Test of Mathematics Solution Subjective 127 -Graphing relations

This is a Test of Mathematics Solution Subjective 127 (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.

Problem

Find all (x, y) such that sin x + sin y = sin (x+y) and |x| + |y| = 1

Discussion

|x| + |y| =1 is easier to plot. We have to treat the cases separately.

• First quadrant: x +y = 1
• Second quadrant: -x + y = 1 (since |x| = -x when x is negative)
• Fourth Quadrant: x - y =1

Now we work on sin x + sin y = sin (x + y).

This implies $\displaystyle{2 \sin \left ( \frac{x+y} {2} \right ) \cos \left ( \frac{x-y} {2} \right ) = 2 \sin \left ( \frac{x+y} {2} \right ) \cos \left ( \frac{x+y} {2} \right ) }$. Hence we have two possibilities:

• $\displaystyle{\sin \left ( \frac{x+y} {2} \right ) = 0 }$ OR
• $\displaystyle{\cos \left ( \frac{x+y} {2} \right ) - \cos \left ( \frac{x-y} {2} \right ) = 0 }$ or $\displaystyle { \sin \frac{x}{2} \sin \frac{y}{2} } = 0$

The above situations can happen when when

$\displaystyle{ \frac{x +y}{2} = k \pi }$ or $\displaystyle{\frac {x}{2} = k \pi }$ or $\displaystyle{ \frac{y}{2} = k \pi }$, where k is any integer.

Thus we need to plot the class of lines $\displaystyle{ x + y = 2 k \pi }$, $\displaystyle{ x = 2k\pi }$ and $\displaystyle{ y = 2k\pi }$, and consider the intersection points of these lines with the graph of |x| + |y| = 1.

Clearly only for k=0, such intersection points can be found.

Hence required points are (0,1), (0,-1), (1,0), (-1,0), (1/2, -1/2), (-1/2, 1/2).

Chatuspathi

• What is this topic: Graphing
• What are some of the associated concept: Trigonometric Identities
• Where can learn these topics: I.S.I. & C.M.I. Entrance Course of Cheenta, discusses these topics in the ‘Calculus’ module.
• Book Suggestions: Play With Graphs (Arihant Publication)

Test of Mathematics Solution Subjective 115 - Trigonometric Relation

This is a Test of Mathematics Solution Subjective 115 (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.

Problem

If $\displaystyle { \frac{\sin^4 x }{a} + \frac{\cos^4 x }{b} = \frac{1}{a+b} }$ , then show that ${\frac{1}{(a+b)^2} }$

Solution

Put $\sin^2 x = t$.

The given expression reduces to $\displaystyle { \frac{t^2 }{a} + \frac{1+t^2 -2t }{b} = \frac{1}{a+b} }$

$\displaystyle {\Rightarrow (a+b)t^2 -2at +a - \frac{ab}{a+b} = 0 }$
$\displaystyle {\Rightarrow (a+b)^2t^2 -2a(a+b)t +a^2 = 0 }$
$\displaystyle {\Rightarrow ((a+b)t-a)^2=0 }$
$\displaystyle {\Rightarrow \frac{a}{a+b}=t }$
$\displaystyle {\Rightarrow \frac{b}{a+b}=1-t }$

Hence replacing in the required expression we get

$\displaystyle { \frac{\sin^6 x }{a^2} + \frac{\cos^6 x }{b^2} }$
$\displaystyle {=\frac{t^3 }{a^2} + \frac{(1-t)^3 }{b^2}}$
$\displaystyle {=\frac{a^3 }{(a+b)^3a^2} + \frac{b^3 }{(a+b)^3b^2}}$
$\displaystyle {=\frac{a}{(a+b)^3} + \frac{b}{(a+b)^3} }$
$\displaystyle {=\frac{a+b}{(a+b)^3} }$
$\displaystyle {=\frac{1}{(a+b)^2} }$

(Proved)

Chatuspathi:

• What is this topic: Trigonometry
• What are some of the associated concept: Change of Variables
• Where can learn these topics: Cheenta I.S.I. & C.M.I. course, discusses these topics in the ‘Trigonometry’ module.
• Book Suggestions: Trigonometry Volume I by S.L. Loney

Test of Mathematics Solution Subjective 107 - Perpendiculars from Center

This is a Test of Mathematics Solution Subjective 107 (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.

