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FIND THE ANSWER OF THIS QUESTION MAKE A QUESTION BANK OF THIS MEANS RELEVENT QUESTION WHICH CAN COME IN MATH EXAM .IMP CHAPTER AREImportant chapters for exams Matrices and Determinants Trigonometry Differential Calculus Integral Calculus Differential Equations Laplace Transform Analytical geometry MAKE A SHORT QUESTION BANK WHICH CAN BE DONE IN A DAY SOLVE IT ALSO

I've analyzed all your exam papers from Om Sterling Global University - Remedial Mathematics (DPH-106). Below is a complete Question Bank with Full Solutions covering all important chapters - designed to be studied in one day!

📚 REMEDIAL MATHEMATICS - QUESTION BANK WITH SOLUTIONS

(Based on Your Actual Exam Papers - DPH-106)

CHAPTER 1: MATRICES AND DETERMINANTS

Q1. Define Matrix and its Order.

Answer: A Matrix is a rectangular arrangement of numbers (real or complex) arranged in rows and columns, enclosed in brackets.

Order of a matrix = Number of rows × Number of columns

Example:

A = [1  2  3]
    [4  5  6]

This matrix has 2 rows and 3 columns, so its order is 2×3.

Q2. Add a Note on Types of Matrices.

Answer:

Type	Description	Example
Row Matrix	Only 1 row	[1, 2, 3]
Column Matrix	Only 1 column	[1; 2; 3]
Square Matrix	Rows = Columns	2×2, 3×3
Zero Matrix	All elements = 0	[0,0;0,0]
Identity Matrix (I)	Diagonal = 1, rest = 0	[1,0;0,1]
Diagonal Matrix	Non-diagonal = 0	[2,0;0,5]
Symmetric Matrix	A = A^T	-
Skew-Symmetric	A = -A^T	-

Q3. State Transpose of a Matrix with Example.

Answer: The Transpose of matrix A (written as A^T) is obtained by interchanging rows and columns.

Example:

If A = [1  2  3]      Then A^T = [1  4]
       [4  5  6]                  [2  5]
                                  [3  6]

Property: (A^T)^T = A

Q4. Write a Note on Operations on Matrices with Example.

Answer: 1. Addition: Add corresponding elements. (Same order required)

[1 2] + [5 6] = [6  8]
[3 4]   [7 8]   [10 12]

2. Subtraction: Subtract corresponding elements.

3. Scalar Multiplication:

2 × [1 2] = [2 4]
    [3 4]   [6 8]

4. Matrix Multiplication: (A is m×n, B is n×p → AB is m×p)

[1 2] × [5 6] = [(1×5+2×7) (1×6+2×8)] = [19 22]
[3 4]   [7 8]   [(3×5+4×7) (3×6+4×8)]   [43 50]

Q5. Find Adj(A) and A⁻¹ for the matrix:

A = [1  1  1]
    [1  2 -3]
    [2 -1  3]

Step 1 - Find det(A):

det(A) = 1(2×3 - (-3)×(-1)) - 1(1×3 - (-3)×2) + 1(1×(-1) - 2×2)

= 1(6 - 3) - 1(3 + 6) + 1(-1 - 4)

= 1(3) - 1(9) + 1(-5)

= 3 - 9 - 5 = -11

Step 2 - Find Cofactor Matrix:

C₁₁ = +(2×3 - (-3)×(-1)) = +(6-3) = 3 C₁₂ = -(1×3 - (-3)×2) = -(3+6) = -9 C₁₃ = +(1×(-1) - 2×2) = +(-1-4) = -5

C₂₁ = -(1×3 - 1×(-1)) = -(3+1) = -4 C₂₂ = +(1×3 - 1×2) = +(3-2) = 1 C₂₃ = -(1×(-1) - 1×2) = -(-1-2) = 3

C₃₁ = +(1×(-3) - 1×2) = +(-3-2) = -5 C₃₂ = -(1×(-3) - 1×1) = -(-3-1) = 4 C₃₃ = +(1×2 - 1×1) = +(2-1) = 1

Cofactor Matrix:

C = [ 3  -9  -5]
    [-4   1   3]
    [-5   4   1]

Step 3 - Adj(A) = Transpose of Cofactor Matrix:

Adj(A) = [ 3  -4  -5]
         [-9   1   4]
         [-5   3   1]

Step 4 - A⁻¹ = Adj(A) / det(A):

A⁻¹ = (1/-11) × [ 3  -4  -5]
                  [-9   1   4]
                  [-5   3   1]

A⁻¹ = [-3/11   4/11   5/11]
       [ 9/11  -1/11  -4/11]
       [ 5/11  -3/11  -1/11]

Q6. Find Adjoint of Matrix:

A = [ 1   3   4]
    [-1   6   0]
    [ 3   4   2]

det(A) = 1(12-0) - 3(-2-0) + 4(-4-18) = 12 + 6 - 88 = -70

Cofactors: C₁₁ = +(12-0) = 12 C₁₂ = -(-2-0) = 2 C₁₃ = +(-4-18) = -22

C₂₁ = -(6-16) = 10 C₂₂ = +(2-12) = -10 C₂₃ = -(4-9) = 5

C₃₁ = +(0-24) = -24 C₃₂ = -(0+4) = -4 C₃₃ = +(6+3) = 9

Adj(A):

Adj(A) = [ 12   10  -24]
         [  2  -10   -4]
         [-22    5    9]

Q7. Solve by Cramer's Rule: 5x - 7y + z = 11, 6x - 8y - z = 15, 3x + 2y - 6z = 7

Setting up:

D = |5  -7   1|
    |6  -8  -1|
    |3   2  -6|

D = 5[(-8)(-6)-(-1)(2)] - (-7)[(6)(-6)-(-1)(3)] + 1[(6)(2)-(-8)(3)]

= 5[48+2] + 7[-36+3] + 1[12+24]

= 5(50) + 7(-33) + 1(36)

= 250 - 231 + 36 = 55

Dx = |11  -7   1|
     |15  -8  -1|
     | 7   2  -6|

Dx = 11[48+2] - (-7)[(-90+7)] + 1[30+56]

= 11(50) + 7(-83) + 86

= 550 - 581 + 86 = 55

Dy = |5  11   1|
     |6  15  -1|
     |3   7  -6|

Dy = 5[(-90+7)] - 11[(-36+3)] + 1[(42-45)]

= 5(-83) - 11(-33) + (-3)

= -415 + 363 - 3 = -55

Dz = |5  -7  11|
     |6  -8  15|
     |3   2   7|

Dz = 5(-56-30) - (-7)(42-45) + 11(12+24)

= 5(-86) + 7(-3) + 11(36)

= -430 - 21 + 396 = -55

Solutions:

x = Dx/D = 55/55 = 1
y = Dy/D = -55/55 = -1
z = Dz/D = -55/55 = -1

Q8. Solve by Matrix Method: x+y+z=3, 2x-y+z=2, x-2y+3z=2

Matrix form AX = B:

A = [1   1  1]    X = [x]    B = [3]
    [2  -1  1]        [y]        [2]
    [1  -2  3]        [z]        [2]

det(A) = 1(-3+2) - 1(6-1) + 1(-4+1) = -1 - 5 - 3 = -9 ... wait, let me recalculate:

= 1((-1)(3)-(1)(-2)) - 1((2)(3)-(1)(1)) + 1((2)(-2)-(-1)(1))

= 1(-3+2) - 1(6-1) + 1(-4+1)

= 1(-1) - 1(5) + 1(-3) = -9

Using Cramer's rule:

Dx: Replace col 1 with B:

|3  1  1|
|2 -1  1| = 3(-3+2)-1(6-2)+1(-4+2) = 3(-1)-1(4)+1(-2) = -3-4-2 = -9
|2 -2  3|

x = -9/-9 = 1

Dy: Replace col 2 with B:

|1  3  1|
|2  2  1| = 1(6-2)-3(6-1)+1(-4-2) = 4-15-6 = -17
|1  2  3|

Hmm, let me redo: = 1(2×3-1×2)-3(2×3-1×1)+1(2×2-2×1) = 1(4) - 3(5) + 1(2) = 4-15+2 = -9

y = -9/-9 = 1

Dz: Replace col 3 with B:

|1  1  3|
|2 -1  2| = 1(-2+4)-1(4-2)+3(-4+1) = 2-2-9 = -9
|1 -2  2|

z = -9/-9 = 1

Answer: x = 1, y = 1, z = 1

Q9. Resolve (x-5)/[(x-3)(x-4)] into Partial Fractions

Let: (x-5)/[(x-3)(x-4)] = A/(x-3) + B/(x-4)

Multiply both sides by (x-3)(x-4):

x - 5 = A(x-4) + B(x-3)

Put x = 3: 3-5 = A(3-4) → -2 = -A → A = 2

Put x = 4: 4-5 = B(4-3) → -1 = B → B = -1

Answer: (x-5)/[(x-3)(x-4)] = 2/(x-3) - 1/(x-4)

CHAPTER 2: ANALYTICAL GEOMETRY

Q10. Define Quadrant.

Answer: The coordinate axes divide the plane into 4 regions called Quadrants:

Quadrant	x	y	Example
I (First)	+	+	(3, 5)
II (Second)	-	+	(-2, 4)
III (Third)	-	-	(-1, -3)
IV (Fourth)	+	-	(4, -2)

Q11. Define Distance Formula.

Answer: The distance between two points P(x₁, y₁) and Q(x₂, y₂) is:

d = √[(x₂-x₁)² + (y₂-y₁)²]

This is derived from the Pythagorean theorem.

Q12. Find the Distance between (5, 4) and (-5, 6).

Solution:

d = √[(-5-5)² + (6-4)²]

= √[(-10)² + (2)²]

= √[100 + 4]

= √104

= 2√26 ≈ 10.2 units

Q13. Show the four points (0,5), (-2,-2), (5,0) and (7,7) on a graph.

Answer: Plot each point on the coordinate plane:

(0,5) - on Y-axis, 5 units up
(-2,-2) - 2nd quadrant (negative both)... wait, (-2,-2) is in 3rd quadrant
(5,0) - on X-axis, 5 units right
(7,7) - 1st quadrant

Y
8|          *(7,7)
7|
6|
5| *(0,5)
4|
3|
2|
1|
--+--+--+--+--+--+--+--  X
 -3 -2 -1  0  1  2  3  4  5  6  7
        *(5,0)
-1|
-2| *(-2,-2)

These 4 points form a quadrilateral when joined.

Q14. Find Area of Triangle with vertices (0,0), (1,0), (1,1).

Formula: Area = (1/2)|x₁(y₂-y₃) + x₂(y₃-y₁) + x₃(y₁-y₂)|

= (1/2)|0(0-1) + 1(1-0) + 1(0-0)|

= (1/2)|0 + 1 + 0|

= (1/2)|1|

= 1/2 square unit

Q15. Find the equation of a straight line passing through (-4, -2) with slope -8.

Using Point-Slope Form: y - y₁ = m(x - x₁)

y - (-2) = -8(x - (-4))

y + 2 = -8(x + 4)

y + 2 = -8x - 32

y = -8x - 34 or 8x + y + 34 = 0

CHAPTER 3: TRIGONOMETRY

Q16. Prove: cos 52° = cos 68° + cos 172° = 0

Proof: cos 68° + cos 172°

Using sum-to-product: cos A + cos B = 2cos[(A+B)/2]cos[(A-B)/2]

= 2cos[(68+172)/2]cos[(172-68)/2]

= 2cos(120°)cos(52°)

= 2(-1/2)cos(52°)

= -cos(52°)

So: cos 68° + cos 172° = -cos 52°

Therefore: cos 52° + cos 68° + cos 172° = cos 52° - cos 52° = 0 ✓

Q17. If tan A = √3 and tan B = 2-√3, find tan(A-B).

