IB Maths AA SL 5.1 Calculus Question Bank

Practise choosing differentiation or integration methods, interpreting first and second derivatives, finding constants from conditions and solving area, motion and optimisation…

Exam points

differentiate with chain, product or quotient rules before applying tangent, normal or related-rate conditions
use derivative signs and the second derivative to classify intervals, extrema, inflexions and optimal values
integrate a derivative or rate and use boundary data to determine the constant and original function
set up definite integrals with correct intersections and signs to calculate exact areas between curves
link displacement, velocity and acceleration, splitting intervals when direction changes to find distance

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: 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 5.1 Calculus - SL content 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.

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Covered syllabus points

5.1 Calculus - SL content question 1

5.1 Calculus - SL content question 2

Question (a)

Question (b)

Question (c)

Question (d)

5.1 Calculus - SL content question 3