IB Math Analysis & Approaches Questionbank-Topic: SL 3.8 Solving trigonometric equations in a finite interval SL Paper 1

 

https://play.google.com/store/apps/details?id=com.iitianDownload IITian Academy  App for accessing Online Mock Tests and More..

Question

Let \(f(x) = \sqrt 3 {{\rm{e}}^{2x}}\sin x + {{\rm{e}}^{2x}}\cos x\) , for \(0 \le x \le \pi \) . Solve the equation \(f(x) = 0\) .

Answer/Explanation

Markscheme

\({{\rm{e}}^{2x}}\left( {\sqrt 3 \sin x + \cos x} \right) = 0\)     (A1)

\({{\rm{e}}^{2x}} = 0\) not possible (seen anywhere)     (A1)

simplifying

e.g. \(\sqrt 3 \sin x + \cos x = 0\) , \(\sqrt 3 \sin x =  – \cos x\) , \(\frac{{\sin x}}{{ – \cos x}} = \frac{1}{{\sqrt 3 }}\)     A1 

EITHER

\(\tan x = – \frac{1}{{\sqrt 3 }}\)     A1

\(x = \frac{{5\pi }}{6}\)     A2     N4

OR

sketch of \(30^\circ \) , \(60^\circ \) , \(90^\circ \) triangle with sides \(1\), \(2\), \(\sqrt 3 \)     A1

work leading to \(x = \frac{{5\pi }}{6}\)     A1

verifying \(\frac{{5\pi }}{6}\) satisfies equation     A1     N4

[6 marks]

Examiners report

Those who realized \({{\rm{e}}^{2x}}\) was a common factor usually earned the first four marks. Few could reason with the given information to solve the equation from there. There were many candidates who attempted some fruitless algebra that did not include factorization.

Question

Solve \(\cos 2x – 3\cos x – 3 – {\cos ^2}x = {\sin ^2}x\) , for \(0 \le x \le 2\pi \) .

Answer/Explanation

Markscheme

evidence of substituting for \(\cos 2x\)     (M1)

evidence of substituting into \({\sin ^2}x + {\cos ^2}x = 1\)     (M1)

correct equation in terms of \(\cos x\) (seen anywhere)     A1

e.g. \(2{\cos ^2}x – 1 – 3\cos x – 3 = 1\) , \(2{\cos ^2}x – 3\cos x – 5 = 0\)

evidence of appropriate approach to solve     (M1)

e.g. factorizing, quadratic formula

appropriate working     A1

e.g. \((2\cos x – 5)(\cos x + 1) = 0\) , \((2x – 5)(x + 1)\) , \(\cos x = \frac{{3 \pm \sqrt {49}}}{4}\)

correct solutions to the equation

e.g. \(\cos x = \frac{5}{2}\) , \(\cos x = – 1\) , \(x = \frac{5}{2}\) , \(x = – 1\)     (A1)

\(x = \pi \)     A1     N4

[7 marks]

Examiners report

This question was quite difficult for most candidates. A number of students earned some credit for manipulating the equation with identities, but many earned no further marks due to algebraic errors. Many did not substitute for \(\cos 2x\) ; others did this substitution but then did nothing further.

Few candidates were able to get a correct equation in terms of \(\cos x\) and many who did get the equation didn’t know what to do with it. Candidates who correctly solved the resulting quadratic usually found the one correct value of \(x\), earning full marks. 

Question

Let \(f(x) = 6 + 6\sin x\) . Part of the graph of f is shown below.



The shaded region is enclosed by the curve of f , the x-axis, and the y-axis.

Solve for \(0 \le x < 2\pi \)

(i)     \(6 + 6\sin x = 6\) ;

(ii)    \(6 + 6\sin x = 0\) .

[5]
a(i) and (ii).

Write down the exact value of the x-intercept of f , for \(0 \le x < 2\pi \) .

[1]
b.

The area of the shaded region is k . Find the value of k , giving your answer in terms of \(\pi \) .

[6]
c.

Let \(g(x) = 6 + 6\sin \left( {x – \frac{\pi }{2}} \right)\) . The graph of f is transformed to the graph of g.

Give a full geometric description of this transformation.

[2]
d.

