How to Prepare for board Exams-A Few Tips

How to Prepare for board Exams?

In my previous post of this series, we discussed the advantages of CBSE board exams. Now I am suggesting a few tricks about How to Prepare for board exams.
There are a lot of post on the internet that suggests you how to eat, how to sleep and how to manage your stress during board exam but in this post, I shall only discuss the academic tips about How to Prepare for board exams. I am taking class 10 Mathematics as a base subject here but you can apply more or less same tricks on almost all your subjects.

Question Paper Break-Up<img src="how to prepare for board exams".jpg" alt="how to prepare for board exams">

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Definite Integration-Topics in IB Mathematics

Definite Integration

In the previous post, we discussed indefinite integration. Now we shall discuss definite integration

► Definite Integration- We already know that   \int {f\left( x \right){\rm{ }}dx = g\left( x \right) + c}    \leftarrow  this c here is an integral constant. we are not sure about its value. This c is the reason we call this process indefinite integration. But suppose we do our integration between certain limits like:-

\int\limits_a^b {f(x)dx = \left[ {g(x) + c} \right]} _a^b   here a \to  lower limit while b \to  higher limit

\int\limits_a^b {f(x)dx = \left[ {g(b) + c} \right]} - \left[ {g(a) + c} \right]

=g(b)-g(a)

You can clearly see that this function is independent of ‘c’. Means we can be sure about its value so this type of integration is called  Definite Integration.

►Definite Integration of a function f(x) is possible in [a,b] if f(x) is continuous in the given interval

►If f(x), the integrand, is not continuous for a given value of x then it doesn’t mean that g(x), the integral, is also discontinuous for that value of x.

► Definite integration of a function between given limits like     \int\limits_a^b {f\left( x \right)dx} \Rightarrow         Algebraic sum of areas bounded by the given curve f(x) and given lines x=a and x=b. That’s why the answer for definite integration problems is a single number.

► If \int\limits_a^b {f\left( x \right)dx} = 0 that shows a few things:-

(i) The lines between which area is bounded are co-incident(a=b)

(ii) Area covered above the x-axis=Area covered below the x-axis that means positive part of area and negative part of area is equal

(iii) there must be at least one solution/root to f(x) between x=a and x=b(this is something we study in ROLE’S THEOREM in detail)

► If given function f(x) is not continuous at x=c then we should write

\int\limits_a^b {f\left( x \right)dx} = \int\limits_a^{{c^ - }} {f(x)dx} + \int\limits_{{c^ + }}^a {f(x)dx}

► If given function f(x) > or <0 in any given interval (a,b) then  \int\limits_a^b {f\left( x \right)dx} \Rightarrow  >0 or <0 in given interval (a,b)

► If given function f(x)  \ge  g(x) in the given interval (a,b) then    \int\limits_a^b {f(x)dx \ge } \int\limits_a^b {g(x) \ge } dx 

in the given interval

► If we integrate the given function f(x) in the given interval (a,b) then

\int\limits_a^b {f(x)dx \le } \left| {\int\limits_a^b {g(x) \ge } dx} \right| \le \int\limits_a^b {\left| {f(x)} \right|dx}

<img src="definite integration.jpg" alt="definite integration">

Some More Properties of Definite Integration:- Read more

Continuity of functions-IB Maths topics

Continuity of functions-

The word continuous means without any break or gap. Continuity of functions exists when our function is without any break or gap or jump . If there is any gap in the graph, the function is said to be discontinuous.

Graph of functions like sinx,cosx, secx, 1/x etc are continuous (without any gap) while greatest integer function has a break at every point(discontinuous).

1. A function f(x) is said to be continuous at x = c,  if  {\lim }\limits_{x \to c} f(x) = f(c) .

 

symbolically f is continuous at x = c if  {\lim }\limits_{x \to c - h} f(c + h) = {\lim }\limits_{x \to c - h} f(c - h) = f(c).

 

It should be noted that continuity of a function at x = a is meaningful only if the function is defined in the immediate neighborhood of x = a, not necessarily at x = a.

