Fifty Sat Maths subject level 2 questions corresponding to those in sample 1, with detailed solutions and explanations are presented.
When two dice are tossed, there are 6 * 6 = 36 possible outcomes. |
(1,1) , (1,2) , (1,3) , (1,4) , (1,5) , (1,6)
(2,1) , (2,2) , (2,3) , (2,4) , (2,5) , (2,6) (3,1) , (3,2) , (3,3) , (3,4) , (3,5) , (3,6) (4,1) , (4,2) , (4,3) , (4,4) , (4,5) , (4,6) (5,1) , (5,2) , (5,3) , (5,4) , (5,5) , (5,6) (6,1) , (6,2) , (6,3) , (6,4) , (6,5) , (6,6) |
The outcomes that give a sum of 3 or less are | (1,1) , (1 , 2) and (2,1) |
There are | 36 - 3 = 33 possible outcomes that give a sum that is greater than 3 |
The probability that the sum of the two dice is greater than 3 | 33 / 36 = 11 / 12 |
The slope of the line through (2,5) and (4,6) is given by | (6 - 5) / (4 - 2) = 1 / 2 |
The slope of the line through (4,6) and (7,k) is given by | (k - 6) / (7 - 4) = (k - 6) / 3 |
The three points are on the same line and therefore the slopes found above must be equal | 1 / 2 = (k - 6) / 3 |
Solve the above equation for k to find | k = 15/2 |
To find points of intersection of the graphs of y = x2 - 2 x + 7 and y = k x + 5, we need to solve the system of the two equations | y = x2 - 2 x + 7 y = k x + 5 |
Eliminate y between the two equations to obtain | x2 - 2 x + 7 = k x + 5 |
The above equation may be written as | x2 - x(2 + k) + 2 = 0 |
This is a quadratic equation which may have no solutions meaning that the two graphs have no points of intersection, two solutions which means the two graphs have two points of intersection or one solution which means the two graphs are tangent to each other. A quadratic equation have one solution if its discriminant is equal to 0. Hence | (2 + k)2 - 4(1)(2) = 0 |
Solve for k and select the negative solution | k = - 2 - 2 √2 |
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One way to find the coordinates of N is to find the components of vector ON. | Vector ON = Vector OC + Vector CN |
The equation of the circle gives the coordinates of the center C which are | C (3 , 2) |
The components of vector OC are | < 3 , 2 > |
Distance CN is equal to the radius of the circle which is 1. Points C and M have the same y coordinates and therefore CM is parallel to the x axis, hence the components of vector CN are given by | <1 . cos(30o) , 1 . sin(30o)> = < √3 / 2 , 1 / 2> |
Hence the components of vector ON are given by | < 3 , 2 > + < (1 / 2)√3 , 1 / 2 > = < 3 + √3 / 2 , 5/2 > |
The coordinates of point N are given by | ( 3 + √3 / 2 , 5/2 ) |
We first find w = -u - 2v as follows |
w = -u - 2v = - < 2 , 0 >
-2 < - 3 , 1 > = < - 2 , 0> + < 6 , - 2 > = < 4 , - 2 > |
The length of W is given by | √(42 + (-2)2) = 2 √5 |
Let us find the center and radius of the circle | The equation of the circle is in standard form and its center has coordinates (1 , 2) and its radius is equal to 2. |
The distance between the center of the circle and point (-2, -2) is given by | √( (-2 -1)2 + (-2 -2)2 ) = 5 |
Point (-2,-2) is outside the circle. The shortest distance between point (-2,-2) and a point on the circle is given by | 5 - 2 = 3 |
Let us find the slope of the line with equation x + 3y = 2 |
3y = - x + 2 y = -(1 / 3)x + 2/3 slope = -1 / 3 |
The slope of the line with equation -2x + ky = 5 is calculated as follows | k y = 2x + 5 y = (2 / k) x + 5 / k slope = 2 / k |
For two lines to be perpendicular, the product of their slopes must be equal to - 1. Hence | (-1 / 3)*(2 / k) = -1 |
Solve the above equation for k to find | k = 2 / 3 |
(g o f) (x) is the composition of functions defined as follows | (g o f) (x) = g ( f (x) ) |
Substitute f (x) by (x - 1) 2 |
(g o f) (x) = g ( f (x) ) = g( (x - 1) 2 ) = √(x - 1) 2 = | x - 1 | |
NOTE THAT √ x 2 = | x | not x |
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and the solution set of (x - 1)(x + 1) > 0 is given by the interval.
and the solution set of (2 - x)(2 + x) ? 0 is given by the interval.