Also visit: I.S.I. & C.M.I. Entrance Course of Cheenta

Problem:

If a, b and c are the lengths of the sides of a triangle ABC and  if $$p_1 , p_2$$ and $$p_3$$   are the lengths of the perpendiculars drawn from the circumcentre onto the sides BC, CA and AB respectively, then show that

${\displaystyle{\frac{a}{p_1}} + {\frac{b}{p_2}} + {\frac{c}{p_3}} = {\frac{abc}{4{p_1}{p_2}{p_3}}}}$.

Solution

:
Let  O be the circum centre

${\displaystyle{\frac{a}{p_1}}}$ = 2 ${\displaystyle \left({\frac{\frac{a}{2}}{p_1}}\right) = 2 \tan \left({\frac{1}{2}}{\angle{BOC}}\right)}$

Similarly, ${\displaystyle{\frac{b}{p_2}} = 2 \tan \left({\frac{1}{2}}{\angle{AOC}}\right)}$

& ${\displaystyle{\frac{c}{p_3}} = 2 \tan \left({\frac{1}{2}}{\angle{AOB}}\right)}$
Now, ${\displaystyle{\frac{1}{2}}{\angle{BOC}} + {\frac{1}{2}}{\angle{AOC}} + {\frac{1}{2}}{\angle{AOB}}}$ = ${\displaystyle{\frac{1}{2}} ({\angle{BOC}} + {\angle{AOC}} + {\angle{AOB}})}$
= ${\displaystyle{\frac{1}{2}} (360^{\circ})}$ = ${\displaystyle{180^{\circ}}}$

Lemma
For ${\displaystyle{\alpha},{\beta},{\gamma} > 0}$ & ${\displaystyle{\alpha} + {\beta} + {\gamma} = {\pi}}$
${\displaystyle{\tan{\alpha}} + {\tan{\beta}} + {\tan{\gamma}} = {\tan{\alpha}}{\times}{\tan{\beta}}{\times}{\tan{\gamma}}}$
${\displaystyle{\alpha} + {\beta} + {\gamma} = {\pi}}$
${\Rightarrow}$ ${\displaystyle{\alpha + \beta} = {\pi - \gamma}}$
${\Rightarrow}$ ${\displaystyle{\tan (\alpha + \beta)} = {\tan (\pi - \gamma)}}$
${\Rightarrow}$ ${\displaystyle{\frac{\tan{\alpha} + \tan{\beta}}{1 - \tan{\alpha} \tan{\beta}}} = -\tan{\gamma}}$
${\Rightarrow}$ ${\displaystyle{\tan\alpha + \tan \beta} = {\tan{\alpha}. \tan{\beta} .\tan{\gamma}} - \tan\gamma}$
${\Rightarrow}$ ${\displaystyle{\tan{\alpha}} + {\tan{\beta}} + {\tan{\gamma}} = {\tan{\alpha}}{\times}{\tan{\beta}}{\times}{\tan{\gamma}}}$ (lemma proved)

So, ${\displaystyle{\frac{1}{2}}\left({\frac{a}{p_1}} + {\frac{b}{p_2}} + {\frac{c}{p_3}}\right) = \tan({\frac{1}{2}}\angle{AOB}) + \tan({\frac{1}{2}}\angle{BOC}) + \tan({\frac{1}{2}}\angle{COA})}$
= ${\displaystyle{\tan({\frac{1}{2}}\angle{AOB})\times \tan({\frac{1}{2}}\angle{BOC})\times \tan({\frac{1}{2}}\angle{COA})}}$ [ as ${\displaystyle{\angle{AOB} + \angle{BOC} + \angle{COA} = {360}^{\circ}}}$ ]
= ${\displaystyle{\frac{a}{2{p_1}}}\times{\frac{b}{2{p_2}}}\times{\frac{c}{2{p_3}}}}$
= ${\displaystyle{\frac{abc}{8{p_1}{p_2}{p_3}}}}$
$\Rightarrow \frac{a}{p_1} + \frac{b}{p_2} + \frac{c}{p_3} = \frac{abc}{4{p_1}{p_2}{p_3} }$ ( proved )

Finding smallest positive Integer | AIME I, 1996 Problem 10

Try this beautiful problem from the American Invitational Mathematics Examination, AIME I, 1996 based on Finding the smallest positive Integer.

Finding smallest positive Integer - AIME I, 1996

Find the smallest positive integer solution to $$tan19x=\frac{cos96+sin96}{cos96-sin96}$$.