<img src="continuous functions.png" alt="continuous functions">

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How To Solve Limit Problems

How To Solve Limit Problems

 

In my previous post on limits, We have discussed some basic as well as advanced concepts of limits. Here we shall discuss different methods to solve limit questions. Based on the type of function, we can divide all our work into sections-:

Algebraic Limits- Problems of limits that involve algebraic functions are called algebraic limits. They can be further divided into following sections:-

Direct Substitution Method –Suppose we have to find. L = {\lim }\limits_{x \to a} f(x) we can directly substitute the value of the limit of the variable (i.e replace x=a) in the expression.

► If f(a) is finite then L=f(a)

► If f(a) is undefined then L doesn’t exist

► If f(a) is indeterminate  then this method fails

<img src="limit.png" alt="limit">

Example-1:- Find value of   {\lim }\limits_{x \to 2} (x²-5x+6) Read more

Limit, Continuity & Differentiability-IB Maths Topics

Limit of a function

Limit of a function f(x) is said to exist as, x \to a when

 {\lim }\limits_{x \to {a^ + }} f(x) = {\lim }\limits_{x \to {a^ - }} f(x) =   finite quantity.

 

 <img src="limit.png" alt="limit">

Fundamental Theorems On Limits :

Let    {\lim }\limits_{x \to {a^{}}} f(x) = l &   {\lim }\limits_{x \to {a^{}}} f(x) = l   If l & m exists then :

(i) f (x) ± g (x) = l ± m

 

(ii) f(x). g(x) = l. m

 

(iii)  {\lim }\limits_{x \to \infty } \frac{{f(x)}}{{g(x)}} = m  provided  m \ne 0

 

(iv)  {\lim }\limits_{x \to {a^{}}} kf(x) = k {\lim }\limits_{x \to {a^{}}} f(x)   where k is a constant.

 

(v)    {\lim }\limits_{x \to {a^{}}} f[g(x)] = f[ {\lim }\limits_{x \to {a^{}}} g(x)] = f(m)provided f is continuous at        g (x) = m

 

Standard Limits :

(a)  {\lim }\limits_{x \to 0} \frac{{\sin x}}{x} = 1 and {\lim }\limits_{x \to 0} \frac{{\tan x}}{x} = {\lim }\limits_{x \to 0} \frac{{{{\tan }^{ - 1}}x}}{x} = 1 {\lim }\limits_{x \to 0} \frac{{{{\sin }^{ - 1}}x}}{x} = 1 Where x is measured in radians

 

(b)  {\lim }\limits_{x \to 0} {(1 + x)^{\frac{1}{x}}}and {\lim }\limits_{x \to 0} {(1 + \frac{1}{x})^x} both are equal to e

 

(c) {\lim }\limits_{x \to a} f(x) = 1and {\lim }\limits_{x \to a} \theta (x) = \infty  then this will show that  {\lim }\limits_{x \to a} f{(x)^{ {\lim }\limits_{x \to a} \theta (x)}} = {e^{ {\lim }\limits_{x \to a} \theta (x)[f(x) - 1]}}

 

(d)  {\lim }\limits_{x \to a} f(x) = A > 0 and   {\lim }\limits_{x \to a} \theta (x) = B (a finite quantity) then    {\lim }\limits_{x \to a} f{(x)^{ {\lim }\limits_{x \to a} \theta (x)}} = {e^z}

 

where z= ^{ {\lim }\limits_{x \to a} \theta (x)\ln f(x)} = {e^{B\ln A}} = {A^B}

 

(e)  {\lim }\limits_{x \to 0} \frac{{{a^x} - 1}}{x} = \ln a where a>0. In particular  {\lim }\limits_{x \to 0} \frac{{{e^x} - 1}}{x} = 1

 

Indeterminant Forms:

\frac{0}{0},\frac{\infty }{\infty },0 \times \infty ,{0^\infty },{\infty ^0} etc are considered to be indeterminant values

We cannot plot \infty  on the paper. Infinity\infty is a symbol & not a number. It does not obey the laws of elementary algebra.

\infty +\infty =\infty

\infty ×\infty \infty

(a/\infty ) = 0 if a is finite v is not defined

a b =0,if & only if a = 0 or b = 0  and  a & b are finite.

