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 = 16/3Basic idea is max occurs when there is a symmetry.

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.

A polygon has internal angles of measures 90° and 270° only. If it has18 angles of measure 270°, then what is the number of angles with measure 90°?
Sol:Let there be n angles of 90°, thn sum of all angles = (90n) + (18 × 270) = (n + 16) 180
How RHS= (n+16) 180?
Could anyone explain the part.?

Sum of the interior angles of a polygon is: (n  2) × 180° where n is the number of sides of the polygon.
now n  sided polygon has n interior angles. here we have 18 angles that measure to 270 and say y angles that measures to 90.
So n = number of sides = number of interior angles = 18 + y
sum of all angles = (n  2) × 180° = (18 + y  2) * 180° = (y + 16) * 180°Detailed solution :
Let there be x angles of 90°
Also let there be y angles of 90°
so we can say n = x + y
Sum of interior angles = (x + y  2) × 180°
we know, sum of interior angles = x * 90° + y * 270°
x * 90° + y * 270° = (x + y  2) × 180°
we know, y = 18
so x * 90° + 18 * 270° = (x + 16) × 180°
x * 90° = 18 * 270°  16 * 180°
x = (18 * 270°  16 * 180°)/90°
= 54  32 = 22
So there are 22 interior angles with measure 90°
what is the OA ?

Thanks a lot:)

What is the greatest common divisor of the 2010 digit and 2005 digit numbers below?
333.........333(2010 times)
777........777(2005 times)Let A = 333..33 and B = 777..77, then A/3 and B/7 both contains the digit 1 only (2010 times and 2005
times respectively). As HCF(2010, 2005) = 5 that means HCF(A/3, B/7) = 11111.
Now A is also divisible by 7 as 2010 = 6 × 335 (Remember a number formed by repeating same digit 6 times is divisible by 7). So HCF(A, B) = 77777Here, As HCF(2010, 2005) = 5 that means HCF(A/3, B/7) = 11111. ???
Could anyone explain the solution?

In ∆ABC (not drawn to scale), the altitude from A, the angle bisector of ∠BAC, and the
median from A to the midpoint of BC divide ∠BAC into four equal angles. What is the measure in degrees of angle ∠BAC?Here the solution says apply sine rule
What I could get is AB sin3A/4 = AC sin A/4
But nt able to establish relation between AB and AC. How to find the angle BAC then?

@sumitagarwal
A = 3333 ...... (2010 times)
B = 7777 ...... (2005 times)A = 3 * 1111 ..... (2010 times)
B = 7 * 1111 ..... (2005 times)HCF(11111.... (m times), 1111 .... (n times)) = 1111.. (HCF(m,n) times)
A is divisible by 7. B is not divisible by 3
HCF(A, B) = 11111 * 7 = 77777

30,000 is divided into three equal installments.The remaining sum at rate 4 % is added to the intallment amounts.Find the value of each installment.

ABCD is a square with side length 2 cm. It is divided into five rectangles of equal areas, as shown in the figure. The perimeter of the rectangle BEFG is
(a) 51/16
(b) 36/11
(c) 58/15
(d) 47/13

Can anyone explain please.

Side length = 2 cm.
So area of square = 4 cm^2
Given square is divided into 5 rectangles of equal areas. So each rectangle has an area of 4/5 cm^2.
Proceed with this understanding (i.e, if we know one side of rectangle = 2 then other side should be 2/5 to get an area of 4/5) and we get EB = 4/3 cm and BG = 3/5 cm. Figure as belowPerimeter = 58/15 cm (does it match with the OA?)


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@zabeer
Sir one more