Legacy HL question bank

May 2014 TZ1 HL P1 · Question 1 · Section A

Remainder theorem: find a and b

Functions › Polynomials

Legacy HL · Paper 1Easy5 marks

When the polynomial $3x^3 + ax + b$ is divided by $(x-2)$ , the remainder is 2, and when divided by $(x+1)$ , it is 5. Find the value of $a$ and the value of $b$ .

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Substitute $x=2$ : $2a+b+24=2\Rightarrow 2a+b=-22$ M1A1
Substitute $x=-1$ : $-a-b+3=5\Rightarrow -a-b=2$ A1
Attempt to solve simultaneouslyM1
$a=-10,\ b=-2$ A1

May 2014 TZ1 HL P1 · Question 2 · Section A

Four numbers from mean, median, mode

Statistics and Probability › Basic stats

Legacy HL · Paper 1Easy4 marks

Four numbers are such that their mean is 13, their median is 14 and their mode is 15. Find the four numbers.

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Sum of the four numbers divided by 4 is 13M1
Two of the numbers must be 15 (the mode)A1
As the median is 14, one of the middle numbers is 13A1
Fourth number is $52-15-15-13=9$ A1

May 2014 TZ1 HL P1 · Question 3 · Section A

Telescoping logarithm product

Number and Algebra › Exponents and logs

Legacy HL · Paper 1Medium5 marks

Consider $a = \log_2 3 \times \log_3 4 \times \log_4 5 \times \cdots \times \log_{31} 32$ . Given that $a \in \mathbb{Z}$ , find the value of $a$ .

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Rewrite each term as $\log_k(k+1)=\dfrac{\log(k+1)}{\log k}$ M1
Product telescopes: $\dfrac{\log3}{\log2}\times\dfrac{\log4}{\log3}\times\cdots\times\dfrac{\log32}{\log31}$ A1
$=\dfrac{\log32}{\log2}$ A1
$=\dfrac{5\log2}{\log2}$ M1
$=5$ A1

May 2014 TZ1 HL P1 · Question 4 · Section A

Cubic roots and Vieta's formulas

Functions › Polynomials

Legacy HL · Paper 1Medium6 marks

The equation $5x^3 + 48x^2 + 100x + 2 = a$ has roots $r_1$ , $r_2$ and $r_3$ . Given that $r_1+r_2+r_3+r_1r_2r_3=0$ , find the value of $a$ .

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Sum of roots: $r_1+r_2+r_3=-\dfrac{48}{5}$ M1
Product of roots: $r_1r_2r_3=-\dfrac{2-a}{5}$ A1
Set up the given condition: $-\dfrac{48}{5}-\dfrac{2-a}{5}=0$ M1
Multiply through by 5: $-48-(2-a)=0$ A1
Simplify: $a-50=0$ M1
$a=50$ A1

May 2014 TZ1 HL P1 · Question 5 · Section A

Half-angle identities, then a definite integral

Geometry and Trigonometry › Trig equations and identities · Calculus › Basic integration

Legacy HL · Paper 1Medium-Hard8 marks

(a)Use the identity $\cos 2\theta = 2\cos^2\theta - 1$ to prove that $\cos\tfrac12 x = \sqrt{\tfrac{1+\cos x}{2}}$ , $0\le x\le \pi$ .[2 marks]
(b)Find a similar expression for $\sin\tfrac12 x$ , $0\le x\le \pi$ .[2 marks]
(c)Hence find the value of $\displaystyle\int_0^{\pi/2}\left(\sqrt{1+\cos x}+\sqrt{1-\cos x}\right)\,dx$ .[4 marks]

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(a)

$\cos\tfrac12x=\pm\sqrt{\tfrac{1+\cos x}{2}}$ M1
Positive since $0\le x\le\pi\Rightarrow\cos\tfrac12x\ge0$ R1

(b)

Using $\cos2\theta=1-2\sin^2\theta$ with $\theta=\tfrac12x$ M1
$\sin\tfrac12x=\sqrt{\tfrac{1-\cos x}{2}}$ A1

(c)

Integrand: $\sqrt2\cos\tfrac12x+\sqrt2\sin\tfrac12x$ A1
Integrate: $\Big[2\sqrt2\sin\tfrac12x-2\sqrt2\cos\tfrac12x\Big]_0^{\pi/2}$ A1
Evaluate at $x=\tfrac\pi2$ and $x=0$ A1
$=2\sqrt2$ A1

May 2014 TZ1 HL P1 · Question 6 · Section A

Sketch g(x) from the graph of f(x)

Calculus › Basic integration

Legacy HL · Paper 1Medium6 marks

The axes below show the graph of $y=f(x)$ for $-4\le x\le 4$ . Let $g(x)=\displaystyle\int_{-4}^{x} f(t)\,dt$ for $-4\le x\le 4$ .