• is 107
• is 159
• is 840
• cannot be determined from the given information

Key Concepts

Functions

Trigonometry

Integers

AIME I, 1996, Question 10

Plane Trigonometry by Loney

Try with Hints

First hint

$$\frac{cos96+sin96}{cos96-sin96}$$

=$$\frac{sin(90+96)+sin96}{sin(90+96)-sin96}$$

=$$\frac{sin186+sin96}{sin186-sin96}$$

=$$\frac{sin(141+45)+sin(141-45)}{sin(141+45)-sin(141-45)}$$

=$$\frac{2sin141cos45}{2cos141sin45}$$

=tan141

Second Hint

here $$tan(180+\theta)$$=$$tan\theta$$

$$\Rightarrow 19x=141+180n$$ for some integer n is first equation

Final Step

multiplying equation with 19 gives

$$x \equiv 141\times 19 \equiv 2679 \equiv 159(mod180)$$ [since 2679 divided by 180 gives remainder 159]

$$\Rightarrow x=159$$.

Triangle and integers | AIME I, 1995 | Question 9

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1995 based on Triangle and integers.

Triangle and integers - AIME I, 1995

Triangle ABC is isosceles, with AB=AC and altitude AM=11, suppose that there is a point D on AM with AD=10 and $$\angle BDC$$=3$$\angle BAC$$. then the perimeter of $$\Delta ABC$$ may be written in the form $$a+\sqrt{b}$$ where a and b are integers, find a+b.

• is 107
• is 616
• is 840
• cannot be determined from the given information

Key Concepts

Integers

Triangle

Trigonometry

AIME I, 1995, Question 9

Plane Trigonometry by Loney

Try with Hints

First hint

Let x= $$\angle CAM$$

$$\Rightarrow \angle CDM =3x$$

$$\Rightarrow \frac{tan3x}{tanx}=\frac{\frac{CM}{1}}{\frac{CM}{11}}$$=11 [by trigonometry ratio property in right angled triangle]

Second Hint

$$\Rightarrow \frac{3tanx-tan^{3}x}{1-3tan^{2}x}=11tanx$$

solving we get, tanx=$$\frac{1}{2}$$

$$\Rightarrow CM=\frac{11}{2}$$

Final Step

$$\Rightarrow 2(AC+CM)$$ where $$AC=\frac{11\sqrt {5}}{2}$$ by Pythagoras formula

=$$\sqrt{605}+11$$ then a+b=605+11=616.

ISI MStat 2019 PSA Problem 15 | Trigonometry Problem

This is the problem from ISI MStat 2019 PSA Problem 15. First, try it yourself and then go through the sequential hints we provide.

Trigonometry - ISI MStat Year 2019 PSA Question 15

How many solutions does the equation $$cos ^{2} x+3 \sin x \cos x+1=0$$ have for $$x \in[0,2 \pi)$$ ?

• 1
• 3
• 4
• 2

Key Concepts

Trigonometry

Factorization

ISI MStat 2019 PSA Problem 15

Precollege Mathematics

Try with Hints

Factorize and Solve.

$$\cos ^{2} x+3 \sin x \cos x + 1 = (2\cos x + \sin x)(\cos x +\sin x) = 0$$.
$$tanx = -2, tanx = -1$$.
Draw the graph.

So, if you see the figure you will find there are 4 such x for $$x \in[0,2 \pi)$$.

ISI MStat 2016 PSA Problem 9 | Equation of a circle

This is a problem from ISI MStat 2016 PSA Problem 9 based on equation of a circle. First, try the problem yourself, then go through the sequential hints we provide.

Equation of a circle- ISI MStat Year 2016 PSA Question 9

Given $$\theta$$ in the range $$0 \leq \theta<\pi,$$ the equation $$2 x^{2}+2 y^{2}+4 x \cos \theta+8 y \sin \theta+5=0$$ represents a circle for all $$\theta$$ in the interval

• $$0 < \theta <\frac{\pi}{3}$$
• $$\frac{\pi}{4} < \theta <\frac{3\pi}{4}$$
• $$0 < \theta <\frac{\pi}{2}$$
• $$0 \le \theta <\frac{\pi}{2}$$

Key Concepts

Equation of a circle

Trigonometry

Basic Inequality

Answer: is $$\frac{\pi}{4} < \theta <\frac{3\pi}{4}$$

ISI MStat 2016 PSA Problem 9

Precollege Mathematics

Try with Hints

Complete the Square.

We get ,

$$2{(x+\cos \theta)}^2 + 2{(y+ 2\sin \theta)}^2 = (6{\sin \theta}^2-3))$$
$$6{\sin \theta}^2-3 > 0 \Rightarrow {\sin^2 \theta} \geq \frac{1}{2}$$

We are given that $$0 \leq \theta<\pi,$$ . So, $${\sin^2 \theta} \geq \frac{1}{2}$$ $$\Rightarrow \frac{\pi}{4} < \theta <\frac{3\pi}{4}$$.