Let \(g(x) = 6 + 6\sin \left( {x – \frac{\pi }{2}} \right)\) . The graph of f is transformed to the graph of g.

Given that \(\int_p^{p + \frac{{3\pi }}{2}} {g(x){\rm{d}}x}  = k\) and \(0 \le p < 2\pi \) , write down the two values of p.

[3]
e.
Answer/Explanation

Markscheme

(i) \(\sin x = 0\)     A1

\(x = 0\) , \(x = \pi \)     A1A1     N2

(ii) \(\sin x = – 1\)     A1

\(x = \frac{{3\pi }}{2}\)    A1     N1

[5 marks]

a(i) and (ii).

\(\frac{{3\pi }}{2}\)     A1     N1

[1 mark]

b.

evidence of using anti-differentiation     (M1)

e.g. \(\int_0^{\frac{{3\pi }}{2}} {(6 + 6\sin x){\rm{d}}x} \)

correct integral \(6x – 6\cos x\) (seen anywhere)     A1A1

correct substitution     (A1)

e.g. \(6\left( {\frac{{3\pi }}{2}} \right) – 6\cos \left( {\frac{{3\pi }}{2}} \right) – ( – 6\cos 0)\) , \(9\pi  – 0 + 6\)

\(k = 9\pi + 6\)     A1A1     N3

[6 marks]

c.

translation of \(\left( {\begin{array}{*{20}{c}}
{\frac{\pi }{2}}\\
0
\end{array}} \right)\)     A1A1     N2

[2 marks]

d.

recognizing that the area under g is the same as the shaded region in f     (M1)

\(p = \frac{\pi }{2}\) , \(p = 0\)     A1A1     N3

[3 marks]

e.

Question

Solve the equation \(2\cos x = \sin 2x\) , for \(0 \le x \le 3\pi \) .

Answer/Explanation

Markscheme

METHOD 1

using double-angle identity (seen anywhere)     A1

e.g. \(\sin 2x = 2\sin x\cos x\) , \(2\cos x = 2\sin x\cos x\)

evidence of valid attempt to solve equation     (M1)

e.g. \(0 = 2\sin x\cos x – 2\cos x\) , \(2\cos x(1 – \sin x) = 0\)

\(\cos x = 0\) , \(\sin x = 1\)     A1A1

\(x = \frac{\pi }{2}\) , \(x = \frac{{3\pi }}{2}\) , \(x = \frac{{5\pi }}{2}\)     A1A1A1     N4

METHOD 2


     A1A1M1A1

Notes: Award A1 for sketch of \(\sin 2x\) , A1 for a sketch of \(2\cos x\) , M1 for at least one intersection point seen, and A1 for 3 approximately correct intersection points. Accept sketches drawn outside \(\left[ {0,3\pi } \right]\) , even those with more than 3 intersections.

\(x = \frac{\pi }{2}\) , \(x = \frac{{3\pi }}{2}\) , \(x = \frac{{5\pi }}{2}\)     A1A1A1     N4

[7 marks]

Question

Solve \({\log _2}(2\sin x) + {\log _2}(\cos x) =  – 1\), for \(2\pi  < x < \frac{{5\pi }}{2}\).

Answer/Explanation

Markscheme

correct application of \(\log a + \log b = \log ab\)     (A1)

eg\(\,\,\,\,\,\)\({\log _2}(2\sin x\cos x),{\text{ }}\log 2 + \log (\sin x) + \log (\cos x)\)

correct equation without logs     A1

eg\(\,\,\,\,\,\)\(2\sin x\cos x = {2^{ – 1}},{\text{ }}\sin x\cos x = \frac{1}{4},{\text{ }}\sin 2x = \frac{1}{2}\)

recognizing double-angle identity (seen anywhere)     A1

eg\(\,\,\,\,\,\)\(\log (\sin 2x),{\text{ }}2\sin x\cos x = \sin 2x,{\text{ }}\sin 2x = \frac{1}{2}\)

evaluating \({\sin ^{ – 1}}\left( {\frac{1}{2}} \right) = \frac{\pi }{6}{\text{ }}(30^\circ )\)     (A1)

correct working     A1

eg\(\,\,\,\,\,\)\(x = \frac{\pi }{{12}} + 2\pi ,{\text{ }}2x = \frac{{25\pi }}{6},{\text{ }}\frac{{29\pi }}{6},{\text{ }}750^\circ ,{\text{ }}870^\circ ,{\text{ }}x = \frac{\pi }{{12}}\)and \(x = \frac{{5\pi }}{{12}}\), one correct final answer

