Category: IB Mathematics Tutors

IB Elite Tutor provides well trained, well qualified, experienced and result oriented Ib Mathematics Tutors.
Our IB Mathematics Tutors use the latest technology i.e Pdf, images, videos, and presentations to explain difficult concepts of ib Mathematics HL/SL. We do cumulative testing to ensure that all our students get a “7” in exams
Our testing is done on three levels

(i) Chapter end test to check progress in one chapter. This is done at the end of every chapter.

(ii) Unit test to check progress in One chapter. This is done at the end of three chapters.

(i) Full syllabus test to check progress in the full course. This is done at the end of the syllabus.

Complex Numbers

April 17, 2018 admin Leave a comment

Complex Numbers-

Complex numbers come into existence when the square of a number is negative because we know it very well that the square of a number will always be positive doesn’t matter whether the number is positive or negative.

In cases like ${x^2} + 1 = 0$ or $2{x^3} + 5x = 0$ here, if we solve, we find the square of x=-1. We say that x is not real here. Generally, these types of cases are considered as Complex numbers. Complex numbers were first observed by mathematician Girolamo Cardano (1501-1575). In his book Ars Magna, he discussed the mechanics of complex numbers in details and thus he started Complex Algebra.

Standard form of Complex Numbers-

Complex numbers are defined as expressions of the form a + ib where a,b $\in$ R & i = $\sqrt { - 1}$

It is denoted by Z i.e. z= a + ib.

‘a’ is called as real part of z= (Re z)

and ‘b’ is called as imaginary part of z =(Im z).

i or IOTA- iota is a unique symbol. it’s the ninth letter of Latin alphabet. It’s used to denote imaginary numbers whose square root is -1.
Click here to download the book “An Imaginary Tale The Story of i” a very interesting book on iota by Paul J. Nahin. Read more

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Principle of Mathematical Induction

April 8, 2018 admin Leave a comment

Principle of Mathematical Induction:-

we know that the first positive even integer is 2 = 2 $\times$ 1
the second positive even integer is 4 = ${\rm{ }}2 \times 2$

the third positive even integer is 6 = 2 $\times$ 3
the fourth positive even integer is 8 = 2 $\times$ 4

………………………………………………….

If we move using the same pattern then nth positive integer=2n Read more

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Permutations and Combinations-algebra tutors

November 1, 2017 admin Leave a comment

Permutations and Combinations(part-2)

In my previous post, we discussed the fundamental principle of counting and various methods of permutations. In this post, I shall discuss combinations in details.

Meaning of Combination- If we are given a set of objects and we want to select a few objects out of this set, then we can do it by many different ways. These ways are known as combinations.
Example- If we are given three balls marked as B, W and R and we want to select two balls then we can select like this- BW, BR, WR.

These are known as the combination of this selection.

Combination of n different objects taken r at a time when repetition is not allowed– If repetition is not allowed the number of ways of selecting r objects out of a group of n objects is called ${}^n{c_r}$

${}^n{c_r}$ = $\frac{{n!}}{{r!\left( {n - r} \right)!}}$

In latest notation system ${}^n{c_r}$ is also known as C(n;r) or $\left( \begin{array}{l} n\\ \\ r \end{array} \right)$

Properties of $\left( \begin{array}{l} n\\ \\ r \end{array} \right)$ – It’s a very useful and interesting Mathematical tool. It has following properties.

(i) ${}^n{c_r}$ = ${}^n{c_{n - r}}$

(ii) ${}^n{c_n} = {}^n{c_0} = 1$

(iii) ${}^n{c_r} + {}^n{c_{r - 1}} = {}^{n + 1}{c_r}$ known as Pascal’s law

(iv) r. ${}^n{c_r} = n{}^{n - 1}{c_{r - 1}}$

(v) $\frac{{{}^n{c_r}}}{{{}^n{c_{r - 1}}}} = \frac{{n - r + 1}}{r}$

(vi) If n is even then we should put r=n/2 for maximum value of ${}^n{c_r}$ and if n is odd then ${}^n{c_r}$ is greatest when r= $\frac{{{n^2} - 1}}{4}$

(vii) In the expansions of ${(1 + x)^n}$ if we put x=1 then

${}^n{c_0} + {}^n{c_1} + {}^n{c_2} + ....... + {}^n{c_n} = {2^n}$

${}^n{c_0} + {}^n{c_2} + {}^n{c_4} + ......... = {2^{n - 1}}$

${}^n{c_1} + {}^n{c_3} + {}^n{c_5} + ......... = {2^{n - 1}}$ Read more

Definite Integration-Topics in IB Mathematics

October 8, 2017 admin One comment

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

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

Some More Properties of Definite Integration:- Read more

Indefinite Integration-Topics in IB Mathematics

October 5, 2017 admin One comment

Indefinite Integration

After a long series on differentiation and ‘Application of derivatives‘, we shall now discuss Indefinite Integration. It consists of two different words indefinite and integration.
First of all, we shall learn about Integration.

