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Question
The diagram shows a shaded region bounded by the yaxis, the line y = −1 and the part of the curve y = x^{2} + 4x + 3 for which x ≥ −2.
i. Express y=x^{2}+4x+3 in the form y=(x+a)^{2}+b, where a and b are constants. Hence, for x ≥ −2, express x in terms of y.
ii. Hence, showing all necessary working, find the volume obtained when the shaded region is rotated through 360^{O} about the yaxis.
Solution
i.
We have the expression;
We use method of “completing square” to obtain the desired form. We complete the square for the terms which involve .
We have the algebraic formula;
For the given case we can compare the given terms with the formula as below;
Therefore, we can deduce that;
Hence, we can write;
To complete the square, we can add and subtract the deduced value of ;
Therefore, we can write;
Now we express in terms of .
ii.
Expression for the volume of the solid formed when the shaded region under the curve is rotated completely about the yaxis is;
Therefore, for the given case;
We have found in (i) that;
Hence, for to ;
Rule for integration of is:
Rule for integration of is:
Rule for integration of is:
Rule for integration of is:
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Question
Relative to an origin O, the position vectors of points A, B, C and D, shown in the diagram, are given by are given by


and 


i. Show that AB is perpendicular to BC.
ii. Show that ABCD is a trapezium.
iii. Find the area of ABCD, giving your answer correct to 2 decimal places.
Solution
i.
We are required to show that AB is perpendicular to BC i.e., .
If and & , then and are perpendicular.
It is evident that first we need to show that scalar/dot product of vectors and equals ZERO.
First, we find .
A vector in the direction of is;
For the given case;
Next, we find .
Now we find the scalar/dot product of vector and .
The scalar or dot product of two vectors and in component form is given as;