\(x = \frac{{25\pi }}{{12}},{\text{ }}\frac{{29\pi }}{{12}}\) (do not accept additional values)     A2     N0

[7 marks]

Question

Let \(f(x) = 15 – {x^2}\), for \(x \in \mathbb{R}\). The following diagram shows part of the graph of \(f\) and the rectangle OABC, where A is on the negative \(x\)-axis, B is on the graph of \(f\), and C is on the \(y\)-axis.

N17/5/MATME/SP1/ENG/TZ0/06

Find the \(x\)-coordinate of A that gives the maximum area of OABC.

Answer/Explanation

Markscheme

attempt to find the area of OABC     (M1)

eg\(\,\,\,\,\,\)\({\text{OA}} \times {\text{OC, }}x \times f(x),{\text{ }}f(x) \times ( – x)\)

correct expression for area in one variable     (A1)

eg\(\,\,\,\,\,\)\({\text{area}} = x(15 – {x^2}),{\text{ }}15x – {x^3},{\text{ }}{x^3} – 15x\)

valid approach to find maximum area (seen anywhere)     (M1)

eg\(\,\,\,\,\,\)\(A’(x) = 0\)

correct derivative     A1

eg\(\,\,\,\,\,\)\(15 – 3{x^2},{\text{ }}(15 – {x^2}) + x( – 2x) = 0,{\text{ }} – 15 + 3{x^2}\)

correct working     (A1)

eg\(\,\,\,\,\,\)\(15 = 3{x^2},{\text{ }}{x^2} = 5,{\text{ }}x = \sqrt 5 \)

\(x =  – \sqrt 5 {\text{ }}\left( {{\text{accept A}}\left( { – \sqrt 5 ,{\text{ }}0} \right)} \right)\)     A2     N3

[7 marks]

Question

The first two terms of an infinite geometric sequence are u1 = 18 and u2 = 12sin2 θ , where 0 < θ < 2\(\pi \) , and θ ≠ \(\pi \).

Find an expression for r in terms of θ.

[2]
a.i.

Find the possible values of r.

[3]
a.ii.

Show that the sum of the infinite sequence is \(\frac{{54}}{{2 + {\text{cos}}\,\left( {2\theta } \right)}}\).

[4]
b.

Find the values of θ which give the greatest value of the sum.

[6]
c.
Answer/Explanation

Markscheme

valid approach     (M1)

eg   \(\frac{{{u_2}}}{{{u_1}}},\,\,\frac{{{u_1}}}{{{u_2}}}\)

\(r = \frac{{12\,{{\sin }^2}\,\theta }}{{18}}\left( { = \frac{{2\,{{\sin }^2}\,\theta }}{3}} \right)\)      A1 N2

[2 marks]

a.i.

recognizing that sinθ is bounded      (M1)

eg    0 sin2 θ ≤ 1, −1 ≤ sinθ ≤ 1, −1 < sinθ < 1

0 < r ≤ \(\frac{2}{3}\)      A2 N3

Note: If working shown, award M1A1 for correct values with incorrect inequality sign(s).
If no working shown, award N1 for correct values with incorrect inequality sign(s).

[3 marks]

a.ii.

correct substitution into formula for infinite sum       A1

eg  \(\frac{{18}}{{1 – \frac{{2\,{\text{si}}{{\text{n}}^2}\,\theta }}{3}}}\)

evidence of choosing an appropriate rule for cos 2θ (seen anywhere)         (M1)

eg   cos 2θ = 1 − 2 sin2 θ

correct substitution of identity/working (seen anywhere)      (A1)

eg   \(\frac{{18}}{{1 – \frac{2}{3}\left( {\frac{{1 – {\text{cos}}\,2\theta }}{2}} \right)}},\,\,\frac{{54}}{{3 – 2\left( {\frac{{1 – {\text{cos}}\,2\theta }}{2}} \right)}},\,\,\frac{{18}}{{\frac{{3 – 2\,{\text{si}}{{\text{n}}^2}\,\theta }}{3}}}\)

correct working that clearly leads to the given answer       A1

eg  \(\frac{{18 \times 3}}{{2 + \left( {1 – 2\,{\text{si}}{{\text{n}}^2}\,\theta } \right)}},\,\,\frac{{54}}{{3 – \left( {1 – {\text{cos}}\,2\theta } \right)}}\)

\(\frac{{54}}{{2 + {\text{cos}}\left( {2\theta } \right)}}\)    AG N0

[4 marks]

b.

