Discussion Room : Quant

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Question asked by @deepalis727
Find the maximum value of (x  6)^2 (11  x)^3 for 6 < x < 11.

If you are comfortable with derivatives then you can check Part 1. Else, jump to Part 2 :)
Part 1:
f(x) = g(x)h(x)
f'(x) = g'(x)h(x) + h'(x)g(x) where f'(x) is the derivative of f(x).
So if f(x) = (ax + b)^m * (px + q)^n then
f'(x) = ma(ax + b)^(m1)(px + a)^n  np(px + q)^(n1)(ax + b)^m
Fox max of f(x), f'(x) = 0
ma(ax + b)^(m1)(px + q)^n + np(px + q)^(n1)(ax + b)^m = 0
ma(ax + b)^(m1)(px + q)^n =  np(px + q)^(n1)(ax + b)^m
ma(ax + b)^(m1)/(ax+b)^m = np(px + q)^(n1)/(px + q)^n
ma/(ax + b) = np/(px+q)Part 2:
If f(x) = (ax + b)^m * (px + q)^n then max/min occurs at values of x for which (ax + b)/ma =  (px + q)/np
So here, f(x) = (x  6)^2 (11  x)^3
max occurs at values of x such that (x  6)/(2 * 1) = (11  x)/(3 * 1)
3x  18 = 22  2x
5x = 40
x = 8max of (x6)^2 * (11x)^3 happens at x = 8 and max value = 2^2 * 3^3 = 108
So, if we want to find for what of x for maximizing f(x) = (4x  3)^5 * (18  2x)^4
We need to find x such that (4x  3)/20 = (18  2x)/8
32x  24 = 360  40x
72x = 384
x = 384/72 = 16/3
So max for f(x) occurs at x = 16/3solution from @DeekondaSaikrishna
max occurs when there is a symmetry
so (x6)/2 = (11x)/3
so at x =8
therefore max value =2^2 * 3^3