Expansion of function like Binomial expansion, exponential & logarithmic expansion, expansion of sinx , cosx , tanx should be remembered by heart & are given below:

(i)  ex =1+x/1!+x3/3!+x4/4!……\infty

 

(ii)  ax=1+(xloga)/1!+ (xloga)2/2!+ (xloga)3/3!+ (xloga)4/4!+……….where a > 0

 

(iii)   ln(1-x)=x-x2/2+x3/3-x4/4……….    where -1 < x  1

 

(iv)  ln(1-x)=-x-x2/2-x3/3-x4/4……….     where  -1 x < 1

 

(v )  \sin x = x - \frac{{{x^3}}}{{3!}} + \frac{{{x^5}}}{{5!}} - \frac{{{x^7}}}{{7!}}.......

 

(vi) \cos x = 1 - \frac{{{x^2}}}{{2!}} + \frac{{{x^4}}}{{4!}} - \frac{{{x^6}}}{{6!}}.......

 

(v)  \tan x = x + \frac{{{x^3}}}{3} + \frac{{2{x^5}}}{{5!}} - ..........

 

In next post, I will discuss various types of limit problems, their solutions and L’ Hospital’s rule.In the meantime, you can solve these basic questions from this PDF. This PDF is for beginners only. I will post difficult and higher level questions in the next post on this topic

 LimitsExercises.pdf

In my second post on limits, you can learn how to solve different types of questions on limits
Here is the link

http://ibelitetutor.com/blog/how-to-solve-limit-problems/

 

How to solve trigonometric problems based on complimentary anngles?(concept-3)

IB Maths tutors give great importance to Trigonometry.

Trigonometry is one of the fascinating branches of Mathematics. It deals with the relationships among the sides and angles of a triangle.Word trigonometry was originated from the Greek word, where, ‘TRI‘ means Three‘GON‘ means sides and the ‘METRON’ means to measure. It’s an ancient and probably most widely used branch Mathematics. For basic learning, IB Maths Tutors divide trigonometry in two part:-

1. Trigonometry based on right triangles

2. Trigonometry based on non-right triangles.

Here, we are discussing trigonometry based on non-right triangles only.

In the third article of this series, we will discuss problems based on complementary angles

In the third article of this series, we will discuss problems based on complementary angles

<img src="right triangle.jpg" alt="right triangle">

In this right triangle Sin A=BC/AC & Cos C=BC/AC   clearly: Sin A=Cos C  In the given triangle A+C=90° so we can write C=(90°-A). This gives us freedom to write Sin A=Cos (90°-A) similarly we can write these relationships     Read more

How to solve basic problems in trigonometry?(concept-2)

Mathematics tutors give great importance to Trigonometry

Trigonometry is one of the fascinating branches of Mathematics. It deals with the relationships among the sides and angles of a triangle.Word trigonometry was originated from the Greek word, where, ‘TRI‘ means Three‘GON‘ means sides and the ‘METRON’ means to measure. It’s an ancient and probably most widely used branch Mathematics. For basic learning, I am dividing trigonometry in two part:-

1 Trigonometry based on right triangles

1 Trigonometry based on non-right triangles.

In this post, we will discuss problems based on trigonometric ratios of a few specific angles like 0°,30°, 45°, 60° and 90°. Mathematics tutors use different tricks to form this table. I will discuss my tricks in a separate post

<img src="trigonometric table.jpg" alt="trigonometric table"> Read more

How to solve basic problems in trigonometry?(concept-1)

Mathematics tutors give great importance to Trigonometry

Trigonometry is one of the fascinating branches of Mathematics. It deals with the relationships among the sides and angles of a triangle.Word trigonometry was originated from the Greek word, where, ‘TRI‘ means Three‘GON‘ means sides and the ‘METRON’ means to measure. It’s an ancient and probably most widely used branch Mathematics. For basic learning, I am dividing trigonometry in two part:-

1 Trigonometry based on right triangles

1 Trigonometry based on non-right triangles.

In this post, I will only discuss Trigonometry based on right triangles.

In a right triangle, there are three sides hypotenuse (the longest side), adjacent side(base) and the opposite side(perpendicular).

