Maths

P will be the centroid of triangle ABC.
The centroid P is (x? +x? +x? )/3, (y? +y? +y? )/3).
The coordinates of P are given as (17/6, 8/3).
The coordinates of Q are not given, but a calculation is shown.
PQ = √ (24/6)² + (9/3)²) = √ (4² + 3²) = √ (16+9) = √25 = 5.
This implies the coordinates of Q are such that the difference in coordinates with P leads to this result. For example if P= (x? , y? ) and Q= (x? , y? ), then x? -x? =4 and y? -y? =3.

Var (1, 2, ., n) = (Σn²/n) - (Σn/n)² = 10.
(n (n+1) (2n+1)/6n) - (n (n+1)/2n)² = 10.
(n+1) (2n+1)/6 - (n+1)/2)² = 10.
(n+1)/12 * [2 (2n+1) - 3 (n+1)] = 10.
(n+1)/12 * (4n+2 - 3n-3) = 10.
(n+1) (n-1)/12 = 10.
n² - 1 = 120 ⇒ n² = 121 ⇒ n = 11.

Var (2, 4, ., 2m) = Var (2* (1, 2, ., m) = 2² * Var (1, 2, ., m) = 16.
4 * Var (1, 2, ., m) = 16.
Var (1, 2, ., m) = 4.
Using the formula from above: (m²-1)/12 = 4.
m² - 1 = 48 ⇒ m² = 49 ⇒ m = 7.
m + n = 7 + 11 = 18.

f (x) + f (x + 1) = 2 (1)
replace x with x + 1: f (x + 1) + f (x + 2) = 2 (2)
(2) - (1) ⇒ f (x + 2) = f (x)
∴ f (x) is periodic with period 2.
I? = ∫? f (x)dx = 4 ∫? ² f (x)dx.
I? = ∫? ³ f (x)dx = ∫? f (u-1)du. Let u = x+1.
I? = ∫? f (x-1)dx = 2 ∫? ² f (x-1)dx.
From (1), f (x-1) + f (x) = 2.
I? + 2I? = 4∫? ² f (x)dx + 2 (2∫? ² f (x-1)dx) = 4∫? ² f (x)dx + 4∫? ² (2 - f (x)dx
= 4∫? ² (f (x) + 2 - f (x)dx = 4∫? ² 2 dx = 4 [2x] from 0 to 2 = 16.

y (x) = ∫? (2t² - 15t + 10)dt
dy/dx = 2x² - 15x + 10.
For tangent at (a, b), slope is m = dx/dy = 1 / (dy/dx) = 1 / (2a² - 15a + 10).
Given slope is -1/3.
2a² - 15a + 10 = -3
2a² - 15a + 13 = 0 (The provided solution has 2a²-15a+7=0, suggesting a different problem or a typo)
Following the image: 2a² - 15a + 7 = 0
(2a - 1) (a - 7) = 0
a = 1/2 or a = 7.
a = 1/2 Rejected as a > 1. So a = 7.
b = ∫? (2t² - 15t + 10)dt = [2t³/3 - 15t²/2 + 10t] from 0 to 7.
6b = [4t³ - 45t² + 60t] from 0 to 7 = 4 (7)³ - 45 (7)² + 60 (7) = 1372 - 2205 + 420 = -413.
|a + 6b| = |7 - 413| = |-406| = 406.

The function f (x) is non-differentiable at x=1, 3, 5.
Σ f (f (x) = f (f (1) + f (f (3) + f (f (5).
Assuming f (x) is defined such that f (1)=1, f (3)=1, f (5)=1 (based on context of absolute value functions).
Then Σ f (f (x) = f (1) + f (1) + f (1) = 1 + 1 + 1 = 3.

Two drawn cards are spades. There are 50 cards left.
The missing card could be a spade or not a spade.
P (missing card is spade) = 11/50 (since 11 spades remain out of 50 cards).
P (missing card is not spade) = 1 - 11/50 = 39/50

Digits are 1, 3, 5, 7, 9. We need to form a 6-digit number where exactly one digit is repeated.

Choose the digit to be repeated:? C? ways.

Choose the positions for these two repeated digits:? C? ways.

Arrange the remaining 4 distinct digits in the remaining 4 places:? P? = 4! ways.
Total numbers =? C? *? C? * 4! = 5 * 15 * 24 = 1800.
The solution in the image 5/2 (6!) seems to follow a different logic which is unclear. 5 * (6!/2) = 5 * 360 = 1800. This logic is: choose one of 5 digits to repeat. Arrange the 6 digits, and since two are identical, divide by 2!