Maths

Solve sin? ¹ (3x/5) + sin? ¹ (4x/5) = sin? ¹x.
Using the formula sin? ¹a + sin? ¹b = sin? ¹ (a√ (1-b²) + b√ (1-a²):
sin? ¹ ( (3x/5)√ (1 - (4x/5)²) + (4x/5)√ (1 - (3x/5)²) ) = sin? ¹x
(3x/5) * √ (1 - 16x²/25) + (4x/5) * √ (1 - 9x²/25) = x
x * [ (3/5) * √ (25-16x²)/5 + (4/5) * √ (25-9x²)/5 - 1 ] = 0
So, x = 0 is one solution.
For the other part:
3√ (25-16x²) + 4√ (25-9x²) = 25
Let's check integer solutions. If x = 1:
3√ (9) + 4√ (16) = 33 + 44 = 9 + 16 = 25. So x = 1 is a solution.
If x = -1:
3√ (9) + 4√ (16) = 25. So x = -1 is a solution.
The solutions are x = 0, 1, -1.

Kindly go through the solution

Kindly go through the solution 

The shortest distance D between two skew lines is given by the formula:
D = | (a? - a? ) ⋅ (b? x b? )| / |b? x b? |
Line L? : (x-1)/2 = (y-2)/3 = (z-4)/4
Line L? : (x-2)/3 = (y-4)/4 = (z-5)/5

Here, a? = I + 2j + 4k, b? = 2i + 3j + 4k
a? = 2i + 4j + 5k, b? = 3i + 4j + 5k

a? - a? = I + 2j + k
b? x b? = | I j k |
| 2 3 4 |
| 3 4 5 |
= I (15-16) - j (10-12) + k (8-9) = -i + 2j - k

D = | (i + 2j + k) ⋅ (-i + 2j - k)| / √ (-1)² + 2² + (-1)²)
= |-1 + 4 - 1| / √ (1 + 4 + 1)
= 2 / √6

The equation of the plane is given as x + y + z = 42. It is also mentioned that x³ + y³ + z³ = 3xyz.
From the identity, if x³ + y³ + z³ - 3xyz = 0, then x + y + z = 0 or x = y = z.
Given the expression:
3 + (x³ + y³ + z³ - 3xyz) / (xyz)²

Since x³ + y³ + z³ = 3xyz, the expression simplifies to:
3 + 0 = 3

The problem is to evaluate the integral:
I = ∫? ¹? [x] * e^ [x] / e^ (x-1) dx, where [x] denotes the greatest integer function.

The solution breaks the integral into a sum of integrals over unit intervals:
I = ∑? ∫? ¹ n * e? / e^ (x-1) dx
= ∑? n * e? ∫? ¹ e^ (1-x) dx
= ∑? n * e? [-e^ (1-x)] from n to n+1
= ∑? n * e? [-e? - (-e¹? )]
= ∑? n * e? (e¹? - e? )
= ∑? n * e? * e? (e - 1)
= (e - 1) ∑? n
= (e - 1) * (0 + 1 + 2 + . + 9)
= (e - 1) * (9 * 10 / 2)
= 45 (e - 1)

P (x) = f (x³) + xg (x³) is divisible by x²+x+1. The roots of x²+x+1=0 are the complex cube roots of unity, ω and ω².
P (ω) = f (ω³) + ωg (ω³) = f (1) + ωg (1) = 0 — (I)
P (ω²) = f (ω²)³) + ω²g (ω²)³) = f (1) + ω²g (1) = 0 — (II)
Subtracting (II) from (I): (ω - ω²)g (1) = 0. Since ω ≠ ω², we must have g (1) = 0.
Substituting g (1)=0 into (I) gives f (1) = 0.
We need to find P (1) = f (1³) + 1*g (1³) = f (1) + g (1).
P (1) = 0 + 0 = 0.

We want to evaluate S = ∑ (r=1 to 10) r! (r³ + 6r² + 2r + 5).
We can rewrite the polynomial r³ + 6r² + 2r + 5 as (r³+6r²+11r+6) - 9r - 1.
Note that (r+1) (r+2) (r+3) = r³+6r²+11r+6.
So the term is r! [ (r+1) (r+2) (r+3) - 9r - 1] = (r+3)! - (9r+1)r!
Rewrite 9r+1 as 9 (r+1) - 8.
The term is (r+3)! - [9 (r+1)-8]r! = (r+3)! - 9 (r+1)! + 8r!
Let T? = (r+3)! - 9 (r+1)! + 8r! This does not form a simple telescoping series.
Following the OCR's final calculation, the sum simplifies to 13! + 12! - 8 (11!).
= 11! (13*12 + 12 - 8) = 11! (156 + 4) = 160 (11!).