Preparing for your exams (Year 12)

Hello and welcome to this week's column! We continue with our series on helping to prepare you all for your year-end Term 3 examinations. The tips and techniques we discuss here apply to all the students of all grades. The study techniques can also be applied for any subject and not just for maths.

Maths is one of those subjects which you can easily spend hours studying but end up none the wiser. However much you have studied, if you cannot solve the problem on day of the test, you are lost.

Some more tips to master your exam:

 Practice, practice and more practice – Every problem will have a different approach to the solution, so practice as many questions as possible.

 Review your errors - During your practice, work through the entire process to arrive at the solution. Review your errors – it will help you to correct your mistakes and get the right answer to the problem.

 Create a Math Dictionary – Yes! Create one if possible. You can browse through just the night before your exam and it might save a lot of time in looking for the formulas.

 Don't memorize the steps – Understand each and every step you do to solve the problems and please don't just cram.

 Work with smaller and easier numbers – Sometimes, questions with decimals and fractions might be confusing. To make the questions easy to understand, just try and replace the given decimals and fractions with simple whole numbers. This will definitely help you to understand better and solve.

The past years math paper shows that the weightage of the questions in Algebra and Calculus is more than the other strands. Go through the previous year papers and try to concentrate on these areas.

Revising Algebra:

Example 1: For what values of p will the quadratic equation have equal roots?

Solution: For a quadratic equation to have equal roots, its discriminant must be equal to 0. Discriminant = b2-4ac

In the given question a=1;b=p+4;c=p+7 b2-4ac= (p+4)2-4 x (p+7)x1=0 p2+8p+16-4p-28=0 p2+4p-12=0

Factorising,we get,(p+6)(p-2)=0 p=-6 0r p=2

Example 2: Simplify 4x/7-4y ÷ 12y/7x Solution: Remember the BEDMAS rule and do the division first

4x/7-4y x 7x/(12y 3) = 4x/7- 7x/3 = (4x*3-7x*7)/(3*7)

= (12x-49x)/21

= (-37x)/21

Example 3: Multiple Choice Question:

If the function h(x)= -x3- 3x2+bx+5 has a remainder of −2 when divided by x + 2, what is the value of b ?

A. −6 B. −2 C. D. 6

This question needs some working to be done. To find the remainder of this division you have sustitute the value of x by -2.

-(-2)3-3(-2)2-2b+5= -2 8-12-2b+5=-2

13-12-2b= -2

-2b= -3 b=3/2

Arithmetic and Geometric Progression: To find the nth term in a given A.P. tn=a+(n-1)d where a is the first term and d is the common difference

The sum of the first n terms is Sn= n/2 (2a+(n-1) d)

If we know the value of the last term l instead of the common di erence d then we can write the sum as Sn= n/2 (a+l)

The sum of first n natural numbers is given by Sn= (n(n+1))/2

To find the nth term in a G.P. tn=a (r(n-1))

The sum of the terms of a geometric progression gives a geometric series. If the starting value is a and the common ratio is r then the sum of the first n terms is Sn= (a(1-rn))/(1-r) provided that r is not equal to 1.

The sum to infinity of a geometric progression with starting value a and common ratio r is given by S∞= a/(1-r)

Example 4: A sequence is given as 27, 9, 3, 1, . . . (i) State why the sequence is a geometric sequence.

(ii) Find the 8th term.

(iii) Calculate the sum of all the terms of this sequence.

(i) The sequence given in the question is a GEOMETRIC PROGRESSION because this sequence goes from one term to the next by dividing or multiplying by the same value.

(ii) The formula to find the 8th term is a (r^(8-1) ) a=27 r= second term ÷first term= 9/27= 1/3 replacing the values of a and r in the formula,we get,

8th term=27 (1/3)7

27 can be written as 3^3 because 27=3x3x3 8th term= 33 x 1/37

Simplifying we get 8th term= 1/34 = 1/81

(iii) The sum of all terms is nothing but the sum to infinity

S∞= a/(1-r)

S∞= 27/(1-1/3)

S∞= 27/(2⁄3)

=27 x 3/2

= 81/2=40.5

Some basic formulas for revision.

Power Rule : d/dx xn=nxn-1

Derivative of Exponential:

If y=ex then dy/dx= ex. In other words, the derivative of ex is always equal to ex

More examples:

If y= e(1/2x),then dy/dx= 1/2 e(1/2 x)

If y=e(-x),then dy/dx= -ex

Derivative of a Constant

The derivative of a Constant is 0.

For any constant c, and any differentiable function y = c f(x), dy/dx=c d/dx

Sum and Difference Rules

For any two functions f(x) and g(x), d/dx (f(x) ± g(x))=d/dx f(x) ± d/dx g(x).

Product Rule

For any two functions f(x) and g(x), d/dx (f(x).g(x))=g(x).f’(x)+f(x).g’(x)

Quotient Rule

For any two functions f(x) and g(x), d/dx (f(x))/(g(x))= (g(x).f^’ (x)-f(x).g’(x))/ g(x)2 We end our column here today. Try to identify your weaker areas in every subject and concentrate on them. Have a proper schedule and work accordingly. Remember you are going to have a longer vacation after your exams. If you work hard and do the exams very well, you can have a really good summer vacation!