Left: y = f(x). Right: blank axes for sketching y = g(x).

(a)State the value of $x$ at which $g(x)$ is a minimum.[1 mark]
(b)On the second set of axes, sketch the graph of $y=g(x)$ .[5 marks]

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(a)

The minimum of $g$ occurs where $f$ changes sign from negative to positive: $x=1$ A1

(b)

Point $(-4,0)$ A1
Point $(0,-4)$ A1
Minimum at $x=1$ in approximately the correct placeA1
Point $(4,0)$ A1
Correct shape, continuous through $x=0$ A1

May 2014 TZ1 HL P1 · Question 7 · Section A

Cosine rule in an equilateral triangle

Geometry and Trigonometry › Sectors and triangles

Legacy HL · Paper 1Easy-Medium5 marks

The triangle $ABC$ is equilateral with side 3 cm. The point $D$ lies on $[BC]$ such that $BD=1$ cm. Find $\cos D\hat{A}C$ .

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$AD^2=3^2+1^2-2(3)(1)\cos60^\circ$ M1
$AD^2=7$ A1
$\cos D\hat AC=\dfrac{AD^2+AC^2-DC^2}{2\cdot AD\cdot AC}$ M1
$=\dfrac{7+9-4}{2\sqrt7(3)}$ A1
$=\dfrac{2}{\sqrt7}$ A1

May 2014 TZ1 HL P1 · Question 8 · Section A

Acceleration from v(s) via chain rule

Calculus › Kinematics

Legacy HL · Paper 1Medium6 marks

A body is moving in a straight line. When it is $s$ metres from a fixed point $O$ , its velocity is $v=-\dfrac{1}{s^2}$ , $s>0$ . Find the acceleration of the body when it is 50 cm from $O$ .

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$v=-s^{-2}\Rightarrow\dfrac{dv}{ds}=2s^{-3}=\dfrac2{s^3}$ M1A1
$a=v\dfrac{dv}{ds}=-\dfrac1{s^2}\times\dfrac2{s^3}=-\dfrac2{s^5}$ M1A1
At $s=0.5$ m: $a=-\dfrac{2}{(0.5)^5}=-64\text{ m s}^{-2}$ M1A1

May 2014 TZ1 HL P1 · Question 9 · Section A

Implicit differentiation with arctan

Calculus › Implicit differentiation

Legacy HL · Paper 1Hard9 marks

A curve has equation $\arctan x^2 + \arctan y^2 = \dfrac{\pi}{4}$ .

(a)Find $\dfrac{dy}{dx}$ in terms of $x$ and $y$ .[4 marks]
(b)Find the gradient of the curve at the point where $x=\dfrac{1}{\sqrt2}$ and $y<0$ .[5 marks]

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(a)

Differentiate implicitly (method)M1
LHS: $\dfrac{4x}{1+x^4}$ A1
RHS: $\dfrac{4y^3}{1+y^4}\dfrac{dy}{dx}$ A1
Rearrange: $\dfrac{dy}{dx}=-\dfrac{x(1+y^4)}{y^3(1+x^4)}$ A1

(b)

$y^2=\tan\left(\tfrac\pi4-\arctan\tfrac12\right)$ M1
Expand using the tangent subtraction formulaM1
$=\dfrac{1-\tfrac12}{1+\tfrac12}=\dfrac13$ A1
$y=-\dfrac1{\sqrt3}$ A1
Substitute into (a): $\dfrac{dy}{dx}=\dfrac{4\sqrt6}{9}$ (accept equivalent forms)A1

May 2014 TZ1 HL P1 · Question 10 · Section A

Find cos 4x from sin x + cos x

Geometry and Trigonometry › Trig equations and identities

Legacy HL · Paper 1Medium-Hard6 marks

Given that $\sin x + \cos x = \dfrac{2}{3}$ , find $\cos 4x$ .

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$(\sin x+\cos x)^2=\sin^2x+2\sin x\cos x+\cos^2x=\tfrac49$ M1A1
Using $\sin^2x+\cos^2x=1$ : $1+2\sin x\cos x=\tfrac49$ M1
Using $2\sin x\cos x=\sin2x$ : $\sin2x=-\tfrac59$ M1
$\cos4x=1-2\sin^22x$ M1
$=1-2\left(\tfrac{25}{81}\right)=\tfrac{31}{81}$ A1

May 2014 TZ1 HL P1 · Question 11 · Section B

Full curve analysis of f(x) = ln(x)/x

Calculus › Basic differentiation

Legacy HL · Paper 1Hard21 marks

Consider the function $f(x)=\dfrac{\ln x}{x}$ , $x>0$ . The sketch shows $y=f(x)$ and its tangent at $A$ , where the curve crosses the $x$ -axis; $B$ is the maximum point; $C$ is the point of inflection.

