ALGEBRA QUESTIONS WITH SOLUTIONS

1. (a) Polynomial Remainder Problem

When P(x) is divided by (x-2) the remainder is 2 and when divided by (x-3) the remainder is 5. Find the remainder when divided by (x-2)(x-3).

Solution:

Given:
P(2) = 2
P(3) = 5

When dividing by quadratic (x-2)(x-3), remainder will be linear: R(x) = ax + b

Thus:
P(2) = 2a + b = 2
P(3) = 3a + b = 5

Subtract first equation from second: a = 3

Substitute back: 2(3) + b = 2 ⇒ b = -4

Remainder is R(x) = 3x - 4

1. (b) Prove d/dx(xⁿ) = nxⁿ⁻¹ by induction

Proof by Mathematical Induction:

Base Case (n=1):
d/dx(x¹) = 1 = 1x⁰ ✓

Inductive Hypothesis: Assume true for n=k:
d/dx(xᵏ) = kxᵏ⁻¹

Inductive Step (n=k+1):
d/dx(xᵏ⁺¹) = d/dx(x·xᵏ) = xᵏ + x·kxᵏ⁻¹ (Product Rule)
= xᵏ + kxᵏ = (k+1)xᵏ = (k+1)x⁽ᵏ⁺¹⁾⁻¹ ✓

Thus by induction, the formula holds for all positive integers n.

2. (b) Matrix Induction Proof

Prove [1 -1; 0 1]ⁿ = [1 -n; 0 1] for n ∈ ℤ⁺ using induction

Proof by Induction:

Base Case (n=1):
[1 -1; 0 1]¹ = [1 -1; 0 1] ✓

Inductive Hypothesis: Assume true for n=k:
[1 -1; 0 1]ᵏ = [1 -k; 0 1]

Inductive Step (n=k+1):
[1 -1; 0 1]ᵏ⁺¹ = [1 -1; 0 1]ᵏ × [1 -1; 0 1]
= [1 -k; 0 1] × [1 -1; 0 1]
= [1·1+(-k)·0 1·(-1)+(-k)·1; 0·1+1·0 0·(-1)+1·1]
= [1 -(k+1); 0 1] ✓

Thus by induction, the formula holds for all positive integers n.

3. (a) Sum of Series

Evaluate ∑[r=1→∞] 2/[r(r+1)(r+2)]

Solution:

Use partial fractions decomposition:

2/[r(r+1)(r+2)] = A/r + B/(r+1) + C/(r+2)

Solving gives: A = 1, B = -2, C = 1

Thus the sum becomes:

∑[1/r - 2/(r+1) + 1/(r+2)]

Write out terms (telescoping series):

(1 - 1 + 1/2) + (1/2 - 2/3 + 1/3) + (1/3 - 2/4 + 1/4) + ...

Most terms cancel, leaving:

Sum = 1/2

4. (a) Exponential Equation

If aˣ = kʸ = (a/k)ᵐ where a≠1, show that y = mx/(m-x)

Proof:

Given three equal expressions:
aˣ = kʸ = (a/k)ᵐ = aᵐk⁻ᵐ

From aˣ = aᵐk⁻ᵐ ⇒ k⁻ᵐ = aˣ⁻ᵐ

From kʸ = aᵐk⁻ᵐ ⇒ kʸ⁺ᵐ = aᵐ

Substitute k from first result (k = a⁽ᵐ⁻ˣ⁾/ᵐ):

[a⁽ᵐ⁻ˣ⁾/ᵐ]ʸ⁺ᵐ = aᵐ

Simplify exponents:

a⁽ᵐ⁻ˣ⁾⁽ʸ⁺ᵐ⁾/ᵐ = aᵐ

Since bases equal, exponents equal:

(m-x)(y+m)/m = m

Solve for y:

(m-x)(y+m) = m²

y+m = m²/(m-x)

y = m²/(m-x) - m = [m² - m(m-x)]/(m-x) = mx/(m-x)

5. (a) Binomial Approximation

If p is very small such that its fourth and higher powers are neglected, show that:

(1+p)¹ᐟ⁴ + (1-p)¹ᐟ⁴ ≈ 2 + (5p²)/16

Proof:

Using binomial expansion (1±p)ⁿ ≈ 1 ± np + n(n-1)p²/2 for small p:

(1+p)¹ᐟ⁴ ≈ 1 + (1/4)p + (1/4)(-3/4)p²/2 = 1 + p/4 - 3p²/32

(1-p)¹ᐟ⁴ ≈ 1 - (1/4)p + (1/4)(-3/4)p²/2 = 1 - p/4 - 3p²/32

Add them together:

(1+p)¹ᐟ⁴ + (1-p)¹ᐟ⁴ ≈ (1+1) + (1/4-1/4)p + (-3/32-3/32)p²

≈ 2 - (6/32)p² = 2 - (3/16)p²

For p=1/16:

≈ 2 - 3/(16×256) ≈ 1.99927

6. (a) Sum of Series

Find the sum of 5/(1×2×3) + 8/(2×3×4) + 11/(3×4×5) + ...

Solution:

General term: aₙ = (3n+2)/[n(n+1)(n+2)]

Use partial fractions decomposition:

(3n+2)/[n(n+1)(n+2)] = A/n + B/(n+1) + C/(n+2)

Solving gives: A = 1, B = 1, C = -2

Thus the sum becomes:

∑[1/n + 1/(n+1) - 2/(n+2)]

Write out terms (telescoping series):

(1 + 1/2 - 2/3) + (1/2 + 1/3 - 2/4) + (1/3 + 1/4 - 2/5) + ...

Most terms cancel, leaving:

Sum = 1 + 1/2 + 1/2 = 2

7. (a) Inequality Solution

Solve (x-1)(x+2)/[(x+1)(x-3)] < 0

Solution:

Critical points: x = -2, -1, 1, 3

Test intervals:

Interval	(x-1)	(x+2)	(x+1)	(x-3)	Overall
x < -2	-	-	-	-	+ / + = +
-2 < x < -1	-	+	-	-	- / + = -
-1 < x < 1	-	+	+	-	- / - = +
1 < x < 3	+	+	+	-	+ / - = -
x > 3	+	+	+	+	+ / + = +

Solution where expression < 0:

x ∈ (-2, -1) ∪ (1, 3)

Note: x ≠ -1, 3 (undefined)

Key Algebra Concepts Covered:

• Polynomial division and remainder theorem
• Mathematical induction proofs
• Matrix operations and properties
• Series summation and telescoping series
• Exponential and logarithmic equations
• Binomial approximations
• Rational inequalities