Maths

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.

f (x) + g (x) = √x + √1-x. The domain requires x ≥ 0 and 1-x ≥ 0, so x ≤ 1. Domain is [0,1].

f (x) - g (x) = √x - √1-x. Domain is [0,1].

f (x)/g (x) = √x / √1-x. Requires x ≥ 0 and 1-x > 0, so x < 1. Domain is [0,1).

g (x)/f (x) = √1-x / √x. Requires 1-x ≥ 0 and x > 0. Domain is (0,1].

The common domain for all these functional forms to be considered is (0,1).

A line passes through (1,3). Its equation is y - 3 = m (x - 1) or y = mx + (3-m).
The angle θ between this line and the line y = 3√2x - 1 (with slope m? = 3√2) is given by tanθ = √2.

tanθ = | (m - m? )/ (1 + m*m? )|
√2 = | (m - 3√2) / (1 + m*3√2)|

This gives two cases:

Case 1 (+ve):
√2 = (m - 3√2) / (1 + 3√2m)
√2 (1 + 3√2m) = m - 3√2
√2 + 6m = m - 3√2
5m = -4√2
m = -4√2 / 5

Case 2 (-ve):
-√2 = (m - 3√2) / (1 + 3√2m)
-√2 (1 + 3√2m) = m - 3√2
-√2 - 6m = m - 3√2
7m = 2√2
m = 2√2 / 7

Consider the series:
1/ (3²-1) + 1/ (5²-1) + 1/ (7²-1) + . + 1/ (201)²-1)

The general term T? can be written for the r-th term starting with r=1 for 3, r=2 for 5.
T? = 1/ (2r+1)² - 1) = 1/ (2r+1-1) (2r+1+1) = 1/ (2r * (2r+2) = 1/4 * 1/ (r (r+1)
T? = 1/4 * (1/r - 1/ (r+1)

The sum of the first n terms is:
S? = Σ T? = 1/4 * Σ (1/r - 1/ (r+1) from r=1 to n
S? = 1/4 * [ (1 - 1/2) + (1/2 - 1/3) + . + (1/n - 1/ (n+1) ]
S? = 1/4 * (1 - 1/ (n+1)

The last term is (201)²-1, so 2r+1 = 201, which gives r = 100. So, n=100.
S? = 1/4 * (1 - 1/101) = 1/4 * (100/101) = 25/101.

We need to evaluate the sum:
Σ (100 - r) (100 + r) for r from 0 to 99.

Σ (100² - r²) for r from 0 to 99
= Σ 100² - Σ r² for r from 0 to 99
= 100 * 100² - [99 (99+1) (2*99+1)]/6
= 100³ - [99 * 100 * 199]/6
= 100³ - (1650 * 199)

Comparing this with (100)³ – 199β, we get:
β = 1650
If we consider a comparison form α = 3β, this part of the solution seems to have a typo, but based on the final calculation, Slope = α/β = 1650 / 3 = 550 seems to be intended.