Resources by Casio
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This collection of videos is produced by Casio in conjunction with Mathstouch Ltd. The collection consists of material found in A2 level mathematics courses. Each resource consists of at least two videos. The first video explains how to solve a mathematical problem using familiar mathematical techniques. The second...
This collection of videos is produced by Casio in conjunction with Mathstouch Ltd. The collection consists of material found in AS level mathematics courses. Each resource consists of at least two videos. The first video explains how to solve a mathematical problem using familiar mathematical techniques. The second...
These collections of videos are produced by Casio in conjunction with Mathstouch Ltd. The collection consists of two sub-collections: the first covering material found in AS level mathematics courses, the second featuring material found in the A2 course. Each resource consists of at least two videos. The first...
Evaluating the Definite Integral of a Polynomial
The mathematical solution explains how to use calculus to find the definite integral of an expression. The first step is to write the expression using index form using, where appropriate, negative indices. The expression is then integrated and the limits used to evaluate the value of the integral.
The...
Find the Value of the First and Second Derivative of a Function in x, at Specified Values of x
The mathematical solution explains how to find the value of the first and second derivatives of a given function at specific values for x. The function is presented in the form of algebraic fractions which needs to be converted to index notation and the chain rule used to find the differentials.
...Finding All the Solutions of Trigonometric Equations
The mathematical solution explains how to solve the equation: sin(x+150) = 1/sqrt2 for x between 0⁰ and 360⁰. To begin with, the principle values for (x+150) are found. The fact that a sine curve is periodic is used to find the solution set for (x+150) based upon these two initial values. Subtracting 150 from each...
Finding the Area Bounded Between the x Axis and a Parabola
Given a quadratic graph which cuts the x axis at two points, the mathematical solution explains how to find the area bounded between the curve and the x axis. The points of intersection with the x axis are located by equating the function to zero, factorising and solving the resulting equation. The required...
Finding the Minimum of a Curve and Evaluating a Definite Integral
The mathematical solution explains how to find the coordinates of the minimum point of y = 8x + 1/x, for x > 0 and how to find the area between the curve, the x axis and the lines x = 1 and x = 6 using integration. The x value of the minimum point is found by writing the function in index form,...
Finding the Minimum Value and Sketching a Quadratic Function
The mathematical solution explains how to find the minimum value and sketch the curve of y = 4x2 + 12x + 10. The first method uses calculus, differentiating the function and equating the differential to zero. The resulting equation is solved to find the x value of the minimum point. This value is then...
Finding the Points of Intersection of a Linear and a Quadratic
The mathematical solution explains how to find the points of intersection of a linear and a quadratic function by solving the equations simultaneously. By rearranging the linear equation and equating to form a quadratic equation, the x values of the intersection are found by solving the equation using...