Integration is the reverse process of differentiation so we can also call it as antiderivative. There is one more name for it, that is Primitive.
If f & g are functions of x such that g'(x) = f(x) then the function g is called a Primitive Or Antiderivative Or Integral of f(x) w.r.t. x and is written symbolically as:-

$\int {f\left( x \right){\rm{ }}dx = g\left( x \right) + c}$

If $\frac{d}{{dx}}\left\{ {f(x) + c} \right\} = f'(x)$

then $\int {f'\left( x \right){\rm{ }}dx = f\left( x \right) + c}$ here c is just an arbitrary constant. Value of c is not definite that’s why we call it Indefinite Integration.

Techniques Of Integration-: There are a few important techniques used to solve problems based on integration

(i) Substitution or Change of Independent Variable- If the derivative of a function is given in the question, then we should use the method of substitution to integrate that question. Read more

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Increasing and Decreasing Functions

October 2, 2017 admin Leave a comment

Increasing and decreasing functions

This is my third post in the series of “Applications of derivatives”. The previous two were based on “Tangent and Normal” and “Maxima and Minima”.In this post, we shall learn about increasing and decreasing functions. That is one more application of derivatives.

Increasing and Decreasing Functions- We shall first learn about increasing functions

Increasing Function-

(a) Strictly increasing function- A function f (x) is said to be a strictly increasing function on (a, b) if x₁< x2 $\Rightarrow$ f(x₁) < f (x₂) for all x_l, x₂ $\in$ (a, b).Thus, f(x) is strictly increasing on (a, b) if the values of f(x) increase with the increase in the values of x.Refer to the graph in below-given figure $\Downarrow$

IB Mathematics Tutors

IB Mathematics HL SL-Maxima and Minima

September 28, 2017 admin 3 comments

In my previous post, we discussed how to find the equation of tangents and normal to a curve. There are a few more Applications of Derivatives in IB Mathematics HL SL, ‘Maxima and Minima’ is one of them.

Maxima and Minima:-

1. A function f(x) is said to have a maximum at x = a if f(a) is greater than every other value assumed by f(x) in the immediate neighbourhood of x = a. Symbolically

$\left. \begin{array}{l} f(a) > f(a + h)\\ f(a) > f(a - h) \end{array} \right] \Rightarrow x = a$ gives maxima for a sufficiently small positive h.

Similarly, a function f(x) is said to have a minimum value at x = b if f(b) is least than every other value assumed by f(x) in the immediate neighbourhood at x = b. Symbolically

$\left. \begin{array}{l} f(b) > f(b + h)\\ f(b) > f(b - h) \end{array} \right]$ If x = b gives minima for a sufficiently small positive h.

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Applications of Derivatives in IB Mathematics

September 18, 2017 admin Leave a comment

Applications of Derivatives in IB

Mathematics-

In my previous post, we discussed how to find the derivative of different types of functions as well as the geometrical meaning of differentiation. Here we are discussing Applications of Derivatives in IB Mathematics
There are many different fields for the Applications of Derivatives. We shall discuss a few of them-

Slope and Equation of tangents to a curve- If We draw a tangent to a curve y=f(x) at a given point $({x_1},{y_1})$ , then

The gradient of the curve at given point=the gradient of the tangent line at given point

and we already discussed that slope or gradient of the tangent at given point $({x_1},{y_1})$

m= ${\frac{{dy}}{{dx}}_{at({x_1},{y_1})}}$

= $f'$ ( ${x_1}$ )

Finally to find the equation of tangent we use the slope-point form of equation

$y - {y_1} = m(x - {x_1})$

The major part of this concept is also discussed in the previous post. We should also remember following points while solving these types of questions.

(i) If two lines are parallel to each other, their slopes are always equal
i.e ${m_1} = {m_2}$
(ii) If two lines are perpendicular to each other, the product of their slopes is always -1

${m_1}.{m_2} = - 1$

(iii) If a line is passing through two points $({x_1},{y_1})$ and $({x_2},{y_2})$ then, slope of the line

$m = \frac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}$

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Limit, Continuity & Differentiability-IB Maths Topics

September 3, 2017 admin Leave a comment

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.

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) e^x =1+x/1!+x³/3!+x⁴/4!…… $\infty$

(ii) a^x=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/

Quadratic equations and Quadratic Functions

July 8, 2017 admin 2 comments

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

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 the 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 the 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

quadratic equation and function.pdf

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