A/L මැත්ස් තනියම ඉගෙන ගන්නේ කොහොමද

Lovtus

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    A/L Combined Mathematics syllabus

    Unit 1: Algebra and Functions

    • Functions
    • Sequences and Series
    • Complex Numbers
    • Matrices
    Unit 2: Trigonometry

    • Trigonometric Functions and Identities
    • Solution of Triangles
    • Inverse Trigonometric Functions
    Unit 3: Calculus

    • Limits and Continuity
    • Differentiation
    • Applications of Differentiation
    • Integration
    • Applications of Integration
    Unit 4: Vectors and Geometry

    • Vector Algebra
    • Scalar and Vector Products
    • Lines and Planes
    • Three-dimensional Geometry
    Unit 5: Mechanics

    • Kinematics
    • Forces and Laws of Motion
    • Work, Energy and Power
    • Circular Motion
    • Simple Harmonic Motion
     

    07sanjeewakaru

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    ලංකාවෙ අඩා...!
    උගේ පොත් පට්ට.. එවලින් තමයි කට්ටිය ගණන් කොපි කරන්නේ. බුක් ෆෙයාර් යද්දී අරන් බලලා තියනව. ගත්තේ නෑ කරල තිබ්බ නිසා.
    Issara hena adui oya poth...Ramanujan pawa pawichchi karapu pothak..1890 wage liyala thiyenne...
     

    2by2

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    කොනක ඉදන් බලන්න ගන්න ඕනෙ බන් නැත්නම් එපා වෙයි උබට.. මුලින්ම සමීකරණ යි ශේෂ ප්‍රමෙයයි බලපන්.. එකෙන් ඉගෙන ගන්න පුළුවන් ඕනෑම level එකක සමීකරණයක් විසදන විදිහ..එතනින් පස්සෙ උබට කැමති path එකක්‌ තොර ගන්න ඕනෙ.. ජ්‍යාමිතිය, වීජ ගණිතය, සම්භාවිතාව සහ සංඛ්‍යානය, ව්‍යවහාරික ගණිතය,.. ඔයින් එක පැත්තක් ඉගෙන ගන්න පටන් ගන්න පුළුවන්.. එකක මැදට අවට පස්සෙ අනිත් ඒවාට අත ගහන්න පුළුවන්

    සිංහල පොත් නම් තියෙනවා kmds ජයතිලකගේ.. මගේ පරණ lap එකෙ beginner level කඩ්ඩ පොත් වගයක් තිබ්බා වගේ මතකයි දාන්නම් පුලුවන් වුණොත් 🥲
    ------ Post added on Mar 27, 2023 at 11:15 AM
    මගෙ දෙයියො pdf කරලා අපිටත් දෙන්න 🙏
     

    Lovtus

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    1.1) Linear function

    A linear function is a type of function in which the graph of the function is a straight line. Linear functions have the general form:

    f(x) = mx + b

    where m and b are constants, known as the slope and y-intercept, respectively.

    The slope of a linear function represents the rate of change of the function with respect to its input variable. It is defined as the change in the output variable divided by the change in the input variable, or:

    m = (y2 - y1) / (x2 - x1)

    where (x1, y1) and (x2, y2) are any two points on the line. The slope can also be interpreted as the amount by which the output variable changes for every unit change in the input variable.

    The y-intercept of a linear function represents the value of the function when the input variable is zero. It is the point where the line intersects the y-axis.
    --
    Eg.
    Problem 1: Given the linear function f(x) = 2x + 1, find the slope and y-intercept of the function.

    Solution: The slope of the function is the coefficient of x, which is 2. The y-intercept of the function is the constant term, which is 1. Therefore, the slope of the function is 2 and the y-intercept is 1.

    Problem 2: A car rental company charges a flat fee of $30 per day, plus an additional $0.25 per mile driven. Write a linear function that gives the total cost C as a function of the number of miles driven x.