Since ;
For the given case;
Since scalar/Dot product is vector and equals ZERO, the two vectors are perpendicular.
ii.
We are required to show that ABCD is a trapezium.
It is evident that ABCD will be a trapezium if AB is parallel to DC as we already have shown that AB is perpendicular to BC.
Therefore, we need to show that is parallel to .
The vectors and are parallel if, and only if, they are scalar multiples of one another:⃑
where is a nonzero real number.
Hence, we need to show that;
We have found in (ii) that;
Next, we find .
A vector in the direction of is;
For the given case;
Hence, and are parallel and, therefore, ABCD is a trapezium.
iii.
We are required to find the area of ABCD.
We have demonstrated in (ii) that ABCD is a trapezium.
Expression for area of trapezium for which height is and length of parallel sides are and ;
For the given case, since and are perpendicular, as shown in (i);
Hence;
We have found in (i) and (ii) that;
We need magnitudes of all , and .
Expression for the length (magnitude) of a vector is;
Therefore;
Hence;
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Question
A curve for which passes through the point (2,3).
i. Find the equation of the curve.
ii. Find .
iii. Find the coordinates of the stationary point on the curve and, showing all necessary working, determine the nature of this stationary point.
Solution
i.
We can find equation of the curve from its derivative through integration;
We are given that;
Therefore, to find the equation of the curve;
Rule for integration of is:
Rule for integration of is:
Rule for integration of is:
If a point lies on the curve , we can find out value of . We substitute values of and in the equation obtained from integration of the derivative of the curve i.e. .
We are given that curve passes through the point (2,3).
Therefore, substituting coordinates of the point in above found equation of the curve, we can find value of c.
Therefore, equation of the curve is;
ii.
We are required to find second derivative.
Second derivative is the derivative of the derivative. If we have derivative of the curve as , then expression for the second derivative of the curve is;
We are given;
Therefore;
Rule for differentiation of is:
Rule for differentiation of is:
Rule for differentiation of is:
iii.
Coordinates of stationary point on the curve can be found from the derivative of equation of the curve by equating it with ZERO. This results in value of xcoordinate of the stationary point on the curve.
We are given derivative of the equation of the curve;
To find the coordinates of the stationary point on the curve;
One possible values of implies that there is only one stationary point on the curve at this value of .
To find ycoordinate of the stationary point on the curve, we substitute value of xcoordinate of the stationary point on the curve (found by equating derivative of equation of the curve with ZERO) in the equation of the curve.
We have found equation of the curve in (i);
Substituting ;
Hence coordinates of stationary points are .
Once we have the coordinates of the stationary point of a curve, we can determine its nature, whether minimum or maximum, by finding 2^{nd} derivative of the curve.
We have already found second derivative of the equation of the curve in (ii) as;
We substitute of the stationary point in the expression of 2^{nd} derivative of the curve and evaluate it;
If or then stationary point (or its value) is minimum.
If or then stationary point (or its value) is maximum.
Substituting ;
Since , the stationary point is minimum.
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Question
The diagram shows a sector OAC of a circle with centre O. Tangents AB and CB to the circle meet at B. The arc AC is of length 6 cm and angle radians.
i. Find the length of OA correct to 4 significant figures.
ii. Find the perimeter of the shaded region.
iii. Find the area of the shaded region.
Solution
i.
We are required to find length of OA which is radius of the circle of which we are given;
Expression for length of a circular arc with radius and angle rad is;
Therefore, for the given case;
ii.
We are required to find the perimeter of the shaded region.
It is evident from the diagram that;
We are given that;
We need to find AB & BC.
Consider the diagram below.
Let us consider the righttriangle BOA.
Expression for trigonometric ratio in righttriangle is;
Therefore;
If two tangents to a circle meet at a point external to the circle, a line segments from the point of intersection of tangents to the center of the circle bisects the angle of arc made between tangent points on the circle.
Therefore, in the given case;
Hence;
Hence;
Next, we need to find BC.
If two tangents to a circle meet at a point external to the circle, the line segments (tangents) from respective tangent point to the point of intersection of tangents are equal in length.
Therefore, in the given case;
Hence;
Hence;
iii.
We are required to find the area of the shaded region.
It is evident from the diagram that;
First, we find area of triangle AOB.
Expression for the area of the triangle is;
Consider the diagram below.
As found in (i) and (ii);
If two tangents to a circle meet at a point external to the circle, a line segments from the point of intersection of tangents to the center of the circle bisects the angle of arc made between tangent points on the circle and triangle AOB and triangle BOC are congruent right triangles.
Therefore, in the given case;
Next, we find area of sector AOC.
Expression for area of a circular sector with radius and angle rad is;
Finally, we can find area of shaded region.
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Question
The functions f and g are defined by
for
for .
i. Find the range of f and the range of g.
ii. Find an expression for fg(x), giving your answer in the form , where a, b and c are integers.
iii. Find an expression for , giving your answer in the same form as for part (ii).
Solution
i.
The given function is for .
Therefore, domain of is;
For range of we substitute extreme value(s) of the domain in the function to get the extreme values of the range of the function.
It is evident that variable quantity in the given function is . It can assume any value based on . For the minimum possible value of the will be minimum and consequently will assume the maximum possible value.
Therefore, if ;
Since, , .
Similarly, for the maximum possible value of the will be maximum and consequently will assume the maximum possible value which will be always greater than ZERO.
Hence, the rage of for is;
The given function is for .
Therefore, domain of is;
For range of we substitute extreme value(s) of the domain in the function to get the extreme values of the range of the function.
It is evident that variable quantity in the given function is . It can assume any value based on .
For the minimum possible value of the will be maximum and consequently will assume the maximum possible value.
Therefore, if ;
However, since ;
Similarly, for the maximum possible value of the will be minimum and consequently will assume the minimum possible value.
Hence, the rage of for is;
ii.
We are given;
for
for
We are required to find the rage of function .
First, we find expression for ;
iii.
Next, we find .
We have found above that;
We write it as;
To find the inverse of a given function we need to write it in terms of rather than in terms of .
Interchanging ‘x’ with ‘y’;
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Question
A straight line has gradient and passes through the point (0, −2). Find the two values of for which the line is a tangent to the curve y = x^{2 }− 2x + 7 and, for each value of , find the coordinates of the point where the line touches the curve.
Solution
We are given equation of the curve as;
It is given that line has gradient and passes through the point P (0, 2).
We are required to find the two values of for which the line is a tangent to the curve.
The slope of a curve at a particular point is equal to the slope of the tangent to the curve at the same point;
Therefore, at points where line is tangent to the curve, the gradient of both will be the same.
We are already given gradient of the line as .
We need to find the expression for gradient of the curve.
Gradient (slope) of the curve is the derivative of equation of the curve. Hence gradient of curve with respect to is:
We are given equation of the curve as;
Therefore;
Rule for differentiation of is:
Rule for differentiation of is:
Rule for differentiation of is:
Since ;
Now with gradient of the line and coordinates of a point on the line (0, 2) we can write equation of the line.
SlopeIntercept form of the equation of the line;
Where is the slope of the line.
Since line is tangent to the curve, the line and the curve intersect each other.
If two lines (or a line and a curve) intersect each other at a point then that point lies on both lines i.e. coordinates of that point have same values on both lines (or on the line and the curve). Therefore, we can equate coordinates of both lines i.e. equate equations of both the lines (or the line and the curve).
Equation of the line is;
Equation of the curve is;
Equating both equations;
Two values of x indicate that there are two intersection points.
With xcoordinate of point of intersection of two lines (or line and the curve) at hand, we can find the ycoordinate of the point of intersection of two lines (or line and the curve) by substituting value of xcoordinate of the point of intersection in any of the two equations.
We choose equation of the line;
For 
For 