ISI MStat 2019 PSA Problem 12 | Domain of a function

This is a beautiful problem from ISI MStat 2019 PSA problem 12 based on finding the domain of the function. We provide sequential hints so that you can try.

Domain of a function- ISI MStat Year 2019 PSA Question 12

What is the set of numbers $$x$$ in $$(0,2 \pi)$$ such that $$\log \log (\sin x+\cos x)$$ is well-defined?

• $$[\frac{\pi}{8},\frac{3 \pi}{8}]$$
• $$(0,\frac{\pi}{2})$$
• $$(0,\frac{ \pi}{4}]$$
• $$(0,\pi) \cup (\frac{3 \pi}{2}, 2 \pi)$$

Key Concepts

Domain

Basic inequality

Trigonometry

Answer: is $$(0,\frac{\pi}{2})$$

ISI MStat 2019 PSA Problem 12

Pre-college Mathematics

Try with Hints

$$logx$$ is defined for $$x \in (0,\infty)$$.

$$sinx+cosx > 0$$.
$$log(sinx+cosx) > 0 \Rightarrow sinx + cosx > 1$$
$$sin(x+\frac{\pi}{4}) > \frac{1}{\sqrt{2}}$$
For $$y$$ in $$(0,2 \pi)$$ , $$siny > \frac{1}{\sqrt{2}} \iff \frac{\pi}{4} < y < \frac{3\pi}{4 }$$

Hence we have $$0< x < \frac{\pi}{2 }$$ .

Parallelogram Problem | AIME I, 1996 | Question 15

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1996 based on Parallelogram.

Parallelogram Problem - AIME I, 1996

In parallelogram ABCD , Let O be the intersection of diagonals AC and BD, angles CAB and DBC are each twice as large as angle DBA and angle ACB is r times as large as angle AOB. Find the greatest integer that does not exceed 1000r.

• is 107
• is 777
• is 840
• cannot be determined from the given information

Key Concepts

Integers

Trigonometry

Algebra

AIME I, 1996, Question 15

Geometry Vol I to IV by Hall and Stevens

Try with Hints

First hint

Let $$\theta= \angle DBA$$

$$\angle CAB=\angle DBC=2 \theta$$

or, $$\angle AOB=180-3\theta, \angle ACB=180-5\theta$$

or, since ABCD parallelogram, OA=OC

Second Hint

by sine law on $$\Delta$$ABO, $$\Delta$$BCO

$$\frac{sin\angle CBO}{OC}$$=$$\frac{sin\angle ACB}{OB}$$

and $$\frac{sin\angle DBA}{OC}=\frac{sin\angle BAC}{OB}$$

here we divide and get $$\frac{sin2\theta}{sin\theta}$$=$$\frac{sin(180-5\theta)}{sin 2\theta}$$

$$\Rightarrow sin^{2}{2\theta}=sin{5\theta}sin{\theta}$$

Final Step

$$\Rightarrow 1-cos^{2}2\theta=\frac{cos4\theta-cos6\theta}{2}$$

or, $$4 cos^{3}2\theta-4cos^{2}2\theta -3cos2\theta+3=(4cos^{2}2\theta-3)(cos2\theta-1)=0 [using cos3\theta=4cos^{3}\theta-3cos\theta]$$

or, $$cos 2\theta=\frac{\sqrt{3}}{2}$$

or, $$\theta$$=15

$$[1000r]=[1000\frac{180-5\theta}{180-3\theta}]=[\frac{7000}{9}]$$=777.

Trigonometry & natural numbers | PRMO 2017 | Question 11

Try this beautiful problem from the Pre-RMO, 2017 based on Trigonometry & natural numbers.

Trigonometry & natural numbers - PRMO 2017

Let f(x) =$$sin\frac{x}{3}+cos\frac{3x}{10}$$ for all real x, find the least natural number x such that $$f(n\pi+x)=f(x)$$ for all real x.

• is 107
• is 60
• is 840
• cannot be determined from the given information

Key Concepts

Trigonometry

Least natural number

Functions

PRMO, 2017, Question 11

Plane Trigonometry by Loney

Try with Hints

First hint

here f(x) =$$sin\frac{x}{3}+cos\frac{3x}{10}$$

Second Hint

period of$$sin\frac{x}{3}$$ is $$6\pi$$

period of $$cos\frac{3x}{10}$$ is $$\frac{20\pi}{3}$$

Final Step

Lcm=$$\frac{60\pi}{3}$$ $$\Rightarrow n=60$$.