabus concepts and methods

Covered syllabus points

SL 5.1—Limits and derivative concept SL 5.2—Increasing and decreasing functions SL 5.3—Derivative of powers SL 5.4—Tangents and normals SL 5.5—Integration as anti-differentiation SL 5.6—Differentiation rules SL 5.7—Second derivative and graph behaviour SL 5.8—Extrema, optimization and inflexion SL 5.9—Kinematics SL 5.10—Indefinite integration SL 5.11—Definite integrals and areas

[Maximum number: 2]

Let $f(x)=\frac{2 x-6}{1-x}$ , for $x \neq 1$ .

Find $\lim _{x \rightarrow \infty} f(x)$ .

valid approach
eg recognizing that $\lim _{x \rightarrow \infty}$ is related to the horizontal asymptote, table with large values of x, their y value from (a)(iii), L'Hopital's rule $\lim _{x \rightarrow \infty} f(x)=-2$

[Maximum number: 15]

The derivative of a function f is given by $f^{\prime}(x)=\frac{2 x+2}{x^{2}+2 x+2}$ , for $x \in \mathbb{R}$ .

(a)

Hence, find the values of x for which f is increasing.

[ 1 ]

$\quad x>-1($ accept $x \geq-1)$

(b)

Write down the value of x for which $f^{\prime}(x)=0$ .

[ 1 ]

$\quad x=-1$

(c)

Show that $f^{\prime \prime}(x)=\frac{-2 x^{2}-4 x}{\left(x^{2}+2 x+2\right)^{2}}$ .

[ 4 ]

attempt to use the quotient rule

f^{\prime \prime}(x)=\frac{2\left(x^{2}+2 x+2\right)-(2 x+2)(2 x+2)}{\left(x^{2}+2 x+2\right)^{2}}

Note: Award A1 for first term in numerator, A1 for second term in numerator (including the minus sign). If the numerator is correct, but the denominator is incorrect or missing, award M1A1A0.

\begin{aligned} & f^{\prime \prime}(x)=\frac{2 x^{2}+4 x+4-4 x^{2}-8 x-4}{\left(x^{2}+2 x+2\right)^{2}} \text { OR } f^{\prime \prime}(x)=\frac{2 x^{2}+4 x+4-\left(4 x^{2}+8 x+4\right)}{\left(x^{2}+2 x+2\right)^{2}} \\ & f^{\prime \prime}(x)=\frac{-2 x^{2}-4 x}{\left(x^{2}+2 x+2\right)^{2}} \end{aligned}

(d)

Hence, justify that the value of x found in part (b)(i) corresponds to a local minimum point on the graph of f.

It is given that $f(2)=3+\ln 10$ .

[ 2 ]

substituting x=-1 into $f^{\prime \prime}(x)$ , clearly leading to positive numerator A1

\begin{aligned} & f^{\prime \prime}(-1)=\left(\frac{-2+4}{1}\right)(=2) \\ & f^{\prime \prime}(-1)>0 \end{aligned}

therefore this is a local minimum point AG

Note: Do not award AOR1.

(e)

It is given that $f(2)=3+\ln10$ .

Find an expression for f(x).

[ 4 ]

recognition to integrate $f^{\prime}(x)$

\begin{aligned} & f(x)=\int \frac{2 x+2}{x^{2}+2 x+2} \mathrm{~d} x \text { OR } \int \frac{1}{u} \mathrm{~d} u \text { with } u=x^{2}+2 x+2 \\ & f(x)=\ln \left(x^{2}+2 x+2\right)(+c) \end{aligned}

using initial conditions $x=2, y=3+\ln 10$ to find c

f(2)=\ln (10)+c=3+\ln (10) \Rightarrow c=3

f(x)=\ln \left(x^{2}+2 x+2\right)+3

(f)

Find the equation of the normal to the graph of f at $(2,3+\ln 10)$ .

[ 3 ]

$f^{\prime}(2)=\frac{6}{10}$
attempt to take the negative reciprocal of their $f^{\prime}(2)$
gradient of normal $=-\frac{10}{6}$

y-(\ln 10+3)=-\frac{5}{3}(x-2) \quad\left(y=-\frac{5}{3} x+\frac{19}{3}+\ln 10\right) \quad\left(\frac{y-\ln 10-3}{x-2}=-\frac{5}{3}\right)

[Maximum number: 6]

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.

