Limits, differentiation rules, tangents & normals, optimisation – IB Maths exam questions with mark scheme

1CoreNon-calculator4 marks

Let $f(x)=4x^3-\dfrac{6}{x}+\sqrt{x}$ for $x>0$.

(a) Find $f'(x)$. [3]

(b) Find the value of $f'(1)$. [1]

$12x^2$ (derivative of $4x^3$)A1
$+\dfrac{6}{x^2}$ (from $-6x^{-1}$; accept $6x^{-2}$)A1
$+\dfrac{1}{2\sqrt{x}}$ (accept $\tfrac12x^{-1/2}$)A1
$f'(1)=12+6+\tfrac12=\dfrac{37}{2}$ (accept 18.5)A1
Answer: (a) $f'(x)=12x^2+\dfrac{6}{x^2}+\dfrac{1}{2\sqrt x}$; (b) $\dfrac{37}{2}$
Examiner tip: Rewrite $6/x$ and $\sqrt{x}$ as powers first; the sign on the $6/x^2$ term is the usual slip (two negatives make a positive).

2StandardNon-calculator6 marks

The curve $C$ has equation $y=x^3-3x^2+2$. The point $P$ on $C$ has $x$-coordinate 3.

(a) Find the equation of the tangent to $C$ at $P$, giving your answer in the form $y=mx+c$. [4]

(b) Find the equation of the normal to $C$ at $P$, giving your answer in the form $ax+by+d=0$ where $a,b,d\in\mathbb{Z}$. [2]

$y=2$ at $x=3$, so $P=(3,2)$B1
Attempts $\dfrac{dy}{dx}=3x^2-6x$ (at least one term correct)M1
Gradient at $P$ is $27-18=9$A1
$y-2=9(x-3)$, so $y=9x-25$A1
Normal gradient $-\dfrac19$ (negative reciprocal) used with $P$: $y-2=-\dfrac19(x-3)$M1
$x+9y-21=0$ (or any integer multiple)A1
Answer: (a) $y=9x-25$; (b) $x+9y-21=0$
Examiner tip: Find the $y$-coordinate of $P$ before anything else, and clear the fraction in (b) because the question demands integer coefficients.

3StandardNon-calculator5 marks

Differentiate with respect to $x$:

(a) $y=x^2e^{3x}$, giving your answer in a fully factorised form; [2]

(b) $y=\dfrac{\sin 2x}{x}$. [3]

Product rule: $2x\,e^{3x}+x^2\cdot 3e^{3x}$M1
$x\,e^{3x}(2+3x)$A1
Quotient rule form $\dfrac{x\cdot(\ldots)-\sin 2x\cdot 1}{x^2}$M1
Numerator correct: $x\cdot2\cos2x-\sin2x\cdot1$ (derivative of $\sin2x$ is $2\cos2x$)A1
$\dfrac{dy}{dx}=\dfrac{2x\cos 2x-\sin 2x}{x^2}$A1
Answer: (a) $xe^{3x}(3x+2)$; (b) $\dfrac{2x\cos2x-\sin2x}{x^2}$
Examiner tip: The chain rule factor 2 from $\sin 2x$ and factor 3 from $e^{3x}$ are the marks most often dropped.

4HardCalculator10 marks

A closed cylindrical tin has radius $r$ cm and height $h$ cm. Its volume is $1000\ \text{cm}^3$.

(a) Show that the total surface area $A\ \text{cm}^2$ of the tin is given by $A=2\pi r^2+\dfrac{2000}{r}$. [3]

(b) Find the value of $r$ for which $A$ is a minimum, giving your answer to 3 significant figures. [4]

(d) Find the minimum surface area, to 3 significant figures. [1]

$\pi r^2h=1000$, so $h=\dfrac{1000}{\pi r^2}$B1
$A=2\pi r^2+2\pi rh$ with $h$ substitutedM1
$A=2\pi r^2+\dfrac{2000}{r}$ (AG, must show the cancellation of $\pi r$)A1
Differentiates: $\dfrac{dA}{dr}=4\pi r-\dfrac{2000}{r^2}$ (attempt, at least one term correct)M1
Both terms correctA1
Sets $\dfrac{dA}{dr}=0$ and rearranges to $r^3=\dfrac{500}{\pi}$M1
$r=5.42$ cmA1
$\dfrac{d^2A}{dr^2}=4\pi+\dfrac{4000}{r^3}$M1
Positive for $r>0$ (so at $r=5.42$), hence minimumR1
$A_{\min}=554\ \text{cm}^2$A1
Answer: (a) shown; (b) $r=5.42$ cm; (c) $A''>0$ so minimum; (d) $554\ \text{cm}^2$
Examiner tip: In a 'show that' part every step must appear; in (c) say why the second derivative is positive, not only that it is.

5StandardNon-calculatorHL only6 marks

(a) Using differentiation from first principles, show that if $f(x)=x^2+3x$ then $f'(x)=2x+3$. [4]

(b) Find $\displaystyle\lim_{x\to2}\dfrac{x^2-4}{x-2}$. [2]

Uses $f'(x)=\displaystyle\lim_{h\to0}\dfrac{f(x+h)-f(x)}{h}$ with $f(x+h)=(x+h)^2+3(x+h)$ substitutedM1
Numerator simplified correctly: $f(x+h)-f(x)=2xh+h^2+3h$A1
Divides by $h$: $\displaystyle\lim_{h\to0}(2x+h+3)$ (limit notation required on this line, or a clear statement that $h\to0$ is still to be taken)A1
Correct limiting step: as $h\to0$, $2x+h+3\to2x+3$, so $f'(x)=2x+3$ AG. Award only if the limit is taken after cancelling $h$; do not award if $h$ is set to $0$ before dividing.A1
Factorises numerator: $\dfrac{(x-2)(x+2)}{x-2}=x+2$ for $x\neq2$ (accept L'Hôpital's rule: $\displaystyle\lim_{x\to2}\dfrac{2x}{1}$)M1
Limit $=4$A1
Answer: (a) shown; (b) $4$
Examiner tip: Write $\lim_{h\to0}$ on every line until you actually let $h\to0$; examiners penalise dropped limit notation.

6HardNon-calculator5 marks

The line $y=2x+k$, where $k\in\mathbb{R}$, is a tangent to the curve $y=x^3-x$. Find all possible values of $k$.

$\dfrac{dy}{dx}=3x^2-1$M1
Sets gradient equal to the line's gradient: $3x^2-1=2$M1
$x=1$ or $x=-1$A1
Uses the point on the curve ($y=0$ at both $x=\pm1$) in $y=2x+k$M1
$k=-2$ and $k=2$ (both required)A1
Answer: $k=-2$ or $k=2$
Examiner tip: There are two tangent points; stopping after $x=1$ loses the final mark.

Book a free introductory session and we'll find exactly which step is costing you.