Formula: tan(A-B) = (tan A - tan B)/(1 + tan A · tan B)

tan A - tan B = √3 - (2-√3) = √3 - 2 + √3 = 2√3 - 2

tan A · tan B = √3(2-√3) = 2√3 - 3

1 + tan A · tan B = 1 + 2√3 - 3 = 2√3 - 2

tan(A-B) = (2√3-2)/(2√3-2) = 1

Therefore: A - B = 45°

CHAPTER 4: DIFFERENTIAL CALCULUS

Q18. Differentiate y = log(x) · tan(x) with respect to x.

Using Product Rule: d/dx[u·v] = u·v' + v·u'

Let u = log x → du/dx = 1/x Let v = tan x → dv/dx = sec²x

dy/dx = log(x) · sec²x + tan(x) · (1/x)

dy/dx = log(x)·sec²x + tan(x)/x

Q19. If y = x⁸ - 12x⁵ + 5x³ - 12, find d²y/dx².

First derivative: dy/dx = 8x⁷ - 60x⁴ + 15x²

Second derivative: d²y/dx² = 56x⁶ - 240x³ + 30x

Q20. Find dy/dx if y = 3x² + 7x⁷ and y = x⁴ + sin x.

Part (a): y = 3x² + 7x⁷

dy/dx = 6x + 49x⁶

Part (b): y = x⁴ + sin x

dy/dx = 4x³ + cos x

CHAPTER 5: DIFFERENTIAL EQUATIONS

Q21. Add a Note on Differential Equation.

Answer: A Differential Equation is an equation that contains derivatives of a dependent variable with respect to one or more independent variables.

Examples:

dy/dx = 2x + 3 (1st order)
d²y/dx² + 5(dy/dx) + 6y = 0 (2nd order)
2xydx + (x²+y²)dy = 0

Types:

Ordinary DE (ODE): One independent variable
Partial DE (PDE): Two or more independent variables

Q22. Define Order and Degree of a Differential Equation with 3 Examples.

Definitions:

Order = the order of the highest derivative present
Degree = the power of the highest order derivative (after clearing fractions/radicals)

Equation	Order	Degree
dy/dx = 5x + 3	1	1
(d²y/dx²)³ + dy/dx = 0	2	3
d³y/dx³ + 2(dy/dx)² = x	3	1

Q23. Solve the Differential Equation: 2xy·dx + (x² + y²)dy = 0

Rearranging: 2xy·dx = -(x²+y²)dy

dx/dy = -(x²+y²)/(2xy)

This is a homogeneous equation. Let x = vy → dx/dy = v + y·dv/dy

v + y·dv/dy = -(v²y²+y²)/(2vy²) = -(v²+1)/(2v)

y·dv/dy = -(v²+1)/(2v) - v = -(v²+1+2v²)/(2v) = -(3v²+1)/(2v)

Separating variables: 2v·dv/(3v²+1) = -dy/y

Integrating both sides: (1/3)ln(3v²+1) = -ln y + C

ln(3v²+1) = -3ln y + C

(3v²+1) = k/y³

Substituting back v = x/y: (3x²/y² + 1) = k/y³

3x² + y² = ky ← General Solution

Q24. Solve: 4d²y/dx² + 4dy/dx - 3y = e^(2x)

Auxiliary Equation: 4m² + 4m - 3 = 0

(2m+3)(2m-1) = 0

m = 1/2 or m = -3/2

Complementary Function (C.F.): y_c = C₁e^(x/2) + C₂e^(-3x/2)

