IB Questionbank

User interface language: English | Español

Date	May 2022	Marks available	5	Reference code	22M.1.SL.TZ2.5
Level	Standard Level	Paper	Paper 1 (without calculator)	Time zone	Time zone 2
Command term	Find	Question number	5	Adapted from	N/A

Question

Find the least positive value of $x$ for which $\cos (\frac{x}{2} + \frac{π}{3}) = \frac{1}{\sqrt{2}}$ .

Markscheme

determines $\frac{π}{4}$ (or $45 °$ ) as the first quadrant (reference) angle (A1)

attempts to solve $\frac{x}{2} + \frac{π}{3} = \frac{π}{4}$ (M1)

Note: Award M1 for attempting to solve $\frac{x}{2} + \frac{π}{3} = \frac{π}{4}, \frac{7 π}{4} (, \dots)$

$\frac{x}{2} + \frac{π}{3} = \frac{π}{4} \Rightarrow x < 0$ and so $\frac{π}{4}$ is rejected (R1)

$\frac{x}{2} + \frac{π}{3} = 2 π - \frac{π}{4} (= \frac{7 π}{4})$ A1

$x = \frac{17 π}{6}$ (must be in radians) A1

[5 marks]

Examiners report

This question proved to be a struggle for many candidates, and some candidates made no attempt here. While a good number of candidates recognized the reference angle of $\frac{π}{4}$ , this led to a final answer of $x = - \frac{π}{6}$ , which many left as their final answer. In other cases, some candidates heeded the requirement that $x$ must be a positive value, however they gave an incorrect final answer of $x = \frac{11 π}{6}$ . Few candidates correctly rejected their initial reference angle of $\frac{π}{4}$ and correctly solved an equation using $\frac{x}{2} + \frac{π}{3} = \frac{7 π}{4}$ .

Syllabus sections

Topic 3— Geometry and trigonometry » SL 3.5—Unit circle definitions of sin, cos, tan. Exact trig ratios, ambiguous case of sine rule

Show 28 related questions

16N.2.AHL.TZ0.H_7b:
Find the difference between the areas of the two possible triangles ABC.
19N.1.AHL.TZ0.H_9b:
Show that ${\text{B}}\mathop {\text{A}}\limits^ \wedge {\text{D}} = 75^\circ$ .
19M.2.SL.TZ2.S_3a:
Find BD.
19M.2.SL.TZ2.S_3b:
The area of triangle ABD is 18.5 cm². Find the possible values of θ.
19M.1.SL.TZ2.S_9a:
Find the value of ${\text{tan}}\,\theta$ .
22M.1.SL.TZ1.4b:
Solve the equation $(f \circ g) (x) = 2 \cos 2 x$ where $0 \leq x \leq π$ .
22M.1.SL.TZ1.9b.i:
Show that $\int_{0}^{m} f (x) d x = \frac{32}{7} \sin^{7} m$ , where $m$ is a positive real constant.
22M.1.SL.TZ1.9b.ii:
It is given that $\int_{m}^{\frac{π}{2}} f (x) d x = \frac{127}{28}$ , where $0 \leq m \leq \frac{π}{2}$ . Find the value of $m$ .
22M.1.AHL.TZ1.9b:
Hence, given that $arg (z_{1} z_{2}) = \frac{π}{4}$ , find the value of $b$ .
22M.1.AHL.TZ2.7:
By using the substitution $u = sec x$ or otherwise, find an expression for $\int_{0}^{\frac{π}{3}} {sec}^{n} x \tan x d x$ in terms of $n$ , where $n$ is a non-zero real number.
22M.2.AHL.TZ2.7b:
Using l’Hôpital’s rule, show algebraically that the value of the limit is $- \frac{1}{4}$ .
18M.2.SL.TZ1.S_6:
Triangle ABC has a = 8.1 cm, b = 12.3 cm and area 15 cm². Find the largest possible perimeter of triangle ABC.
18N.2.SL.TZ0.S_7a:
Let SR = $x$ . Use the cosine rule to show that ${x^2} - \left( {76\,{\text{cos}}\,43^\circ } \right)x + 420 = 0$ .
16N.1.AHL.TZ0.H_13d:
Hence or otherwise solve the equation $\sin x + \sin 3x = \cos x$ in the interval $0 < x < \pi$ .
19M.1.AHL.TZ1.H_4a:
Show that ${\text{sin}}\,\theta = \frac{{\sqrt {15} }}{4}$ .
19N.1.AHL.TZ0.H_9c.i:
By considering triangle ${\text{ABD}}$ , show that ${\text{B}}{{\text{D}}^2} = 5{r^2} - 2{r^2}q\sqrt 6$ .
19N.1.AHL.TZ0.H_9d:
Use your answers to part (c) to show that ${\text{cos}}\,75^\circ = \frac{1}{{\sqrt 6 + \sqrt 2 }}$ .
22M.2.AHL.TZ2.7a:
Show that a finite limit only exists for $k = \frac{π}{4}$ .
19N.1.AHL.TZ0.H_9c.ii:
By considering triangle ${\text{CBD}}$ , find another expression for ${\text{B}}{{\text{D}}^2}$ in terms of $r$ and $q$ .
17N.1.SL.TZ0.S_4a:
Show that ${\text{AC}} = 7{\text{ cm}}$ .
16N.2.AHL.TZ0.H_7a:
Use the cosine rule to find the two possible values for AC.
16N.1.AHL.TZ0.H_13a:
Find the value of $\sin \frac{\pi }{4} + \sin \frac{{3\pi }}{4} + \sin \frac{{5\pi }}{4} + \sin \frac{{7\pi }}{4} + \sin \frac{{9\pi }}{4}$ .
19N.1.AHL.TZ0.H_9a:
Given that ${\text{cos}}\,75^\circ = q$ , show that ${\text{cos}}\,105^\circ = - q$ .
17M.1.SL.TZ1.S_3:
The following diagram shows triangle PQR.

Find PR.
19M.1.AHL.TZ1.H_4b:
Find the two possible values for the length of the third side.
17N.1.SL.TZ0.S_4b:
The shape in the following diagram is formed by adding a semicircle with diameter [AC] to the triangle.

Find the exact perimeter of this shape.
18N.2.SL.TZ0.S_7b:
Hence or otherwise, find the total distance along the road where the signal from the tower can reach cellular phones.
21N.2.SL.TZ0.3:
Consider a triangle $ABC$ , where $AC = 12, CB = 7$ and $B \hat{A} C = 25 °$ .

Find the smallest possible perimeter of triangle $ABC$ .

Hide related questions

Topic 3— Geometry and trigonometry » SL 3.8—Solving trig equations

Topic 3— Geometry and trigonometry

Question

Markscheme

Examiners report

Syllabus sections

View options