CIRCULAR MOTION
Concept 1 : Circular Motions In Common Admission Test (CAT) Is Mostly Based On Two Or Three Objects Moving Around A Circular Track. In Such Cases The Relative Speed Becomes Effective Speed Of The Bodies. Let Us Assume The Three Objects Are Moving With A Speed Of A, B & C On A Track Of Length L. (A > B > C)
 If Two Bodies Are Moving In The Same Direction, Their Relative Speed Is (AB).
 If Two Bodies Are Moving In The Opposite Direction, Their Relative Speed Is (A+B)
■Number Of Distinct Points At Which Body Meets On A Track  If They Are Moving In Same Direction, They Will Meet At (AB) Distinct Points, Where A And B Are Coprimes
 If They Are Moving In Opposite Directions, They Will Meet At (A+B) Distinct Points, Where A And B Are Coprimes
 All These Points Will Be Equidistant From Each Other.
●Note : If They Aren't Coprimes Then Cancel The Common Part , Make Them Coprime (A:B) And Then Use The Above Points.
Ex 1 : If Two Bodies Are Moving In The Opposite Direction At Speed 7 And 19, Then They Will Meet At (7+19) = 26 Distinct Points On The Track.
Ex 2 : If Two Bodies Are Moving In The Same Direction At Speed 6 And 15, (6:15 :: 2 : 5)Then They Will Meet At (52) = 3 Distinct Points On The Track.
■Concept 1 Post 2 : Number Of Distinct Points 3 Bodies Will Meet On A Circular Track .
●Step 1: Find Out Pairwise Distinct Meeting Points.
●Step 2 : The Overall Answer Will Be The Highest Common Factor Of The Pairwise Values.
■Example: Suppose A, B And C Are Running With Speeds Of 7, 9 And 13 On A Circular Track. A And B Are Running Clockwise, Whereas C Is Running AntiClockwise.
●A And B Will Have 9  7 = 2 Distinct Meeting Points.
●A And C Will Have 7 + 13 = 20 Distinct Meeting Points.
●B And C Will Have 9 + 13 = 22 Distinct Meeting Points.
■A, B And C Will Have HCF (2, 20, 22) = 2 Distinct Meeting Points.
They Will Meet At 2 Distinct Points On The Track■Concept Post 2 : Time When Two Objects Meets For The First Time When They Are Running Around A Circular Track In The Same Direction.
●Let Us Assume There Are Two Persons (A) And (B) .
Speed Of A = 20 M/S.
Speed Of B = 10 M/S.
Length Of Track = 1000 M.
●Try To Visualise, They Will Meet The First Time When A Has Covered 1000m Distance More Than B. Hence The Relative Distance Is 1000.
●Now Let's Assume A And B To Be One Single Body, And Hence Everything Considered Will Be Relative,
Relative Distance = 1000
Relative Speed = 2010 = 10
●Time Taken = Relative Distance/Relative Speed = 1000/10 = 100 S
Hence They Will Meet For The First Time After 100 Seconds.■Concept 3 : Time When Three Objects Meets For The First Time When They Are Running Around A Circular Track In The Same Direction.
●Now, Let Us Assume There Are 3 Persons A, B And C Running In The Same Directions On 1000 M Track .
Speed Of A = 30 M/S
Speed Of B = 20 M/S
Speed Of C = 10 M/S
●Again The Concept Of Relativity, The Time When They Will Meet For The First Time = Time Taken By The Fastest Runner To Take One Round Over Other Runners.
●Time Taken By A To Take One Round Over B , 1000/(3020) = 100 S
●Time Taken By A To Take One Round Over C, 1000/(3010) = 50 S
●Now The Time After Which They All Three Will Meet = LCM(100,50) = 100.
Hence All Three Persons Will Meet After 100 Seconds.■Concept 4 : Time When They Will Meet For The First Time At The Starting Point.
●Time When They Will Meet First Time At The Starting Point = LCM(Time Taken By Each Individual To Take One Round) .
●Let's Assume There Are Three Persons, A, B And C With Respective Speed Of 30, 20 And 10 M/S
●Length Of Track = 1000 M .
Time Taken By A = 1000/30
Time Taken By B = 1000/20
Time Taken By C = 1000/10
●LCM(100/3, 50, 100) = LCM(100,50,100)/HCF(3,1,1) = 100/1 = 100 S.
●Hence They All Will Meet First Time At The Starting Point After 100 S.■Concept 6 : Application Of AMGM In TSD
●Using AM & HM In Proportionality Relations,Quite Often In Questions We Find That The Given Speeds (Or Time Taken) Are In An Arithmetic Progression. And If Distance Covered At The Speeds Is Constant, Then Time Taken (Or Speeds) Will Be Inversely Proportional I.E. They Will Be In Harmonic Progression.
●For Those Who Have Forgotten, The Arithmetic Mean Of A And B Is(A + B)/2 And The Harmonic Mean Of A And B Is 2ab/(A+B)
Though It Requires A Little Trained Eyes To Identify The Above, It Will Be Useful If You Keep A Watch For It. See The Following Data To Realise That Either Time Taken Or Speeds Are In An Arithmetic Progression.
E.G.: 1. If I Travel At 15 Kmph, I Reach Office At 10 Am,If I Travel At 10 Kmph, I Reach Office At 10:30 Am. At What Speed Should I Travel So That I Reach Office At 10:15. Assume I Leave Home At Same Time And Take The Same Route. ( Use Am/Hm  No Other Method)
Solution . Leaving At Same Time And Reaching At 10 Am, 10:15 Am And 10:30 Am Suggests That The Time Travelled Are In AP. Thus, Speeds Are In HP And Required Speed Is The
HM Of 10 & 15
I.E.2 *10 *15/25 = 12 Kmphcredit  iquanta

TRIANGLE
A triangle is a polygon with three edges and three vertices. It is one of the basic shapes in geometry. A triangle with vertices A, B, and C is denoted.
Area: ½ × base × height
Perimeter: sum of side lengths of the triangle
Number of vertices: 3
Number of edges: 3
Internal angle: 60° (for equilateral)
Sum of interior angles: 180°
Line of symmetry: 3TYPES OF TRIANGLE
Three types of triangles:
Acute (all angles less than 90°),
Right Angle (one angle is 90°),
Obtuse (one angle is more than 90°).IMPORTANT  Angle opposite to the larger side is always greater than angle opposite to smaller side.
Sum of two sides is greater than third.
So, in a ∆ABC
AB + BC > AC,
AB + AC > BC
And AC + BC > AB~ For acute angle triangle:  AB² + BC² > AC², AB² + AC² > AB² and BC² + AC² > AB²
For right angle triangle:  AB² + BC² = AC², where AC is hypotenuse
For obtuse angle triangle:  AB² + BC² < AC², where AC is the largest side~ Sine Rule : if a, b,c are the sides of a triangle with opposite angles A,B,C then
a/SinA = b/SinB = c/SinC = 2R. ( R = circum radius )~Cosine Rule
The Cosine Rule can be used in any triangle where you are trying to relate all three sides to one angle.
Finding Sides
If you need to find the length of a side, you need to know the other two sides and the opposite angle.
formula:
a^2 = b^2 + c^2 – 2bc cos(A)
Side a is the one you are trying to find. Sides b and c are the other two sides, and angle A is the angle opposite side a .
In case of right triangle A=90 and Cos90=0
So a^2=b^2+c^2 which is pythagoras theorem~SIMILARITY OF TRIANGLES
AAA (angle angle angle)
All three pairs of corresponding angles are the same.
SSS in same proportion (side side side)
All three pairs of corresponding sides are in the same proportion
SAS (side angle side)
Two pairs of sides in the same proportion and the included angle equal.AREA OF TRIANGLE
Area = 1/2 * bh
Area = 1/2* ab SinC = 1/2* ac SinB = 1/2* bc SinA
Area = r*s ( r= inradius and s= semi perimeter ) semi perimeter = sum of sides /2
Area = abc/4R ( R= circum radius )RIGHT ANGLE TRIANGLE , ISOSCELES TRIANGLE , EQUILATERAL TRIANGLE
IMP The incenter and circumcentre lies at a point that divides height in the ratio 2 : 1. ( incenter = inradius center , circumcentre = circum radius center )
~ N sided POLYGON
Sum of interior angles of nsided polygon= (n2) x 180°
each angle = (n2) x 180°/n ( n= number of sides )
Sum of exterior angle of any polygon = 360°.
Number of diagonals in any n sided Polygon = n(n3)/2.CENTROID
The centroid of a triangle is the intersection of the three medians of the triangle (each median connecting a vertex with the midpoint of the opposite side). It lies on the triangle's Euler line, which also goes through various other key points including the orthocenter and the circumcenter.
The centroid divides each median in the ratio 2:1INCENTRE AND CIRCUMCENTRE
The incenter of a triangle is the center of its inscribed circle. It has several important properties and relations with other parts of the triangle, including its circumcenter, orthocenter, area, and more.Centre of the circle circumscribed about a triangle . Radius of which is called circumradius.
ORTHOCENTER
The orthocenter is the point where the three altitudes of a triangle intersect. A altitude is a perpendicular from a vertex to its opposite side.IMPORTANT  PTOLEMY’S THEOREM
AC * BD=AB * CD + BC * AD