The domain of f is given by
We first need to write the given equation in standard form by completing square | x 2 + y 2 - 8 y - 48 = 0 |
Write the term (y 2 - 8 y) as a square plus a constant : add and subtract 16 | y 2 - 8 y = y 2 - 8 y
+ 16 - 16 = (y - 4) 2 - 16
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Substitute by in the given equation and write the equation of the circle in standard form | x 2 +
(y - 4) 2
- 16 -
48 = 0 or x 2 + (y - 4) 2 = 64 |
Hence the radius R of the circle is equal to 8. The area is given | Pi R 2 = 64 Pi |
The points of intersections of the parabola and the circle are solutions to both equations simultaneously and therefore to find these points we need to solve the system of equations defined by | y2 = x + 2 x2 + y2 = 4 |
Substitute y2 by x + 2 in the second equation and rewrite the equation obtained in standard form | x2 + x + 2 = 4 x2 + x - 2 = 0 |
Solve the quadratic equation obtained | x = 1 and x = -2 |
We now substitute the values of x obtained in the equation y2 = x + 2 to find y | For x = 1 , and y2 = 1 + 2 = 3 y = + or - √3 for x = -2 , and y2 = -2 + 2 = 0 hence y = 0 |
The y coordinates of the points of intersection are | - √3 , 0 and √3 |
The zeros of function f are found by solving the equation | 1/2 - sin(3x + Pi/3) = 0 |
Rearrange the equation to obtain | sin(3x + Pi/3) = 1 / 2 |
In order to solve the trigonometric equation above, we need to solve the two equations below. Note that because sine functions are periodic, there is an infinite number of solutions. | 3x + Pi/3 = Pi / 6 + 2 k Pi and 3x + Pi/3 = 5 Pi / 6 + 2 k Pi for k any integer |
We now solve the above equations. | First equation: 3x + Pi/3 = Pi / 6 + 2 k Pi
3x = - Pi / 6 + 2 k Pi x = - Pi / 18 + 2 k Pi / 3 Second equation: 3x + Pi/3 = 5 Pi / 6 + 2 k Pi 3x = Pi / 2 + 2 k Pi x = Pi / 6 + 2 K Pi / 3 |
We now need to find the smallest positive solution. | For k = 0 we have two solutions: x = - Pi / 6 and Pi / 6 and hence the smallest zero is equal to Pi / 6 |
We first use the factors to write polynomial P as follows | P(x) = k (x - 1)(x - 3)(x + 1) , k is a nonzero real number |
Other possible factors of P(x) are | (x - 1)(x - 3) = x2 - 4 x + 3
(x - 1)(x + 1) = x2 - 1 and (x - 3)(x + 1) = x2 - 2x -3 |
Of those listed in I, II and III, II and III are also factors of P(x). |
Let V1 be the volume of the small cylinder and V2 the volume of the large cylinder where | V1 = (52 Pi) * 20 V2 = (102 Pi) * 20
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The volume V of the region bewteen the two cylinders in the difference between the two volumes | V = V2 - V1 = 1500 Pi cm3
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The standard deviation involves the square root of the sum of(xi - m)2 where xi are the data values and m is the mean. If each data value is multiplied by k then the mean will be k times. Hence the new standard deviation will involve the square root of the sum of (k xi - k m)2 |
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100 |
∑ (3 + k) |
k=1 |
100 |
∑ (3 + k) = 4 + 5 + 6 +...+ 103 |
k=1 |
100 |
∑ (3 + k) = 100*(4 + 103)/2 = 5350 |
k=1 |
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Solution
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Find a, b and c.