Test of Mathematics Solution Subjective 127 -Graphing relations

Test of Mathematics at the 10+2 Level

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.

Screen Shot 2015-11-19 at 8.45.54 PM

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:

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.

Screen Shot 2015-11-19 at 9.13.56 PM

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

Test of Mathematics Solution Subjective 115 - Trigonometric Relation

Test of Mathematics at the 10+2 Level

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:

Test of Mathematics Solution Subjective 107 - Perpendiculars from Center

Test of Mathematics at the 10+2 Level

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

Check the Answer


Answer: is 159.

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]

\(\Rightarrow x=159\).

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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.

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

Key Concepts


Integers

Triangle

Trigonometry

Check the Answer


Answer: is 616.

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.

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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

Check the Answer


Answer: is 4

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.

Trigonometry problem graph - ISI MStat 2019 PSA Problem 15
Fig:1

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

Outstanding Statistics Program with Applications

Outstanding Statistics Program with Applications

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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

Check the Answer


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 2016 PSA Problem 9
Outstanding Statistics Program with Applications

Outstanding Statistics Program with Applications

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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

Check the Answer


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 \( 0< x < \frac{\pi}{2 } \) .

ISI MStat 2019 PSA Problem 12
Outstanding Statistics Program with Applications

Outstanding Statistics Program with Applications

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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

Check the Answer


Answer: is 777.

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

Parallelogram Problem

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.

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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

Check the Answer


Answer: is 60.

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=\(\frac{60\pi}{3}\) \(\Rightarrow n=60\).

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