Hence, line and the curve intersect at;



Now, we need to find the values of at these points.
We have found above that;
For 
For 






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Question
i. Given that , show, without using a calculator, that .
ii. Hence, showing all necessary working, solve the equation
for
Solution
i.
We are given that;
Since ;
We have the trigonometric identity;
From this we can write;
Therefore;
Let ;
Now we have two options.





Since ;



We know that;
Hence, only possibility is;
ii.
We are required to solve the equation;
For .
As demonstrated in (i) can be reduced to;
Therefore, need to solve;
Let , then;
We are required to solve the given equation for .
Therefore, interval for can be found as follows.
Multiplying entire inequality with ;
Subtracting from entire inequality;
Hence, we need to solve following for ;
Using calculator;
We utilize the symmetry property of to find other solutions (root) of :
Properties of 

Domain 

Range 

Odd/Even 

Periodicity 



Translation/ Symmetry 






Hence;
Therefore;
Since given interval is , all other solutions of the equations we utilize the periodicity property of :
For the given case,




For 


For 


Now;













As demonstrated above;
Therefore, only desired solutions of are;


Since ;






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Question
The line y = ax + b is a tangent to the curve y = 2x^{3} − 5x^{2} − 3x + c at the point (2, 6). Find the values of the constants a, b and c.
Solution
We are given equation of the line as;
We are given equation of the curve as;
It is given that line is tangent to the curve at point P(2,6).
The slope of a curve at a particular point is equal to the slope of the tangent to the curve at the same point;
First, we find slope of the line.
SlopeIntercept form of the equation of the line;
Where is the slope of the line.
Therefore, in the given equation of the line , the slope of the line is;
Next, we find slope of the curve at point P(2,6).
Gradient (slope) of the curve is the derivative of equation of the curve. Hence gradient of curve with respect to is:
We are given equation of the curve;
Therefore;
Rule for differentiation of is:
Rule for differentiation of is:
Rule for differentiation of is:
To find slope of the tangent to the curve at point we need gradient of the curve at the same point.
Gradient (slope) of the curve at the particular point is the derivative of equation of the curve at that particular point.
Gradient (slope) of the curve at a particular point can be found by substituting x coordinates of that point in the expression for gradient of the curve;
Therefore, to find the gradient at the curve at the point we substitute in;
Now we have slope of the curve as;
We can equate slope of the tangent to the curve at this point of the curve i.e., point ;
We are given equation of the line as;
We can substitute the value of ;
Since this line is tangent to the curve at point P(2,6), the point lies on both the curve and the line.
If a point lies on the line or the curve, the coordinates of the point must satisfy the equation of the line or the curve.
Therefore, we substitute coordinates of the point P(2,6) in equation of the line;
We also substitute coordinates of the point P(2,6) in equation of the curve;
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Question
The line y = ax + b is a tangent to the curve y = 2x^{3} − 5x^{2} − 3x + c at the point (2, 6). Find the values of the constants a, b and c.
Solution
We are given equation of the line as;
We are given equation of the curve as;
It is given that line is tangent to the curve at point P(2,6).
The slope of a curve at a particular point is equal to the slope of the tangent to the curve at the same point;
First, we find slope of the line.
SlopeIntercept form of the equation of the line;
Where is the slope of the line.
Therefore, in the given equation of the line , the slope of the line is;
Next, we find slope of the curve at point P(2,6).
Gradient (slope) of the curve is the derivative of equation of the curve. Hence gradient of curve with respect to is:
We are given equation of the curve;
Therefore;
Rule for differentiation of is:
Rule for differentiation of is:
Rule for differentiation of is:
To find slope of the tangent to the curve at point we need gradient of the curve at the same point.
Gradient (slope) of the curve at the particular point is the derivative of equation of the curve at that particular point.
Gradient (slope) of the curve at a particular point can be found by substituting x coordinates of that point in the expression for gradient of the curve;
Therefore, to find the gradient at the curve at the point we substitute in;
Now we have slope of the curve as;
We can equate slope of the tangent to the curve at this point of the curve i.e., point ;
We are given equation of the line as;
We can substitute the value of ;
Since this line is tangent to the curve at point P(2,6), the point lies on both the curve and the line.
If a point lies on the line or the curve, the coordinates of the point must satisfy the equation of the line or the curve.
Therefore, we substitute coordinates of the point P(2,6) in equation of the line;
We also substitute coordinates of the point P(2,6) in equation of the curve;
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Question
An increasing function, , is defined for x > n, where n is an integer. It is given that . Find the least possible value of n.
Solution
We are given derivative of the function as;
We are also given that it is an increasing function.
To test whether a function is increasing or decreasing at a particular point , we take derivative of a function at that point.
If , the function is increasing.
If , the function is decreasing.
If , the test is inconclusive.
Since we are given that function is increasing;
Therefore;
To find the set of values of x for which , we solve the following equation to find critical values of ;
Now we have two options;