    Solution: The flat fee of $30 is the y-intercept, and the additional charge of $0.25 per mile is the slope. Therefore, the linear function that gives the total cost C as a function of the number of miles driven x is:

    C(x) = 0.25x + 30

    Problem 3: The temperature at a certain location is decreasing at a rate of 2 degrees Fahrenheit per hour. Write a linear function that gives the temperature T as a function of time t in hours, assuming the initial temperature is 70 degrees Fahrenheit.

    Solution: The initial temperature of 70 degrees is the y-intercept, and the rate of decrease of 2 degrees per hour is the slope. Therefore, the linear function that gives the temperature T as a function of time t is:

    T(t) = -2t + 70

    Problem 4: A cell phone plan charges a flat fee of $30 per month, plus an additional $0.10 per minute of talk time. Write a linear function that gives the monthly cost C as a function of the number of minutes of talk time x.

    Solution: The flat fee of $30 is the y-intercept, and the additional charge of $0.10 per minute is the slope. Therefore, the linear function that gives the monthly cost C as a function of the number of minutes of talk time x is:

    C(x) = 0.10x + 30
     
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    Lovtus

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    1.2) Quadratic function.

    A quadratic function is a type of function in which the highest degree of the variable is two. The general form of a quadratic function is:

    f(x) = ax^2 + bx + c

    where a, b, and c are constants. The graph of a quadratic function is a parabola, which can either be concave up or concave down, depending on the sign of the leading coefficient a.

    The vertex of the parabola is the point where the function reaches its minimum or maximum value, and is given by the formula:

    x = -b / 2a

    y = f(-b / 2a)

    The axis of symmetry is a vertical line that passes through the vertex, and is given by the equation:

    x = -b / 2a

    Quadratic functions can be used to model a variety of phenomena, such as the trajectory of a projectile, the shape of a bridge arch, or the growth of a population.

    Here are some sample problems involving quadratic functions:

    Problem 1: Find the vertex, axis of symmetry, and y-intercept of the quadratic function f(x) = x^2 - 4x + 5.

    Solution: To find the vertex, we first need to find the x-coordinate, which is given by x = -b / 2a = -(-4) / 2(1) = 2. To find the y-coordinate, we substitute x = 2 into the function: f(2) = 2^2 - 4(2) + 5 = 1. Therefore, the vertex is (2, 1). The axis of symmetry is x = 2. To find the y-intercept, we set x = 0: f(0) = 0^2 - 4(0) + 5 = 5. Therefore, the y-intercept is (0, 5).

    Problem 2: A ball is thrown upward from the ground with an initial velocity of 32 feet per second. The height h (in feet) of the ball at time t (in seconds) is given by the quadratic function h(t) = -16t^2 + 32t. Find the maximum height of the ball and the time it takes to reach that height.

    Solution: The maximum height of the ball is reached at the vertex of the parabola. To find the vertex, we use the formula x = -b / 2a = -32 / 2(-16) = 1. The time it takes to reach the maximum height is 1 second. To find the maximum height, we substitute t = 1 into the function: h(1) = -16(1)^2 + 32(1) = 16. Therefore, the maximum height of the ball is 16 feet.
     
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    Lovtus

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    1.3) exponential function.

    An exponential function is a type of function in which the independent variable appears as an exponent. The general form of an exponential function is:

    f(x) = a^x

    where a is a constant and x is the independent variable. The value of a determines the growth rate or decay rate of the function.

    Exponential functions are widely used in various fields, such as finance, economics, biology, physics, and engineering. Some examples of exponential phenomena are population growth, compound interest, radioactive decay, and bacterial growth.

    Here are some sample problems involving exponential functions:

    Problem 1: Write the exponential function y = 3(2)^x in the form y = a(b)^x.

    Solution: We can write 2 as (2/3)^(-1), so:

    y = 3(2)^x y = 3[(2/3)^(-1)]^x y = 3(2/3)^(-x) y = 3(3/2)^x

    Therefore, the function can be written as y = 3(3/2)^x, which is in the form y = a(b)^x.