                             <img src="right triangle.png" alt="right triangle"> Read more

Quadratic equations and Quadratic Functions

Many IB mathematics tutors consider quadratic equations as a very important topic of ib maths. There are following ways to solve a quadratic equation

► Factorization method

►complete square method 

► graphical method

► Quadratic formula method

<img src="ib mathematics tutors.png" alt="ib mathematics tutors>

The quadratic formula is the strongest method to solve a quadratic equation. In this article, I will use …… steps to prove the quadratic formula

Given equation: ax²+bx+c=0

Step-1: transfer constant term to right side

ax²+bx=-c

Step-2: divide both sides by coefficient of x²

                                      x²+bx/a=-c/a
Step-3: write (coefficient of x/2)²     that is (b/2a)²=b²/4a²
Step-4: Add this value to both sides
                                x²+bx/a+b²/4a² =-c/a²+b²/4a²
                                (x+b/2a)²=b²-4ac/4a²
now, take square root on both sides

                                  x+b/2=±√b²-4ac/a²

                                   x=-(b/2a)±√b²-4ac/2a

                                   x = \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}

This formula is known as quadratic formula, we can put values of a, b and c  from any equation and find the value of x (the variable) by directly using this formula.

IB Mathematics tutors can also explain the concept of conjugate roots with the help of quadratic formula. In a quadratic equation,
                                                              ax²+bx+c=0
if a, b and c are all rational numbers and one root of the quadratic equation is a+√b then the second root will automatically become a-√b. that can be understood easily as we use one +ve and one -ve sign in quadratic formula.
These types of roots are called Conjugate Roots.
If  a & b  are  the  roots  of  the  quadratic  equation  ax² + bx + c = 0,  then;
(i)  \alpha {\rm{ }} + {\rm{ }}\beta {\rm{ }} = \frac{{--b}}{a}  (ii)  \alpha {\rm{ }}{\rm{.}}\beta {\rm{ }} = \frac{c}{a}     (iii)  \alpha {\rm{ - }}\beta {\rm{ }} = \frac{{\sqrt D }}{a}
Nature  Of  Roots:
(a) Consider the quadratic equation ax² + bx + c = 0  where a, b, c  \in  R & a \ne 0 then
(i) D > 0   \Leftrightarrow  roots  are  real & distinct  (unequal).
(ii) D = 0  \Leftrightarrow  roots  are  real & coincident  (equal).
(iii) D < 0 \Leftrightarrow  roots  are  imaginary
(B) Consider the quadratic equation ax2+ bx + c = 0 where a, b, c  \in  Q & a \ne 0 then
 If  D > 0  &  is a perfect  square , then  roots  are  rational & unequal.
A quadratic  equation  whose  roots  are  a & b  is  (x – a)(x – b) = 0  i.e.   x2 – (a + b) x + a b = 0 i.e.
                          x2 – (sum of  roots) x +  product  of  roots = 0
Consider  the  quadratic  expression , y = ax² + bx + c  , a, b, c  \in  R & a \ne 0  then
(i) The graph between x, y  is always a  parabola.  If a > 0  then the shape of the parabola is concave upwards &  if a < 0  then the shape of the parabola is concave downwards.
Common  Roots  Of  2  Quadratic  Equations  [Only  One  Common  Root]-   Let  \alpha
be  the  common  root  of  ax² + bx + c = 0  &  a’x2 + b’x + c’ = 0
Therefore    \;a{\alpha ^2}{\rm{ }} + {\rm{ }}b\alpha {\rm{ }} + {\rm{ }}c{\rm{ }} = {\rm{ }}0{\rm{ }}\;and{\rm{ }}\;a{\alpha ^2} + {\rm{ }}b\alpha {\rm{ }} + {\rm{ }}c{\rm{ }} = {\rm{ }}0
 If we solve above pair by cramer’s rule we get

                                            \frac{{{\alpha ^2}}}{{bc' - cb'}} = \frac{\alpha }{{ac' - c'a}} = \frac{1}{{ab' - a'b}}

This will give us                    \alpha  = \frac{{bc' - cb'}}{{ac' - c'a}} = \frac{{ac' - c'a}}{{ab' - a'b}}

                                            {(ac' - c'a)^2} = (ab' - a'b)(bc' - cb')

Every pair of the quadratic equation whose coefficients fulfils the above condition will have one root in common.