Hence the critical points on the curve for the given condition are 2 & 4.
Standard form of quadratic equation is;
The graph of quadratic equation is a parabola. If (‘a’ is positive) then parabola opens upwards and its vertex is the minimum point on the graph. If (‘a’ is negative) then parabola opens downwards and its vertex is the maximum point on the graph.
We recognize that given curve , is a parabola opening upwards.
Therefore, conditions for are;
Since we are given that function is increasing, it is only possible if we consider the parabola when .
Hence, least possible value of is 4.
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Question
Find the term independent of in the expansion of .
Solution
Expression for the general term in the Binomial expansion of is:
In the given case:
Hence;
Since we are looking for the coefficient of the term independent of i.e. , so we can equate;
Hence, substituting ;
Becomes;
Hence coefficient of the term independent of i.e. is .
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Question
The diagram shows part of the curve with equation and the tangent to the curve at the point A. The xcoordinate of A is 4.
i. Find the equation of the tangent to the curve at A.
ii. Find, showing all necessary working, the area of the shaded region.
iii. A point is moving along the curve. At the point P the ycoordinate is increasing at half the rate at which the xcoordinate is increasing. Find the xcoordinate of P.
Solution
i.
We are required to find the equation of tangent to the given curve at point A.
To find the equation of the line either we need coordinates of the two points on the line (TwoPoint form of Equation of Line) or coordinates of one point on the line and slope of the line (PointSlope form of Equation of Line).
First, we find slope of the tangent to the curve at point A.
The slope of a curve at a particular point is equal to the slope of the tangent to the curve at the same point;
Therefore, we need slope of the curve at point A in order to find slope of the tangent to the curve at point A.
Gradient (slope) of the curve at the particular point is the derivative of equation of the curve at that particular point.
Gradient (slope) of the curve at a particular point can be found by substituting x coordinates of that point in the expression for gradient of the curve;
Therefore, we find of the curve and then find gradient of the curve point A.
We are given that;
Therefore;
Rule for differentiation of is:
Now we find gradient of the curve at point A.
We are given that xcoordinate of A is 4.
Hence;
Since;
The slope of tangent to the curve at point A is;
Next, we find coordinates of a point on the tangent to the curve.
We are given that tangent is drawn to the curve at point A. Hence, point A lies both on the curve and the tangent.
We are given that xcoordinate of A is 4, therefore, we need to find the ycoordinate of the point A.
Corresponding value of y coordinate can be found by substituting value of x in equation of the line or equation of the curve.
Since, point A lies on the curve, we can substitute value of in equation of the curve to fin ;
Hence, .
Now we can write equation of the tangent to the curve at point A.
PointSlope form of the equation of the line is;
ii.
We are required to find the area of the shaded region.
It is evident from the diagram that;
To find the area of region under the curve , we need to integrate the curve from point to along xaxis.
We are given equation of the curve as;
We have found equation of tangent above as;
First, we find area under the tangent.
It is evident from the diagram that for area under the curve the varies from to .
Hence;
Rule for integration of is:
Rule for integration of is:
Rule for integration of is:
Next, we find area under the curve.
It is evident from the diagram that for area under the curve the varies from to .
Hence;
Rule for integration of is:
Finally;
iii.
We are given that at the point P the ycoordinate is increasing at half the rate at which the x coordinate is increasing.
Let the rate of increase xcoordinate be;
Then as per given condition;
We know that;
From (i), we have;
Equating both;
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Question
The functions is defined for , where and are positive constants. The diagram shows the graph of y = f(x).
i. In terms of and state the range of .
ii. State the number of solutions of the following equations.
a)
b)
c)
iii. For the case where and , solve the equation , showing all necessary working.
Solution
i.
We are given;
for
We are required to find the range of .
We know that;
However, we also know that;
Multiplying entire inequality with and replacing angle with ;
Adding to entire inequality;
ii.
a)
We are given equation;
We know that graph of for is as shown below.
However, graph of for is as shown below.
Finally, we can graph for is as shown below.
The graph of intesects the xaxis at two points. Hence, there are 02 solutions of the equation;
ALTERNATE: SOLUTION METHOD
We are given equation;
Let , then;
We are required to solve the given equation for .
Therefore, interval for can be found as follows.
Multiplying entire inequality with ;
Hence, we need to solve following for ;