    Problem 2: A certain species of bacteria doubles every 3 hours. If the initial population is 1000 bacteria, find the population after 6 hours.

    Solution: We can model the population using the exponential function P(t) = 1000(2)^(t/3), where t is the time in hours. To find the population after 6 hours, we substitute t = 6 into the function: P(6) = 1000(2)^(6/3) = 1000(2)^2 = 4000. Therefore, the population after 6 hours is 4000 bacteria.

    Problem 3: A car is purchased for $20,000 and is expected to depreciate by 15% per year. Write an equation to model the value V of the car after t years, and find the value of the car after 4 years.

    Solution: The value of the car after t years can be modeled using the exponential function V(t) = 20000(0.85)^t, where t is the time in years. To find the value of the car after 4 years, we substitute t = 4 into the function: V(4) = 20000(0.85)^4 ≈ 9972. Therefore, the value of the car after 4 years is approximately $9,972.

    1.4) Trigonometric functions

    Trigonometric functions are a set of functions that relate angles to the ratios of the sides of a right triangle. The six basic trigonometric functions are:

    • sine (sin)
    • cosine (cos)
    • tangent (tan)
    • cosecant (csc)
    • secant (sec)
    • cotangent (cot)
    These functions are defined based on the ratios of the sides of a right triangle as follows:

    sinθ = opposite/hypotenuse cosθ = adjacent/hypotenuse tanθ = opposite/adjacent cscθ = hypotenuse/opposite secθ = hypotenuse/adjacent cotθ = adjacent/opposite

    where θ is the angle in radians or degrees.

    Trigonometric functions have many applications in mathematics, physics, engineering, and other fields. They are used to solve problems involving triangles, waves, periodic phenomena, and oscillations, among others.

    Here are some sample problems involving trigonometric functions:

    Problem 1: Find the value of sin 60°.

    Solution: From the definition of sine, we have:

    sin 60° = opposite/hypotenuse

    For a 30-60-90 triangle with hypotenuse 1, the opposite side is √3/2. Therefore:

    sin 60° = √3/2

    Problem 2: Find the value of cos π/3.

    Solution: From the definition of cosine, we have:

    cos π/3 = adjacent/hypotenuse

    For a 30-60-90 triangle with hypotenuse 1, the adjacent side is 1/2. Therefore:

    cos π/3 = 1/2

    Problem 3: Find the value of tan 45°.

    Solution: From the definition of tangent, we have:

    tan 45° = opposite/adjacent

    For a 45-45-90 triangle with hypotenuse 1, the opposite side and adjacent side are equal, so:

    tan 45° = 1

    Problem 4: Find the value of sec 30°.

    Solution: From the definition of secant, we have:

    sec 30° = hypotenuse/adjacent

    For a 30-60-90 triangle with hypotenuse 1, the adjacent side is √3/2. Therefore:

    sec 30° = 2/√3

    Problem 5: Find the value of cot π/6.

    Solution: From the definition of cotangent, we have:

    cot π/6 = adjacent/opposite

    For a 30-60-90 triangle with hypotenuse 1, the adjacent side is √3/2 and the opposite side is 1/2. Therefore:

    cot π/6 = (√3/2)/(1/2) = √3
    ------ Post added on Mar 27, 2023 at 12:26 PM
     
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    Tz4400

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    මම OL ඉවර උන දවසේ ඉදල , AL ජනවාරි පටන්ගන්න කල් තනියම පොත් දෙකකින් ත්‍රිකෝණමිතිය සම්පුර්ණ වගේ ගොඩ දැම්ම..

    අජිත් විමංග විජේසිංහ කියලා පොරකගේ පොත්.. සෑහෙන ගණන් අභ්‍යාස තියනව.

