Write the equation as and factorise to . It is important to factor out sin(2x) here rather than divide by sin(2x) because sin(2x) may be zero and division by zero is not allowed. actually provides some solutions to this equation. Inverting sin gives . See Trigonometric Graphs and Trigonometric Equations. Note that if then .
The remaining solutions are given by solving or . Hence, .
The full set of solutions to the original equation are .
The line has equation . Points and B (to the left of A) lie on such that . The line passes through , is perpendicular to and intersects at the point . The point lies on such that the length of times the length of . Find the area of the quadrilateral .
The best thing to do first is sketch a diagram containing all of the important information given in the question. We can’t begin to find the area of the given quadrilateral without this.
The quadrilateral can be seen here (dashed) and the area is given by half of the area of the rectangle containing it (dotted). Since we know that , the only thing to find is the length which we are told is . We already know the coordinates of C and so we need the coordinates of D. This is found by solving the simultaneous equations and . The latter equation is that of found using standard techniques. See Straight Lines Example 1 part 2. It follows that the coordinates of D are (3.5,1). Using Pythagoras, the length of CD is and so CE is . The area of the rectangle is thus – see Surds. It follows that the area of the given quadrilateral is 45 square units.
Question of the Week – Week 20
Use the first 3 terms of the binomial expansion of (in ascending powers of x) to find an approximation for .
The graph is of a positive cubic. There is a root at x=-1 and a repeated root at x=k where k is positive so on the left of the y-axis. The y-intercept can be found by substituting x=0 into the equation for y.
Proof by contradiction usually starts with an assumption that opposes the statement in the question. Going from there, one hopes to meet a contradiction so that the original assumption is ruled out. In this question, we assume that the is rational. A rational number can be expressed where a and b are integers with no common factors. It follows that
From this last expression we can see that 2 is a factor of and so it must be a factor of a (the factors of a will appear in factors of in pairs). We can write and so or and so 2 is also a factor of and also b. The original assumption that can be written as a rational number a/b where a and b are integers with NO COMMON FACTORS has been violated. Our original assumption must be wrong. We may conclude that is irrational.
This proof is quite complicated so you would be expected to memorise it having seen it before.
In order to solve this, one must attempt to express the given powers as a power of 10. can be written as and we can write as . This gives the left hand side as or . Similarly the right hand side can be written as . Equating left and right hand sides and moving over to one side gives
once factorised. and so the term in brackets is zero. It follows that the value of .
Question of the Week – Week 16
Given that (x-3) is a factor of , find all the exact solutions of where .
If x-3 is a factor then can be written as . Click here to find out more on how to do this. Note that the question could also say instead of x-3 is factor. These are equivalent statements according to Factor Theorem. The quadratic can also be factorised so that . Noting that is the same as if we let . Hence, gives the roots , and . It follows that the solutions are and – see Logs. The third equation has no solution for y.
This one is for the students studying calculus and for those who would like a taste of it. Integration by substitution can be used or the reverse chain rule. You could also write as and guess its integral as follows. We know that differentiates to using the chain rule. Hence, integrates to . For definite integration,
Hence, the integral is inversely proportional to k.
The points P(-3,2), Q(9,10) and R(a,4) lie on a circle. Given that PR is a diameter of the circle, show that a=13.
Begin by drawing a rough sketch of the circle. Since PR is a diameter of the circle, angle PQR is a right-angle and PQ is perpendicular to QR. This means that if the gradient of PQ is m then the gradient of QR =-1/m. The gradient of PQ is . Hence the gradient of QR=-3/2. It follows that . Solving for a gives a=13.
Simplifying the surd gives with corresponding y solutions given by . Note that these y solutions are best found by substituting x into the most basic of the original equations.
Question of the Week – Week 12
find the set of all possible values of c.
Consider , for example, x has to be greater than 3 or less than -3. Similarly, if , 5-c has to be greater than root 12 or less than minus root 12. Subtracting 5 suggests that or . Multiplying both sides of the inequalities by -1 reverses the sign of the inequality (click here for more on inequalities). Finally, we have or .
Question of the Week – Week 11
Find all the values for x such that
First think about reading the logs – see the logarithms for more details. can actually be read as ‘the power of 2 that gives 32’ – we know this to be 5 since . Similarly, and so
Multiply both sides of the equation by to get
Note that there is no rule for simplifying a log that is squared – see log rules. Square rooting gives giving the solutions and .
This question was used in a recent exam and students did not like question 2 – they were not used to seeing two surds in the denominator. The same principles of rationalising the denominator apply – multiplying the top and bottom of the fraction with the same expression where the minus sign is replaced with a plus eliminates the unwanted surds.
Question of the Week – Week 8
where p and q are integers. Find the values of p and q.
When you are confronted with a question that looks unusual, you should always try to make it look as familiar as possible so you can use the usual methods to tackle it. Start by writing:
Rearranging to get all trigonometric terms on one side:
Students often get stuck at this point – if they don’t know their double angle formulae very well. Using we can convert the equation into something solvable:
see the graph if cos noting that if then we need all the solutions for . It follows that the solutions are
Question of the Week – Week 6
The equation , where k is a constant, has no real roots.
The question indicates that the discriminant is negative since we are told there are no roots – see discriminants. It follows that
At this point we cannot divide both sides by k as we do not know what the sign of k is and so don’t know whether to reverse the inequality or not. You could consider k positive and negative separately but it is easier to factorise here:
This quadratic in k can easily be sketched (by considering the roots) to see that the quadratic is only strictly negative between 0 and 3/4 (not inclusive).
Question of the Week – Week 5
Questions that involve a change of base are quite rare but it is still possible that they will come up in your exam. It should be obvious that a change of base is required because the base is 3 on one of the log expressions whereas it is x on the other. The change of base formula is given (in the Edexel formula booklet) as:
Choosing a=x, x=81 and b=3 in this formula gives:
Substituting this into the original expression gives:
The final log expression can be evaluated as the power of 3 that gives 81 and so the final answer is 4.
Question of the Week – Week 4
The first three terms of a geometric sequence are
where k is a constant. Show that
Given that k is not an integer, show that .
Since the sequence is a geometric one the ratio between each of the terms must be the same i.e. . It follows that . Expanding gives and so as required.
Question of the Week – Week 3
Prove that for all positive x and y:
The trick is to spot the square root on the left and hence consider the square of the right. First note that . Secondly, and are both positive and so . By square rooting both sides, it follows that .
Question of the Week – Week 2
Although it may not be obvious at first, the above equation is actually a quadratic. It can be written as
and then as
Letting , this can be factorised to
It follows that or giving the solutions or .
Question of the Week – Week 1
Do you know the correct way to solve the following inequality?
Most students will instantly multiply both sides by but it is important to consider if is positive or not. This is because if is negative, the inequality sign will be reversed.