Maths

Initial mean of 25 observations is 40.
X? = (Σx? )/25 = 40 => Σx? = 25 * 40 = 1000.

A teacher of age 60 retires.
The new sum of ages for the remaining 24 people is 1000 - 60 = 940.

A new teacher of age x joins.
The new sum for 25 people is 940 + x.
The new mean is 39.
(940 + x) / 25 = 39
940 + x = 39 * 25 = 975
x = 975 - 940 = 35.
The new teacher's age is 35.

Circle 1: x² + y² - 10x - 10y + 41 = 0
Center C? = (5,5).
Radius r? = √ (5² + 5² - 41) = √ (25+25-41) = √9 = 3.

Circle 2: x² + y² - 22x - 10y + 137 = 0
Center C? = (11,5).
Radius r? = √ (11² + 5² - 137) = √ (121+25-137) = √9 = 3.

Distance between centers d (C? , C? ) = √ (11-5)² + (5-5)²) = √ (6²) = 6.

Sum of radii r? + r? = 3 + 3 = 6.
Since the distance between the centers is equal to the sum of their radii, the circles touch externally at one point.

Given the equation y = 3 + 1/ (4 + 1/y).
y - 3 = 1 / (4y+1)/y)
y - 3 = y / (4y+1)
(y-3) (4y+1) = y
4y² + y - 12y - 3 = y
4y² - 11y - 3 = y
4y² - 12y - 3 = 0

Using the quadratic formula to solve for y:
y = [-b ± √ (b²-4ac)] / 2a
y = [12 ± √ (-12)² - 4*4* (-3)] / (2*4)
y = [12 ± √ (144 + 48)] / 8
y = [12 ± √192] / 8 = [12 ± 8√3] / 8 = 3/2 ± √3.
y = 1.5 ± √3.

Since y > 0 (from the structure of the equation), both solutions are positive. The solution selects y = 1.5 + √3.

A function f (x) is continuous at x=1, so lim (x→1? ) f (x) = lim (x→1? ) f (x) = f (1).
Assuming a piecewise function like f (x) = { -x, x<1; ax+b, x1 } (structure inferred from derivative).
Continuity at x=1: f (1) = 1. a (1)+b = 1 => a+b=1.

The function is differentiable at x=1. The derivative of f (x) at x=1 from the left is -1. The derivative from the right is a.
So, a = -1. (The image has 2a = -1, which would imply a function like -x and ax²+b). Let's assume f' (x) = 2a for x>1.
2a = -1 => a = -1/2.

From a+b=1, b = 1 - a = 1 - (-1/2) = 3/2.
So, a = -1/2 and b = 3/2.

Evaluate the limit:
L = lim (x→0) [sin? ¹ (x) - tan? ¹ (x)] / 3x³

Using Taylor series expansions around x=0:
sin? ¹ (x) = x + x³/6 + O (x? )
tan? ¹ (x) = x - x³/3 + O (x? )

L = lim (x→0) [ (x + x³/6) - (x - x³/3) ] / 3x³
L = lim (x→0) [ x³/6 + x³/3 ] / 3x³
L = lim (x→0) [ (1/6 + 1/3)x³ ] / 3x³
L = (1/2) / 3 = 1/6

The solution shows 3L = 1/2, which is correct. And 6L = 1, also correct.
The final line 6L+1=2 implies 6L=1, confirming the result.

Evaluate the integral:
∫ (2x-1)cos (√ (4x²-4x+6) / √ (4x²-4x+6) dx
∫ (2x-1)cos (√ (2x-1)²+5) / √ (2x-1)²+5) dx

Let (2x-1)² + 5 = t².
Differentiating both sides:
2 (2x-1)*2 dx = 2t dt
2 (2x-1) dx = t dt
(2x-1) dx = (t/2) dt

Substitute into the integral:
∫ cos (t)/t * (t/2) dt
= 1/2 ∫ cos (t) dt
= 1/2 sin (t) + C
= 1/2 sin (√ (2x-1)²+5) + C
= 1/2 sin (√ (4x²-4x+6) + C

Given the determinant:
| α β γ |
| β γ α | = 0
| γ α β |

The expansion of this determinant is - (α³ + β³ + γ³ - 3αβγ) = 0.
This implies (α+β+γ) (α²+β²+γ²-αβ-βγ-γα) = 0.

From a cubic equation x³ + ax² + bx + c = 0 with roots α, β, γ:
α+β+γ = -a
αβ+βγ+γα = b
αβγ = -c

Substituting into the determinant condition:
(-a) ( (α+β+γ)² - 3 (αβ+βγ+γα) ) = 0
(-a) ( (-a)² - 3b ) = 0
-a (a² - 3b) = 0
a (a² - 3b) = 0

This implies a=0 or a²=3b. If a≠0, then a²=3b, so a²/b = 3.

Given the family of parabolas y² = 4a (x+a).
Differentiate with respect to x:
2y (dy/dx) = 4a
a = (y/2) (dy/dx)

Substitute a back into the original equation:
y² = 4 * (y/2) (dy/dx) * [x + (y/2) (dy/dx)]
y² = 2y (dy/dx) * [x + (y/2) (dy/dx)]
y = 2 (dy/dx) * [x + (y/2) (dy/dx)]
y = 2x (dy/dx) + y (dy/dx)²
y (dy/dx)² + 2x (dy/dx) - y = 0

Given equations:
3x + 4y = 9
y = mx + 1 (assuming this from the substitution)

Substitute y into the first equation:
3x + 4 (mx + 1) = 9
3x + 4mx + 4 = 9
x (3 + 4m) = 5
x = 5 / (3 + 4m)

For x to be an integer, (3 + 4m) must be a divisor of 5, i.e., ±1, ±5.

• 3 + 4m = 1 => 4m = -2 => m = -1/2
• 3 + 4m = -1 => 4m = -4 => m = -1 (Integer value)
• 3 + 4m = 5 => 4m = 2 => m = 1/2
• 3 + 4m = -5 => 4m = -8 => m = -2 (Integer value)

The two integral values for m are -1 and -2.