Diagram not to scale. Recomputed exactly from f(x) = ln(x)/x.

(a)Show that $f'(x)=\dfrac{1-\ln x}{x^2}$ .[2 marks]
(b)Find the coordinates of $B$ , the maximum point.[3 marks]
(c)Find the coordinates of $C$ , the point of inflection.[5 marks]
(d)Find the equation of the tangent to the graph of $f$ at $A$ .[4 marks]
(e)Find the area enclosed by the curve $y=f(x)$ , the tangent at $A$ , and the line $x=e$ .[7 marks]

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(a)

Quotient rule: $f'(x)=\dfrac{x\cdot\tfrac1x-\ln x}{x^2}$ M1
$=\dfrac{1-\ln x}{x^2}$ A1

(b)

$1-\ln x=0$ has solution $x=e$ M1A1
$y=\tfrac1e$ , maximum at $\left(e,\tfrac1e\right)$ A1

(c)

$f''(x)=\dfrac{-x-(1-\ln x)(2x)}{x^4}$ M1A1
$=\dfrac{2\ln x-3}{x^3}$ (A1)
Point of inflexion where $f''(x)=0\Rightarrow x=e^{3/2}$ M1
$y=\tfrac32e^{-3/2}$ , so $C=\left(e^{3/2},\tfrac32e^{-3/2}\right)$ A1A1

(d)

$f(1)=0$ A1
$f'(1)=1$ (A1)
$y=x+c$ , through $(1,0)$ M1
Tangent: $y=x-1$ A1

(e)

Area $=\displaystyle\int_1^e\left((x-1)-\dfrac{\ln x}{x}\right)dx$ M1A1A1
$\displaystyle\int\dfrac{\ln x}{x}dx=\dfrac{(\ln x)^2}{2}$ M1A1
$\displaystyle\int(x-1)dx=\dfrac{x^2}{2}-x$ A1
Evaluating from $1$ to $e$ : $\tfrac12e^2-e$ A1

May 2014 TZ1 HL P1 · Question 12 · Section B

3D square, plane, perpendicular line and reflection

Geometry and Trigonometry › Vectors

Legacy HL · Paper 1Hard22 marks

$O(0,0,0)$ , $A(6,0,0)$ , $B(6,-\sqrt{24},\sqrt{12})$ , $C(0,-\sqrt{24},\sqrt{12})$ .

(a)Show that O, A, B, C form a square.[3 marks]
(b)Find the coordinates of M, the midpoint of [OB].[1 mark]
(c)Show that the plane $\Pi$ containing OABC has equation $y+\sqrt2\,z=0$ .[3 marks]
(d)Find a vector equation of the line L, through M, perpendicular to $\Pi$ .[3 marks]
(e)Find D, the intersection of L with the plane $y=0$ .[3 marks]
(f)Find E, the reflection of D in the plane $\Pi$ .[3 marks]
(g)Find the angle $O\hat{D}A$ , and state what this tells you about solid OABCDE.[6 marks]

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(a)

$|\overrightarrow{OA}|=|\overrightarrow{CB}|=6$ A1
$|\overrightarrow{OC}|=|\overrightarrow{AB}|=6$ A1
$\overrightarrow{OA}\cdot\overrightarrow{OC}=0$ , therefore a squareA1

(b)

$M=\left(3,-\sqrt6,\sqrt3\right)$ A1

(c)

$\overrightarrow{OA}\times\overrightarrow{OC}$ gives a normal vectorM1A1
Showing $d=0$ since the plane passes through the originM1

(d)

$\mathbf r=\left(3,-\sqrt6,\sqrt3\right)+\lambda(0,1,\sqrt2)$ A1A1A1

(e)

Using $y=0$ to find $\lambda$ M1
Substitute $\lambda$ into the equation from (d)M1
$D=(3,0,3\sqrt3)$ A1

(f)

$\lambda$ for $E$ is the negative of $\lambda$ for $D$ M1
$E=(3,-2\sqrt6,-\sqrt3)$ A1A1

(g)

$\overrightarrow{DA}\cdot\overrightarrow{DO}=18$ M1A1
$\cos O\hat DA=\dfrac{18}{36}=\tfrac12$ M1
$O\hat DA=60^\circ$ A1
OABCDE is a regular octahedron, made up of 8 equilateral trianglesA2

Practise the question. Understand the method.

Remainder theorem: find a and b

Four numbers from mean, median, mode

Telescoping logarithm product

Cubic roots and Vieta's formulas

Half-angle identities, then a definite integral

Sketch g(x) from the graph of f(x)

Cosine rule in an equilateral triangle

Acceleration from v(s) via chain rule

Implicit differentiation with arctan

Find cos 4x from sin x + cos x

Full curve analysis of f(x) = ln(x)/x

3D square, plane, perpendicular line and reflection