Using calculator; 



We utilize the symmetry property of to find other solutions (root) of :
Properties of 

Domain 

Range 

Odd/Even 

Periodicity 



Translation/ Symmetry 






Hence;
Therefore;
For 
For 





Hence, we have following solutions.



Since given interval is , for all other solutions of the equations we utilize the periodicity property of :
For the given case,




For 


For 


For 


Now;



















As demonstrated above;
Therefore, only desired solutions of are;


Since ;




b)
We are given equation;
We know that graph of for is as shown below.
However, graph of for is as shown below.
The graph of intersects the xaxis at three points. Hence, there are 03 solutions of the equation;
ALTERNATE: SOLUTION METHOD
We are given equation;
Let , then;
We are required to solve the given equation for .
Therefore, interval for can be found as follows.
Multiplying entire inequality with ;
Hence, we need to solve following for ;
Using calculator;
We utilize the symmetry property of to find other solutions (root) of :
Properties of 

Domain 

Range 

Odd/Even 

Periodicity 



Translation/ Symmetry 






Hence;
Therefore;
Hence, we have following solutions for equation within interval .


Since given interval is , for all other solutions of the equations we utilize the periodicity property of :
For the given case,




For 


For 


Now;













As demonstrated above;
Therefore, only desired solutions of are;



Since ;






c)
We are given equation;
We know that graph of for is as shown below.
However, graph of for is as shown below.
Finally, we can graph for is as shown below.
The graph of intersects the xaxis at four points. Hence, there are 04 solutions of the equation;
ALTERNATE: SOLUTION METHOD
We are given equation;
Let , then;
We are required to solve the given equation for .
Therefore, interval for can be found as follows.
Multiplying entire inequality with ;
Hence, we need to solve following for ;


Using calculator; 



We utilize the symmetry property of to find other solutions (root) of :
Properties of 

Domain 

Range 

Odd/Even 

Periodicity 



Translation/ Symmetry 






Hence;
Therefore;
For 
For 





So, we have following solutions.




Since given interval is , for all other solutions of the equations we utilize the periodicity property of :
For the given case,




For 


For 


For 


For 


Now;

























As demonstrated above;
Therefore, only desired solutions of are;




Since ;








iii.
We are given the function;
for
For and the function becomes;
for
We are required to solve;
Therefore;
Let , then;
We are required to solve the given equation for .
Therefore, interval for can be found as follows.
Multiplying entire inequality with ;
Hence, we need to solve following for ;


Using calculator; 



We utilize the symmetry property of to find other solutions (root) of :
Properties of 

Domain 

Range 

Odd/Even 

Periodicity 



Translation/ Symmetry 






Hence;
Therefore;
For 
For 





So, we have following solutions.