Particular Integral (P.I.): For e^(2x): P.I. = e^(2x)/f(D) where D→2

= e^(2x)/(4(4)+4(2)-3) = e^(2x)/(16+8-3) = e^(2x)/21

General Solution: y = C₁e^(x/2) + C₂e^(-3x/2) + e^(2x)/21

CHAPTER 6: INTEGRAL CALCULUS

Q25. Evaluate ∫(3x² + 2x + 5)dx

= 3x³/3 + 2x²/2 + 5x + C

= x³ + x² + 5x + C

Q26. Evaluate ∫sin(x)dx and ∫cos(x)dx

∫sin(x)dx = -cos(x) + C

∫cos(x)dx = sin(x) + C

Q27. Evaluate ∫e^x dx and ∫(1/x)dx

∫e^x dx = e^x + C

∫(1/x)dx = ln|x| + C

CHAPTER 7: LAPLACE TRANSFORM

Q28. Define Laplace Transform.

Answer: The Laplace Transform of a function f(t) is defined as:

L{f(t)} = F(s) = ∫₀^∞ e^(-st) f(t) dt

where s is a complex number.

Q29. Standard Laplace Transform Formulas (Must Memorize):

f(t)	L{f(t)} = F(s)
1	1/s
t	1/s²
tⁿ	n!/s^(n+1)
e^(at)	1/(s-a)
sin(at)	a/(s²+a²)
cos(at)	s/(s²+a²)
e^(at)sin(bt)	b/[(s-a)²+b²]
e^(at)cos(bt)	(s-a)/[(s-a)²+b²]

Q30. Find L{3t² + 2e^(3t) + sin 4t}

L{3t²} = 3 × 2!/s³ = 6/s³

L{2e^(3t)} = 2/(s-3)

L{sin 4t} = 4/(s²+16)

Answer: L = 6/s³ + 2/(s-3) + 4/(s²+16)

QUICK REVISION SUMMARY TABLE

Chapter	Key Topics to Remember
Matrices	Order, Types, Adj, A⁻¹, Cramer's Rule
Analytical Geometry	Distance formula, Area of triangle, Line equation, Quadrants
Trigonometry	Sum-to-product, tan(A-B) formula
Differential Calculus	Product rule, Chain rule, 2nd derivatives
Differential Equations	Order, Degree, Homogeneous, 2nd order linear
Integral Calculus	Standard integrals, Definite integrals
Laplace Transform	Definition, Standard formulas

Study Tip: Focus on Q5-Q8 (Matrices), Q12-Q15 (Geometry), and Q23-Q24 (DEs) as they appear most frequently in your past papers. Good luck! 🎯

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Section	Content
Title Page	Course info, chapter list table
Chapter 1 - Matrices	Q1-Q9: Definitions, Types, Transpose, Operations, Adj & A⁻¹, Cramer's Rule, Matrix Method, Partial Fractions
Chapter 2 - Analytical Geometry	Q10-Q15: Quadrants, Distance Formula, Area of Triangle, Line Equations, Graph plotting
Chapter 3 - Trigonometry	Q16-Q17: cos identity proof, tan(A-B) formula
Chapter 4 - Differential Calculus	Q18-Q20: Product rule, 2nd derivatives, dy/dx problems
Chapter 5 - Differential Equations	Q21-Q24: Definitions, Order & Degree, Homogeneous DE, 2nd order linear DE
Chapter 6 - Integral Calculus	Q25-Q27: Standard integrals with formula table
Chapter 7 - Laplace Transform	Q28-Q30: Definition, standard table, solved problems
Quick Revision Page	Summary table of all 7 chapters

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Matrices & Determinants - Order, det, Adj, A⁻¹, Cramer's Rule, Partial Fractions	Differential Calculus - All 10 standard derivatives, Product/Quotient/Chain rules	Differential Equations - Order/Degree, Separable, Linear 1st order, 2nd order (CF+PI)
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