Question: State Cauchy's criterion for the existence of limit of a function.
Solution:
A function f(x) defined in the neighborhood of 'a' has a finite limit as x approaches 'a' if and only if for every ε > 0, there exists a δ > 0 such that for all x1, x2 in the deleted neighborhood 0 < |x - a| < δ, the condition |f(x1) - f(x2)| < ε is satisfied.
Question: Define removable discontinuity with an example.
Solution:
A function f(x) has a removable discontinuity at x = a if the limit as x approaches 'a' exists, but is not equal to the value of the function f(a), or if f(a) is undefined.
Example: Consider f(x) = (x² - 4) / (x - 2). At x = 2, the function is undefined (0/0), but the limit exists (equals 4). By defining f(2) = 4, the discontinuity is removed.
Question: If y = sin(m sin⁻¹x), show that (1-x²)y_n+2 - (2n+1)xy_n+1 - (n²-m²)y_n = 0. Find y_n(0).
Solution:
Given y = sin(m sin⁻¹x)
. Differentiating w.r.t x:y₁ = cos(m sin⁻¹x) * m / √(1-x²)
√(1-x²) y₁ = m cos(m sin⁻¹x). Squaring both sides:
(1-x²) y₁² = m² cos²(m sin⁻¹x) = m² [1 - sin²(m sin⁻¹x)] = m²(1 - y²)
Differentiating again: (1-x²) 2y₁y₂ - 2xy₁² = -2m²yy₁
Dividing by 2y₁: (1-x²) y₂ - xy₁ + m²y = 0.
Applying Leibniz's Theorem for n-th differentiation:
[(1-x²) y_n+2 + n(-2x)y_n+1 + n(n-1)/2 * (-2)y_n] - [x y_n+1 + n(1)y_n] + m²y_n = 0
(1-x²) y_n+2 - 2nxy_n+1 - n(n-1)y_n - xy_n+1 - ny_n + m²y_n = 0
(1-x²) y_n+2 - (2n+1)xy_n+1 - (n² - m²)y_n = 0
Question: State Lagrange's mean value theorem.
Solution:
If a function f(x) is continuous in the closed interval [a, b] and differentiable in the open interval (a, b), then there exists at least one point 'c' in (a, b) such that:
f'(c) = [f(b) - f(a)] / (b - a)
Question: Evaluate lim (x → 0) (x - sin x) / x³.
Solution:
This is a 0/0 form. Applying L'Hôpital's Rule:
= lim (x → 0) (1 - cos x) / 3x² [Still 0/0 form]
= lim (x → 0) (sin x) / 6x [Still 0/0 form]
= lim (x → 0) (cos x) / 6
As x → 0, cos x → 1. Hence, the limit is 1/6.
Question: Define homogeneous function of two variables x and y. Find degree of f(x,y) = tan⁻¹(y/x) + sin⁻¹(x/y).
Solution:
A function f(x, y) is homogeneous of degree 'n' if f(tx, ty) = tⁿ f(x, y) for any positive real t.
For the given function: f(tx, ty) = tan⁻¹(ty/tx) + sin⁻¹(tx/ty) = tan⁻¹(y/x) + sin⁻¹(x/y) = t⁰ f(x, y).
The degree of the function is 0.
Question: Prove that points of y² = 4a{x + a sin(x/a)} with tangents parallel to x-axis lie on a parabola.
Solution:
For tangents to be parallel to the x-axis, dy/dx = 0.
Differentiating y² = 4a{x + a sin(x/a)} w.r.t x:
2y (dy/dx) = 4a [1 + a * cos(x/a) * (1/a)] = 4a [1 + cos(x/a)]
Setting dy/dx = 0, we get 1 + cos(x/a) = 0 => cos(x/a) = -1.
Then sin(x/a) = √(1 - cos²(x/a)) = 0.
Substitute sin(x/a) = 0 into the original equation: y² = 4a{x + a(0)} = 4ax.
The locus is the parabola y² = 4ax.
Question: Evaluate ∫ (0 to π/2) cos⁸x dx.
Solution:
Using Wallis Formula for even powers: [(n-1)/n] * [(n-3)/(n-2)] * ... * (1/2) * (π/2)
= (7/8) * (5/6) * (3/4) * (1/2) * (π/2)
= 105π / 768 = 35π / 256.
Question: Evaluate ∫ (0 to π) x / (a² cos²x + b² sin²x) dx.
Solution:
Let I = ∫ (0 to π) x dx / (a² cos²x + b² sin²x). Using ∫ (0 to a) f(x) dx = ∫ (0 to a) f(a-x) dx:
I = ∫ (0 to π) (π-x) dx / (a² cos²(π-x) + b² sin²(π-x)) = ∫ (0 to π) (π-x) dx / (a² cos²x + b² sin²x)
2I = π ∫ (0 to π) dx / (a² cos²x + b² sin²x) = 2π ∫ (0 to π/2) dx / (a² cos²x + b² sin²x)
I = π ∫ (0 to π/2) sec²x dx / (a² + b² tan²x). Let b tan x = t, then b sec²x dx = dt.
I = (π/b) ∫ (0 to ∞) dt / (a² + t²) = (π/b) [ (1/a) tan⁻¹(t/a) ] (0 to ∞) = (π/ab) [π/2 - 0]
I = π² / (2ab).
Question: Find the area of the region bounded by parabolas x² = 4y and y² = 4x.
Solution:
Points of intersection: (x²/4)² = 4x => x⁴/16 = 4x => x⁴ = 64x => x(x³ - 64) = 0.
x = 0 and x = 4. When x = 0, y = 0. When x = 4, y = 4.
Area = ∫ (0 to 4) [√(4x) - (x²/4)] dx = ∫ (0 to 4) [2√x - x²/4] dx
= [2 * (x^1.5 / 1.5) - (x³ / 12)] (0 to 4)
= [ (4/3) * (8) - (64/12) ] = [32/3 - 16/3] = 16/3 sq. units.
Question: Find the total length of the cardioid r = a(1 - cos θ).
Solution:
Length L = ∫ (0 to 2π) √[r² + (dr/dθ)²] dθ.
dr/dθ = a sin θ.
r² + (dr/dθ)² = a²(1 - cos θ)² + a² sin² θ = a²(1 - 2 cos θ + cos² θ + sin² θ) = 2a²(1 - cos θ) = 4a² sin²(θ/2).
L = ∫ (0 to 2π) 2a sin(θ/2) dθ = 2 * ∫ (0 to π) 2a sin(θ/2) dθ [Due to symmetry]
L = 4a [ -2 cos(θ/2) ] (0 to π) = 4a [ -2(0) - (-2(1)) ] = 8a.
Total length = 8a.