    එයාගෙම පොත් වලින් අවකලනය , අනුකලනයත් ගොඩ දැම්ම ක්ලාස් වලට කලින්.
    Poth zika pdf karala denna barida
     
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    DiksonAlex

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    Colombo
    මචන් A/L මැත්ස් තනියම ඉගෙන ගන්නේ කොහොමද ?
    එක්සෑම් මුකුත් කරන්න නෙවෙයි නිකන් හොබී එකක් විදියට මැත්ස් ඉගෙන ගන්න.

    මම A/L බයෝ කලේ ඒක නිසා මැත්ස් ඉගෙන ගන්න ආසයි.

    උබල දන්නා පොත් එහෙම තියේනම් ලින්ක් දාපන්. (DP education නම් එපා ප්ලීස්)
    අඩෝ මටත් ඔහොම හිතුනා. මාත් බයෝ කරේ නොදන්නා කමට. දන්නවනම් මැත්ස් කරනවා. දැනුත් කරන්න ආසයි.
     

    2by2

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    අඩෝ මටත් ඔහොම හිතුනා. මාත් බයෝ කරේ නොදන්නා කමට. දන්නවනම් මැත්ස් කරනවා. දැනුත් කරන්න ආසයි.
    මට නිදිමතායී... 😪
     

    dilann

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    එකමත් එක රටක
    P
    Poth zika pdf karala denna barida
    ඔය ajith vimanga ගේ පොත් 4න් නම් එකක් වගේ මම ගාව ඇත්තේ දැනට. අනිත් ඒවා තව පොත් ගොඩක් ආයිත් කරපු එකෙක්ට දුන්න. ඌයි උගේ මල්ලිලා ඔක්කොම උන් AL කරා .. පොත් එකක්වත් ආයිත් හම්බුනේ නම් නෑ .
    ඒවා පිටු 200ක් වගේ පොත් බන් . net එකේ නම් හොයාගන්න නෑ දැන්. කඩවල් වල ඇති.. ලොකු ගානක් නෑ .

    අර ලෝනි ගේ ත්‍රිකෝණමිතිය වගේ හොයාගන්න පුලුවන් නම් වෙන පොත් ඕන නෑ හැබැයි...
     
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    Tz4400

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    ඔය ajith vimanga ගේ පොත් 4න් නම් එකක් වගේ මම ගාව ඇත්තේ දැනට. අනිත් ඒවා තව පොත් ගොඩක් ආයිත් කරපු එකෙක්ට දුන්න. ඌයි උගේ මල්ලිලා ඔක්කොම උන් AL කරා .. පොත් එකක්වත් ආයිත් හම්බුනේ නම් නෑ .
    ඒවා පිටු 200ක් වගේ පොත් බන් . net එකේ නම් හොයාගන්න නෑ දැන්. කඩවල් වල ඇති.. ලොකු ගානක් නෑ .

    අර ලෝනි ගේ ත්‍රිකෝණමිතිය වගේ හොයාගන්න පුලුවන් නම් වෙන පොත් ඕන නෑ හැබැයි...
    Maths poth wada ganne Nathan denna puluwanda
     
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    2by2

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    1.3) exponential function.

    An exponential function is a type of function in which the independent variable appears as an exponent. The general form of an exponential function is:

    f(x) = a^x

    where a is a constant and x is the independent variable. The value of a determines the growth rate or decay rate of the function.

    Exponential functions are widely used in various fields, such as finance, economics, biology, physics, and engineering. Some examples of exponential phenomena are population growth, compound interest, radioactive decay, and bacterial growth.

    Here are some sample problems involving exponential functions:

    Problem 1: Write the exponential function y = 3(2)^x in the form y = a(b)^x.

    Solution: We can write 2 as (2/3)^(-1), so:

    y = 3(2)^x y = 3[(2/3)^(-1)]^x y = 3(2/3)^(-x) y = 3(3/2)^x

    Therefore, the function can be written as y = 3(3/2)^x, which is in the form y = a(b)^x.