How many integral triplets (x, y, z) satisfy the equation x^2 + y^2 + z^2 = 1855 ?
solution: Note that 1855 = 7 mod 8 while all perfect squares are 0, 1 or 4 mode8. So it is impossible for 3 squares to sum up to 7 mod8. So no solutions are there.
Could anyone please explain how we choose number to take mod with, on both the sides.Here for example, we took mod 8..why not, mod 4? mod 4 would have given 3 a.d 3 mod 4 is possible .So , how do we arrive at a number which poses contradiction.
How we chose 8 here?

Legendre's threesquare theorem states that a natural number (say n) can be represented as the sum of three squares of integers if and only if n is not of the form n = 4^a(8b + 7) for integers a and b.

Suppose real numbers x and y satisfy x^2 + 9y^2  4x + 6y + 4 = 0.what is the maximum value of 4x  9y?
Geometric approach
We know that (x  2)^2 + (3y + 1)^2 = 1 and we need to maximise 4x  9y.
Now, let's replace 3y by Y, so that know equation turns into a circle
(x  2)^2 + (Y + 1)^2 = 1 with center at (2, 1) and radius 1.
And now we need to find the maximum value of 4x  3Y which will be equal to 4x  3Y = k (say) and will be achieved when this line is tangential to the circle.Doubt  what is the logic behind the point that " for the value of expression to be maximum , that line should be tangential to the circle " ..
Could anyone help me with understanding this concept please.

@sumitagarwal
We are trying to maximize the distance between the line and the centre of the circle. It is also known that the circle lies on the circle so it has to be a tangent.Sharing a solution from @kamal_lohia sir
x^2 + 9y^2  4x + 6y + 4 = 0
i.e. {x^2  2(x)^(2) + 2^2} + {(3y)^2 + 2(3y)(1) + 1} = 1
i.e. (x  2)^2 + (3y + 1)^2 = 1Now we want to maximise 4x  9y. Let 4x  9y = k, i.e. x = (k + 9y)/4
So eliminating x, above equation becomes: (k + 9y  8)^2 + 16(3y + 1)^2 = 16
i.e. 225y^2 + {18(k  8) + 96}y + (k  8)^2 = 0
i.e. 225y^2 + (18k  48)y + (k  8)^2 = 0Now as y is a real number, so discriminant of the above equation should be greater than or equal to zero.
So, (18k  48)^2  4(225)(k  8)^2 ≥ 0
i.e. (18k  48)^2  (30k  240)^2 ≥ 0
i.e. (3k  8)^2  (5k  40)^2 ≥ 0
i.e. (2k + 32)(8k  48) ≥ 0
i.e. (k  6)(k  16) ≤ 0
i.e. 6 ≤ k ≤ 16.Thus required maximum value of 'k' is 16.

P and Q are two diametrically opposite points on a ciccular track. A and B start from P at speed of 20 and 40 rounds per hr respectively. C starts from Q at the same time at 10 rounds per hour. If B and C move in clockwise direction and A in anticlockwise direction ,find the time after which they meet for the third time anywhere on the track?
Could anyone help me with the solution of this question?