Since given interval is , for all other solutions of the equations we utilize the periodicity property of :
For the given case,




For 


For 


For 


For 


Now;

























As demonstrated above;
Therefore, only desired solutions of are;




Since ;








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Question
A curve is such that . The curve has stationary points at (−1, 2) and (3, k). Find the values of the constants a, b and k.
Solution
We are given that derivative of the equation of the of the curve is;
We are also given that curve has stationary points at (−1, 2) and (3, k).
A stationary point on the curve is the point where gradient of the curve is equal to zero;
Therefore, at stationary point (−1, 2); 
Therefore, at stationary point (3, k); 










From we can write;
We substitute this value of in following equation;
Substituting this value in ;
Hence;


Next, we are required to find the value of in (3, k).
We are given that (3, k) is a point on the curve.
Corresponding value of y coordinate can be found by substituting value of x in equation of the curve.
Therefore, we need to find equation of the curve.
We can find equation of the curve from its derivative through integration;
We are given that derivative of the equation of the of the curve is;
As we have found above;


This becomes;
Therefore;
Rule for integration of is:
Rule for integration of is:
Rule for integration of is:
If a point lies on the curve , we can find out value of . We substitute values of and
in the equation obtained from integration of the derivative of the curve i.e. .
We are also given that curve passes through the point (1,2).
Substitution of x and y coordinates of point in above equation;
Hence, equation of the curve is;
Now, can find value of by using coordinates of the point .
We substitute these coordinates in equation of the curve.
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Question
The coordinates of two points A and B are (1, 3) and (9, −1) respectively and D is the midpoint of AB. A point C has coordinates (x, y), where x and y are variables.
i.State the coordinates of D.
ii.It is given that CD^{2} = 20. Write down an equation relating x and y.
iii.It is given that AC and BC are equal in length. Find an equation relating x and y and show that it can be simplified to y = 2x − 9.
Solution
i.
We are required to find the coordinates of point D which is the midpoint of AB.
Expressions for coordinates of midpoint of a line joining points and;
xcoordinate of midpoint of the line
ycoordinate of midpoint of the line
We are given that points A and B have coordinates (1, 3) and (9, −1), respectively.
Therefore;
xcoordinate of midpoint of the line
ycoordinate of midpoint of the line
Hence, coordinates of midpoint of AB are D(5, 1).
ii.
We are given that CD^{2}=20.
Expression to find distance between two given points and is:
We are also given that C(x,y) and found in (i) D(5, 1).
Therefore;
iii.
We are given that;
Expression to find distance between two given points and is:
Therefore;
We are given that coordinates of points A and B are (1, 3) and (9, −1) and point C(x,y).
Hence;
Using algebraic identity;
iv.
We have found in (ii) that;
We have also found in (iii) that;
Substituting in above equation;
Now we have two options.




We substitute these values of to find the corresponding values of in equation;
For 
For 






