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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)^(m-1)(px + a)^n - np(px + q)^(n-1)(ax + b)^m

Fox max of f(x), f'(x) = 0

ma(ax + b)^(m-1)(px + q)^n + np(px + q)^(n-1)(ax + b)^m = 0

ma(ax + b)^(m-1)(px + q)^n = - np(px + q)^(n-1)(ax + b)^m

ma(ax + b)^(m-1)/(ax+b)^m = -np(px + q)^(n-1)/(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 = 8

max of (x-6)^2 * (11-x)^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 = 16/3

Basic idea is max occurs when there is a symmetry.

]]>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 three-square 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.

]]>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.

]]>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 = 1

Now 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 = 0

Now 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.

]]>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:)

]]>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) = 77777

Here, As HCF(2010, 2005) = 5 that means HCF(A/3, B/7) = 11111. ???

Could anyone explain the solution?

]]>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?

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

(a) 51/16

(b) 36/11

(c) 58/15

(d) 47/13

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 below

Perimeter = 58/15 cm (does it match with the OA?)

]]>]]>

Sir one more

]]>

58/15.

One side comes out to be 4/3 and another one comes out to be 3/5

One way could be

You take two triplets:-

9,12,15

5,12,13

a=12, b=5, c=9

Area comes out to be 84...by heron's.

If u put these values in options...only third option would give 84 as answer.

Tan(A + B) = (Tan A + Tan B)/(1 - TanA * TanB)

Tan70 = (Tan50 + Tan20)/(1 - Tan50 * Tan20)

Tan70 - Tan70 * Tan50 * Tan20 = Tan50 + Tan20

We know, Tan70 * Tan20 = 1

So, Tan70 - Tan50 = Tan50 + Tan20

Tan70 = Tan20 + 2Tan50. ]]>

a) Twice

b) Thrice

c) Once

d) 5 times

In every meet their combined distance travelled is 100 m.

Let there be "a" meets.

=>[(100a) * 5/(5+3)]=300

Which gives a=4.8

Though it cant be decimal.

But Where am I proceeding wrong with this approach????could anyone help me?

I guess the problem lies when they might be travelling in same direction ..

]]>Sol:--What i tried is as follows:-

A and C

They meet first in L/2(30)L hr

So 3rd time they meet in L/60L+L/30L+L/30L= 5/60 hr

A and B will meet for the third time = 3L/60L hour

B and C :

they meet first at L/2(40-10)=L/60L hour

Third time they meet in:

L/60+ 2L/30=5L/60L

They meet together third time in LCM(5/60,3/60,5/60)

=15/60 hour=15 min

But options are: 5 ,7, 11and 13....

]]>Assumption that they travel 100 m together between every meetings is not correct. Like if A is 10 m from one of the end while B crossed him, then the next meeting would happen once B comes back from that end. They won't cover 100 m together in this scenario right? ]]>