    Problem 2: A certain species of bacteria doubles every 3 hours. If the initial population is 1000 bacteria, find the population after 6 hours.

    Solution: We can model the population using the exponential function P(t) = 1000(2)^(t/3), where t is the time in hours. To find the population after 6 hours, we substitute t = 6 into the function: P(6) = 1000(2)^(6/3) = 1000(2)^2 = 4000. Therefore, the population after 6 hours is 4000 bacteria.

    Problem 3: A car is purchased for $20,000 and is expected to depreciate by 15% per year. Write an equation to model the value V of the car after t years, and find the value of the car after 4 years.

    Solution: The value of the car after t years can be modeled using the exponential function V(t) = 20000(0.85)^t, where t is the time in years. To find the value of the car after 4 years, we substitute t = 4 into the function: V(4) = 20000(0.85)^4 ≈ 9972. Therefore, the value of the car after 4 years is approximately $9,972.

    1.4) Trigonometric functions

    Trigonometric functions are a set of functions that relate angles to the ratios of the sides of a right triangle. The six basic trigonometric functions are:

    • sine (sin)
    • cosine (cos)
    • tangent (tan)
    • cosecant (csc)
    • secant (sec)
    • cotangent (cot)
    These functions are defined based on the ratios of the sides of a right triangle as follows:

    sinθ = opposite/hypotenuse cosθ = adjacent/hypotenuse tanθ = opposite/adjacent cscθ = hypotenuse/opposite secθ = hypotenuse/adjacent cotθ = adjacent/opposite

    where θ is the angle in radians or degrees.

    Trigonometric functions have many applications in mathematics, physics, engineering, and other fields. They are used to solve problems involving triangles, waves, periodic phenomena, and oscillations, among others.

    Here are some sample problems involving trigonometric functions:

    Problem 1: Find the value of sin 60°.

    Solution: From the definition of sine, we have:

    sin 60° = opposite/hypotenuse

    For a 30-60-90 triangle with hypotenuse 1, the opposite side is √3/2. Therefore:

    sin 60° = √3/2

    Problem 2: Find the value of cos π/3.

    Solution: From the definition of cosine, we have:

    cos π/3 = adjacent/hypotenuse

    For a 30-60-90 triangle with hypotenuse 1, the adjacent side is 1/2. Therefore:

    cos π/3 = 1/2

    Problem 3: Find the value of tan 45°.

    Solution: From the definition of tangent, we have:

    tan 45° = opposite/adjacent

    For a 45-45-90 triangle with hypotenuse 1, the opposite side and adjacent side are equal, so:

    tan 45° = 1

    Problem 4: Find the value of sec 30°.

    Solution: From the definition of secant, we have:

    sec 30° = hypotenuse/adjacent

    For a 30-60-90 triangle with hypotenuse 1, the adjacent side is √3/2. Therefore:

    sec 30° = 2/√3

    Problem 5: Find the value of cot π/6.

    Solution: From the definition of cotangent, we have:

    cot π/6 = adjacent/opposite

    For a 30-60-90 triangle with hypotenuse 1, the adjacent side is √3/2 and the opposite side is 1/2. Therefore:

    cot π/6 = (√3/2)/(1/2) = √3
    ------ Post added on Mar 27, 2023 at 12:26 PM
    Copy paste karaata kamak naa source eka daapan 😋😂
     
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    Mr Bones

    Well-known member
  • Mar 13, 2023
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    අඩෝ මටත් ඔහොම හිතුනා. මාත් බයෝ කරේ නොදන්නා කමට. දන්නවනම් මැත්ස් කරනවා. දැනුත් කරන්න ආසයි.
    ඔව් බන් තනියම කරන්න පුළුවන් දැන් අපිට එක්සෑම් ස්ට්‍රෙස් එකක් නැති නිසා. උබත් පොතක් අරන් බලන්න
     
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