The condition that a quadratic function-  
                                           f(x , y) = ax² + 2 hxy + by² + 2 gx + 2 fy + c  may be  resolved  into  two  linear  factors  is  that     abc{\rm{ }} + {\rm{ }}2{\rm{ }}fgh{\rm{ }} - a{f^2} - b{g^2} - c{h^2}{\rm{ }} = {\rm{ }}0{\rm{ }}\; or
                                                                            \left| {\begin{array}{ccccccccccccccc}
a&h&g\\
h&b&f\\
g&f&c
\end{array}} \right| = 0
Reducible Quadratic Equations-These are the equations which are not quadratic in their initial condition but after some calculations, we can reduce them into quadratic equations
(i) If the power of the second term is exactly half to the power of the first term and the third term is a constant, these types of equations can be reduced to quadratic equations.
i.e. x + \sqrt x  - 6 = 0 , {x^6} + {x^3} - 6 = 0 , {e^{2x}} + {e^x} - 6 = 0 ,{a^{2x}} + {a^x} - 6 = 0

{\left( {2x + \frac{3}{{2x}}} \right)^2} + \left( {2x + \frac{3}{{2x}}} \right) - 6 = 0  all these equations can easily be reduced into quadratic equations by applying the method of substitution.

Example-Solve this equation and find x  {\left( {2x + \frac{3}{{2x}}} \right)^2} + \left( {2x + \frac{3}{{2x}}} \right) - 6 = 0 

Ans:

let y=  \left( {2x + \frac{3}{{2x}}} \right) then given equation will become

 

                                   {y^2} + y - 6 = 0  it’s a simple quadratic equation we can be easily factorised it and solve  so y=–3,2

so  \left( {2x + \frac{3}{{2x}}} \right)=-3

                                    \frac{{4{x^2} + 3}}{{2x}} =  - 3

                               \begin{array}{l}
4{x^2} + 3 =  - 6x\\
4{x^2} - 6x + 3 = 0
\end{array}

the final equation can be solved using Quadratic formula and the same process can be repeated for  y=2

(ii) If a variable is added with its own reciprocal, then we get a quadratic equation i.e x + \frac{1}{x} - 6 = 0,   \frac{{2x + 3}}{{x - 2}} + \frac{{x - 2}}{{2x + 3}} = 0 {e^{2x}} + \frac{1}{{{e^x}}} - 6 = 0  {a^{2x}} + \frac{1}{{{a^x}}} - 6 = 0 all these equations can be reduced into quadratic by replacing one term by any other variable.

 

Standard Form of a Quadratic Function-A quadratic function y=ax2+ bx + c can be

reduced into standard form    y = a{(x - h)^2} + k  by method of completing the square. If we

draw the graph of this function we shall get a parabola with vertex (h,k). The parabola will be upward for a>0 and downward for a<0

 

Maximum and Minimum value of a quadratic function- If the function is in the form

y = a{(x - h)^2} + k Then ‘h’ is the input value of the function while ‘k’ is its output.

(i) If a>0 (in case of upward parabola) the minimum value of f is f(h)=k

(ii) If a<0 (in case of downward parabola)the maximum value of f is f(h)=k
If our function is in the form of  y=ax² + bx + c then vertex of the parabola V = \left( { - \frac{b}{{2a}},\frac{D}{{4a}}} \right)
The line passing through vertex and parallel to the y-axis is called the axis of symmetry.
The parabolic graph of a quadratic function is symmetrical about axis of symmetry.

 \Rightarrow  f(x) has a minimum value at vertex if a>0 and {f_{\min }} =  - \frac{D}{{4a}}  at   x =  - \frac{b}{{2a}}

 

 \Rightarrow   f(x) has a maximum value at vertex if a<0 and   {f_{\min }} =  - \frac{D}{{4a}}  at   x =  - \frac{b}{{2a}}

 

In the next post about quadratics, I shall discuss discriminant, nature of roots, relationships between the roots. In the meantime, you can download the pdf and solve practice questions.

 quadratic equation 200 questions.pdf

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 quadratic equation and function.pdf

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