METHOD 1 (using differentiation)

recognizing \(\frac{{{\text{d}}{S_\infty }}}{{{\text{d}}\theta }} = 0\) (seen anywhere)       (M1)

finding any correct expression for \(\frac{{{\text{d}}{S_\infty }}}{{{\text{d}}\theta }}\)       (A1)

eg  \(\frac{{0 – 54 \times \left( { – 2\,{\text{sin}}\,2\,\theta } \right)}}{{{{\left( {2 + {\text{cos}}\,2\,\theta } \right)}^2}}},\,\, – 54{\left( {2 + {\text{cos}}\,2\,\theta } \right)^{ – 2}}\,\left( { – 2\,{\text{sin}}\,2\,\theta } \right)\)

correct working       (A1)

eg  sin 2θ = 0

any correct value for sin−1(0) (seen anywhere)       (A1)

eg  0, \(\pi \), … , sketch of sine curve with x-intercept(s) marked both correct values for 2θ (ignore additional values)      (A1)

2θ = \(\pi \), 3\(\pi \) (accept values in degrees)

both correct answers \(\theta  = \frac{\pi }{2},\,\frac{{3\pi }}{2}\)      A1 N4

Note: Award A0 if either or both correct answers are given in degrees.
Award A0 if additional values are given.

METHOD 2 (using denominator)

recognizing when S is greatest      (M1)

eg 2 + cos 2θ is a minimum, 1−r is smallest
correct working      (A1)

eg  minimum value of 2 + cos 2θ is 1, minimum r = \(\frac{2}{3}\)

correct working      (A1)

eg  \({\text{cos}}\,2\,\theta  =  – 1,\,\,\frac{2}{3}\,{\text{si}}{{\text{n}}^2}\,\theta  = \frac{2}{3},\,\,{\text{si}}{{\text{n}}^2}\theta  = 1\)

EITHER (using cos 2θ)

any correct value for cos−1(−1) (seen anywhere)      (A1)

eg  \(\pi \), 3\(\pi \), … (accept values in degrees), sketch of cosine curve with x-intercept(s) marked

both correct values for 2θ  (ignore additional values)      (A1)

2θ = \(\pi \), 3\(\pi \) (accept values in degrees)

OR (using sinθ)

sinθ = ±1     (A1)

sin−1(1) = \(\frac{\pi }{2}\) (accept values in degrees) (seen anywhere)      A1

THEN

both correct answers \(\theta  = \frac{\pi }{2},\,\frac{{3\pi }}{2}\)       A1 N4

Note: Award A0 if either or both correct answers are given in degrees.
Award A0 if additional values are given.

[6 marks]

c.

Question

The expression \(6\sin x\cos x\) can be expressed in the form \(a\sin bx\) .

Find the value of a and of b .

[3]
a.

Hence or otherwise, solve the equation \(6\sin x\cos x = \frac{3}{2}\) , for \(\frac{\pi }{4} \le x \le \frac{\pi }{2}\) .

[4]
b.
Answer/Explanation

Markscheme

recognizing double angle     M1

e.g. \(3 \times 2\sin x\cos x\) , \(3\sin 2x\)

\(a = 3\) , \(b = 2\)     A1A1     N3

[3 marks]

a.

substitution \(3\sin 2x = \frac{3}{2}\)     M1

\(\sin 2x = \frac{1}{2}\)     A1

finding the angle     A1

e.g. \(\frac{\pi }{6}\) , \(2x = \frac{{5\pi }}{6}\)

\(x = \frac{{5\pi }}{{12}}\)     A1     N2

Note: Award A0 if other values are included.

[4 marks]

b.

Leave a Reply