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Question
The diagram shows a solid figure ABCDEF in which the horizontal base ABC is a triangle right angled at A. The lengths of AB and AC are 8 units and 4 units respectively and M is the midpoint of AB. The point D is 7 units vertically above A. Triangle DEF lies in a horizontal plane with DE, DF and FE parallel to AB, AC and CB respectively and N is the midpoint of FE. The lengths of DE and DF are 4 units and 2 units respectively.
Unit vectors , and are parallel to , and respectively.
i.Find in terms of , and .
ii.Find in terms of , and .
iii.Find in terms of , and .
iv.Use a scalar product to find angle FMN.
Solution
i.
We are required to find in terms of , and .
A vector in the direction of is;
For the given case;
Therefore, we need the position vectors of points and , respectively.
First, we find the position vector of point F.
Let us find the vector . Since this s the position vector of point , we need coordinates of the point . Consider the diagram below.
·It is given that is parallel to . It is evident that distance of point along from the origin is 0 units.
·It is given that is parallel to and we are given that DF is parallel to AC. We are also given that length of DF is 2 units. Hence distance of point along from the origin is 2 units.
·It is given that is parallel to and we are given that point D is 7 units vertically above origin .
Hence distance of point along from the origin is 7 units.
Hence, coordinates of .
Now we can represent the position vector of point as follows;
A point has position vector from the origin . Then the position vector of is denoted by or .
Next, we find the position vector of point M.
Let us find the vector . Since this is the position vector of point , we need coordinates of the point . Consider the diagram below.
·It is given that is parallel to and we are given that length of AB is 8 units and M is the mid point of AB. Hence, distance of point along from the origin is 4 units.
·It is given that is parallel to and it is evident that distance of point along from the origin is 0 units.
·It is given that is parallel to and it is evident that distance of point along from the origin is 0 units.
Hence, coordinates of .
Now we can represent the position vector of point as follows;
A point has position vector from the origin . Then the position vector of is denoted by or .
Now we can find .
ii.
We are required to find in terms of , and .
A vector in the direction of is;
For the given case;
Therefore, we need the position vectors of points and , respectively.
First, we find the position vector of point F.
As demonstrated in (i);
Next, we find the position vector of point N.
Let us find the vector . Since this is the position vector of point , we need coordinates of the point .
We are given that righttriangle DEF lies in a horizontal plane and N is the midpoint of FE. If we can find coordinates of points F and E we can find coordinates of point N.
First, we find coordinates of point F in horizontal plane. As demonstrated in (i), .
Next, we find coordinates of point E in horizontal plane. Consider the diagram below.
·It is given that is parallel to and we are given that DE is parallel to AB and length of DE is 4 units. Hence, distance of point along from the origin is 4 units.
·It is given that is parallel to and it is evident that distance of point along from the origin is 0 units.
Hence, in horizontal plane, coordinates of .
Now we can find coordinates of point N of FE.
To find the midpoint of a line we must have the coordinates of the endpoints of the line.
Expressions for coordinates of midpoint of a line joining points and;
xcoordinate of midpoint of the line
ycoordinate of midpoint of the line
Therefore, coordinates of point N of and .
xcoordinate of midpoint of the line
ycoordinate of midpoint of the line
Hence, horizontal plane coordinates of point N are (2,1).
·It is given that is parallel to and we are given that point D is 7 units vertically above origin .
Hence distance of point along from the origin is 7 units.
Hence, coordinates of .
Now we can represent the position vector of point as follows;
A point has position vector from the origin .
Then the position vector of is denoted by or .
Now we can find .
iii.
We are required to find in terms of , and .
A vector in the direction of is;
For the given case;
Therefore, we need the position vectors of points and , respectively.
As demonstrated in (i) and (ii);
Now we can find .
iv.
We recognize that is angle between and .
Hence, we use scalar/dot product of and to find angle .
The scalar or dot product of two vectors and in component form is given as;


Since ;
Therefore, we need to find and .
We have from (i) and (ii) that;
Now we find the scalar product of and .
The scalar or dot product of two vectors and is number or scalar, where is the angle between the directions of and .
Where





For the given case;
Therefore;
Equating both scalar/dot products we get;
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Question
Two heavyweight boxers decide that they would be more successful if they competed in a lower weight class. For each boxer this would require a total weight loss of 13 kg. At the end of week 1 they have each recorded a weight loss of 1 kg and they both find that in each of the following weeks their weight loss is slightly less than the week before.
Boxer A’s weight loss in week 2 is 0.98 kg. It is given that his weekly weight loss follows an arithmetic progression.
i. Write down an expression for his total weight loss after weeks.
ii. He reaches his 13 kg target during week . Use your answer to part (i) to find the value of .
Boxer B’s weight loss in week 2 is 0.92 kg and it is given that his weekly weight loss follows a geometric progression.
iii. Calculate his total weight loss after 20 weeks and show that he can never reach his target.
Solution
i.
The Boxer A’s weekly weight loss follows an arithmetic progression.
From the given information, we can collect following information about this Arithmetic Progression (A.P).
Weight (Kg) loss at the end of 1^{st} week;
Weight (Kg) loss at the end of 2^{nd} week;
The height measured on 2^{nd} day;
The expression for difference in Arithmetic Progression (A.P) is:
Hence;
We are required to find expression for his total weight loss after weeks which is sum of weight loss during each week for weeks.
Expression for the sum of number of terms in the Arithmetic Progression (A.P) is:
Therefore;
ii.
We are given that Boxer A loses a total of 13 kg weight during week and we are required to find the value of .
It is evident that sum of terms of this Arithmetic Progression (A.P) will be 13.
Expression for the sum of number of terms in the Arithmetic Progression (A.P) is:
Therefore;
Standard form of quadratic equation is;
The solution of quadratic equation is;
Therefore, for the given equation;
Now we have two options.






