IB Maths AA SL 5 Calculus

[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: 13]

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.

[ 3 ]

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

(b)

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}

(c)

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 $f(2)=3+\ln 10$ .

[ 7 ]

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.

(d)

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: 2]

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

Figure for Question 5 Calculus question 3 — IB Maths AA SL

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

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

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

Note: Award A1 for each correct term.

[Maximum number: 9]

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.

Figure for Question 5 Calculus question 4 — IB Maths AA SL

(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 ]

(c)

Find the area of the shaded region.

[ 3 ]

(f) 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

[Maximum number: 6]

(a)

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.

[ 3 ]

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

Note: Award A1 for each term.

(b)

Find h(x).

[ 3 ]

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: 6]

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

Figure for Question 5 Calculus question 6 — IB Maths AA SL

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

(a)

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

$\quad f(x)=10 \mathrm{e}^{-0.201 x}+3$ $f^{\prime}(x)=10 \mathrm{e}^{-0.201 x} \times-0.201 \quad\left(=-2.01 \mathrm{e}^{-0.201 x}\right)$
A1A1A1
N3
Note: Award A1 for $10 \mathrm{e}^{-0.201 x}, \boldsymbol{A} \mathbf{1}$ for x-0.201, A1 for the derivative of 3 is zero.

(b)

Hence, explain why f is a decreasing function.

valid reason with reference to derivative

R1 N1 e.g. $f^{\prime}(x)<0$ , derivative always negative

(c)

Find the area enclosed by the graphs of f and g.

[ 6 ]

finding limits $3.8953 \ldots, 8.6940 \ldots$ (seen anywhere)
evidence of integrating and subtracting functions
correct expression
e.g. $\int_{3.90}^{8.69} g(x)-f(x) \mathrm{d} x, \int_{3.90}^{8.69}\left[\left(-x^{2}+12 x-24\right)-\left(10 \mathrm{e}^{-0.201 x}+3\right)\right] \mathrm{d} x$
area =19.5

QUESTION 10

IB Maths AA SL 5 Calculus Question Bank

Covered syllabus points

Covered syllabus points

5 Calculus question 1

5 Calculus question 2

Question (a)

Question (b)

Question (c)

Question (d)

5 Calculus question 3

5 Calculus question 4

Question (a)

Question (b)

Question (c)

5 Calculus question 5

Question (a)

Question (b)

5 Calculus question 6

Question (a)

Question (b)

Question (c)