(a)

Find $f^{\prime}(x)$ .

[ 2 ]

$f^{\prime}(x)=3 a x^{2}-12 x$

Note: Award A1 for each correct term.

(b)

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

[ 4 ]

setting their derivative equal to 3 (seen anywhere)
e.g. $\quad f^{\prime}(x)=3$
attempt to substitute x=1 into $f^{\prime}(x)$
e.g. $\quad 3 a(1)^{2}-12(1)$
correct substitution into $f^{\prime}(x)$
e.g. $\quad 3 a-12,3 a=15$
a=5
5. Note: All answers must be given in terms of m. If a candidate makes an error that means there is no m in their answer, do not award the final A1 F T mark.

METHOD 1

[Maximum number: 1]

The derivative of a function f is given by $f^{\prime}(x)=\frac{2 x+2}{x^{2}+2 x+2}$ , for $x \in \mathbb{R}$ .

Hence, find the values of x for which f is increasing.

$\quad x>-1($ accept $x \geq-1)$

[Maximum number: 6]

Consider the function $f(x)=\frac{2}{\pi} \sin (3 \pi x)+2$ , where $0 \leq x \leq 2$ . The following diagram shows the graph of f.

(a)

Find the gradient of $L_{2}$ .

[ 3 ]

substituting their $m_{L_{1}}$ into $-\frac{1}{m_{L_{1}}}$

\left(m_{L_{2}}=\right)-6

(b)

Hence, or otherwise, find the equation of $L_{2}$ .

The line $L_{1}$ intersects the graph of f at another point B , where the x-coordinate of B is greater than 1.5 . This is shown in the following diagram.

Figure for Question (b) — IB Maths AA SL

[ 3 ]

[Maximum number: 3]

Find h(x).

recognition that $h(x)=\int h^{\prime}(x)$

\begin{aligned} & h(x)=\frac{1}{4} \mathrm{e}^{4 x}+6 x+c \\ & (h(1.5)=) \frac{1}{4} \mathrm{e}^{4(1.5)}+6(1.5)+c=105 \\ & c=-4.85719 \ldots \\ & h(x)=\frac{1}{4} \mathrm{e}^{4 x}+6 x-4.86\left(=\frac{1}{4} \mathrm{e}^{4 x}+6 x+96-\frac{1}{4} \mathrm{e}^{6}\right) \end{aligned}

[Maximum number: 3]

Let $f(x)=A \mathrm{e}^{k x}+3$ . Part of the graph of f is shown below.

The y-intercept is at (0,13).

Using your value of k, find $f^{\prime}(x)$ .

$\quad f(x)=10 \mathrm{e}^{-0.201 x}+3$ $f^{\prime}(x)=10 \mathrm{e}^{-0.201 x} \times-0.201 \quad\left(=-2.01 \mathrm{e}^{-0.201 x}\right)$
Note: Award A1 for $10 \mathrm{e}^{-0.201 x}, \boldsymbol{A} \mathbf{1}$ for × -0.201, A1 for the derivative of 3 is zero.

[Maximum number: 4]

The derivative of a function f is given by $f^{\prime}(x)=\frac{2 x+2}{x^{2}+2 x+2}$ , for $x \in \mathbb{R}$ .

Show that $f^{\prime \prime}(x)=\frac{-2 x^{2}-4 x}{\left(x^{2}+2 x+2\right)^{2}}$ .

attempt to use the quotient rule

f^{\prime \prime}(x)=\frac{2\left(x^{2}+2 x+2\right)-(2 x+2)(2 x+2)}{\left(x^{2}+2 x+2\right)^{2}}

Note: Award A1 for first term in numerator, A1 for second term in numerator (including the minus sign). If the numerator is correct, but the denominator is incorrect or missing, award M1A1A0.

\begin{aligned} & f^{\prime \prime}(x)=\frac{2 x^{2}+4 x+4-4 x^{2}-8 x-4}{\left(x^{2}+2 x+2\right)^{2}} \text { OR } f^{\prime \prime}(x)=\frac{2 x^{2}+4 x+4-\left(4 x^{2}+8 x+4\right)}{\left(x^{2}+2 x+2\right)^{2}} \\ & f^{\prime \prime}(x)=\frac{-2 x^{2}-4 x}{\left(x^{2}+2 x+2\right)^{2}} \end{aligned}

[Maximum number: 2]

The derivative of a function f is given by $f^{\prime}(x)=\frac{2 x+2}{x^{2}+2 x+2}$ , for $x \in \mathbb{R}$ .