It is evident that more than 85 weeks is not feasible and does not seem viable answer 
Therefore, the 13Kg loss will be achieved during the 16^{th} week. 
iii.
The Boxer B’s weekly weight loss follows a Geometric Progression (G.P).
From the given information, we can collect following information about this Geometric Progression (G.P).
Weight (Kg) loss at the end of 1^{st} week;
Weight (Kg) loss at the end of 2^{nd} week;
Expression for Common Ratio () in a Geometric Progression (G.P) is;
Hence;
We are required to find his total weight loss after 20 weeks which means sum of 20 terms.
Expression for the sum of number of terms in the Geometric Progression (G.P) when is:
Therefore;
Now if we need to show that he can never reach his target of loosing 13Kg weight, we need to show that sum to infinity of this Geometric Progression (G.P) is less than 13.
Expression for the sum to infinity of the Geometric Progression (G.P) when or ;
Therefore;
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Question
The function f is defined by for . The function g is defined by for where and are constants.
i. Find the greatest value of and the least value of which will permit the formation of the composite function gf.
It is now given that the conditions for the formation of gf are satisfied.
ii. Find an expression for gf(x).
iii. Find an expression for gf ^{1}(x).
Solution
i.
We are given;
for
for
The composite function is required to be formed.
For a composite function , the domain D of must be chosen so that the whole of the range of is included in the domain of .
The function , is then defined as , .
Therefore, for the composite function , the domain D of must be chosen so that the whole of the range of is included in the domain of . The function is then defined as , .
Hence, we need to restrict domain of to available range of .
First, we find range of ;
for
Finding range of a function :
· Substitute various values of from given domain into the function to see what is happening to y.
· Make sure you look for minimum and maximum values of y by substituting extreme values of from given domain.
It is evident that will be maximum when is minimum and will be minimum when is maximum.
When ;
When ;
Hence;
Therefore, for the composite function will exist when range of will form the domain of ;
for
for
ii.
We are given;
for
for
We are required to find the expression for .
Therefore;
iii.
We are required to find the expression for .
We have found in (ii) that;
We can write it as;
To find the inverse of a given function we need to write it in terms of rather than in terms of .
Therefore;
Interchanging ‘x’ with ‘y’;
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Question
The diagram shows triangle ABC which is rightangled at A. Angle radians and AC = 8 cm. The points D and E lie on BC and BA respectively. The sector ADE is part of a circle with centre A and is such that BDC is the tangent to the arc DE at D.
i. Find the length of AD.
ii. Find the area of the shaded region.
Solution
i.
We are required to find the length of AD.
To find length of AD we consider the triangle ADC.
We are given that BDC is the tangent to the arc DE at D where A is center of the arc DE.
It is evident that AD is the radius of the sector ADE.
A radius from the center of the circle to the point of tangency is perpendicular to the tangent line.
Therefore, AD is perpendicular to straight line BDC;
Hence, triangle ADC is a righttriangle.
We know that;
Expression for trigonometric ratio in righttriangle is;
We are given that but we need to find .
To find we consider the triangle ABC where .
We know that;
Therefore, for triangle ABC;
We are given that and , therefore;
Hence;
ii.
It is evident from the diagram that;
First, we find area of sector AED.
Expression for area of a circular sector with radius and angle rad is;
For the given sector AED;
As demonstrated in (i) AD is radius of circle/sector with center A.
It is evident from the diagram that .
Therefore, now we find .
Consider triangle ABD.
We know that;
Therefore, for triangle ABD;
As demonstrated in (i);
Since, BDC is a straight line, tangent to arc DE, therefore;
We are given that , therefore;
Hence;
Hence;
Now we can find area of sector AED;
Next, we find area of the triangle ADC.
Expression for the area of the triangle is;
For the triangle ADC.
We are given that .
We need to find DC.
Expression for trigonometric ratio in righttriangle is;
Therefore, for triangle ADC;
As demonstrated in (i);
Therefore;
Hence;
Finally;
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Question
i. In the binomial expansion of , the first three terms are . Find the remaining three terms of the expansion.
ii. Hence find the coefficient of in the expansion of .
Solution
i.
Expression for the Binomial expansion of is:
First rewrite the given expression in standard form.
In the given case:
Hence;
ii.
To find the term with in the expansion of
From (i) we have;
Therefore;
We are interested only in coefficient of which is equal to 0.
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