Hence, justify that the value of x found in part (b)(i) corresponds to a local minimum point on the graph of f.

It is given that $f(2)=3+\ln 10$ .

substituting x=-1 into $f^{\prime \prime}(x)$ , clearly leading to positive numerator A1

\begin{aligned} & f^{\prime \prime}(-1)=\left(\frac{-2+4}{1}\right)(=2) \\ & f^{\prime \prime}(-1)>0 \end{aligned}

therefore this is a local minimum point AG

Note: Do not award AOR1.

[Maximum number: 4]

The velocity of a particle in $\mathrm{ms}^{-1}$ is given by $v=\mathrm{e}^{\sin t}-1$ , for $0 \leq t \leq 5$ .

Find the total distance travelled by the particle in the first five seconds.

recognizing distance is area under velocity curve
valid approach to find the total area
$e g \quad$ area A+ area $\mathrm{B}, \int v \mathrm{~d} t-\int v \mathrm{~d} t, \int_{0}^{3.14} v \mathrm{~d} t+\int_{3.14}^{5} v \mathrm{~d} t, \int|v|$
correct working with integration and limits (accept d x or missing d t )
eg $\quad \int_{0}^{3.14} v \mathrm{~d} t+\int_{5}^{3.14} v \mathrm{~d} t, 3.067 \ldots+0.878 \ldots, \int_{0}^{5}\left|\mathrm{e}^{\sin t}-1\right|$
distance $=3.95(\mathrm{~m})$

[Maximum number: 3]

Find $\int\left(\mathrm{e}^{4 x}+6\right) \mathrm{d} x$ .

It is given that $h^{\prime}(x)=\mathrm{e}^{4 x}+6$ and h(1.5)=105.

$\frac{1}{4} \mathrm{e}^{4 x}+6 x+c$

Note: Award A1 for each term.

[Maximum number: 3]

Consider the function $f(x)=\frac{2}{\pi} \sin (3 \pi x)+2$ , where $0 \leq x \leq 2$ . The following diagram shows the graph of f.

Find the area of the shaded region.

EITHER

Attempt to form the required integral involving subtraction in either order.
$\int_{t_{1}}^{t_{2}}\left|L_{1}-f(x)\right| \mathrm{d} x$
correct expression with limits 1 and their x value at B
$\int_{1}^{1.64856 \ldots}\left|\left(\frac{1}{6} x+\frac{11}{6}\right)-\left(\frac{2}{\pi} \sin 3 \pi x+2\right)\right| \mathrm{d} x$ (or equivalent)

OR
intersection point between A and B is ( 1.34287 . . ., 2.05714 . .. )
subtracting $L_{1}$ and f, in either order, and substituting into area of region formula for one area with correct limits.

Two correct expressions OR correct values
$\int_{1}^{1.34287 \ldots}\left(\left(\frac{1}{6} x+\frac{11}{6}\right)-\left(\frac{2}{\pi} \sin 3 \pi x+2\right)\right) \mathrm{d} x=(0.144618 \ldots)$ AND
$\int_{1.34287 \ldots}^{1.64856 \ldots}\left(\left(\frac{2}{\pi} \sin 3 \pi x+2\right)-\left(\frac{1}{6} x+\frac{11}{6}\right)\right) \mathrm{d} x=(0.108585 \ldots)$

THEN
0.253204...
(area =) 0.253

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Covered syllabus points

Question 1

Question 2

Question (a)

Question (b)

Question (c)

Question (d)

Question (e)

Question (f)

Question 3

Question (a)

Question (b)

Question 4

Question 5

Question (a)

Question (b)

Question 6

Question 7

Question 8

Question 9

Question 10

Question 11

Question 12