SAT MATH: PASSPORT TO ADVANCE MATH
This section requires you to raise your comfort zone. Be ready for conceptual questions in this section. You will need patience and practice to master this section.
#MASTERING POLYNOMIALS
You need to be comfortable with polynomials. These formulas will help you a lot while working with them -
EX) (X3 + 3X2 + B)(X + 2) = X4 + 5X3 + 6X2 +6X + 12
In the above equation, B is a constant. If the equation is true for all value of X find B.
ANS 1) CLASSICAL METHOD
Solve LHS
(X3 + 3X2 + B)(X + 2)
Equating it with RHS
X4 + 5X3 + 6X2 +BX + 2B = X4 + 5X3 + 6X2 +6X + 12
BX = 6X and 2B = 12
→ B = 6
ANS 2) THE METHOD TO DO
Let X = 0(we can put any value to get our answer)
LHS = (X3 + 3X2 + B)(X + 2)
= (0+0+B)(0+2) = 2B
RHS = X4 + 5X3 + 6X2 + 6X + 12
= 0 + 0 + 0 + 0 + 12
= 12
Here putting X = 0 was most favorable but if it wasn't so, you could have assigned it any other value since B is a constant and X is a variable the value of B won't change for any value of X.
Now questions could come which would involve the formulas listed above
EX) Find (7X2 – 3Y) (7X2 + 3Y)
ANS) it is (7X2)2 – (3Y)2
which is equal to 49X4 – 9Y2
The reverse of which could have also been there. Remember time is your greatest enemy in this exam.
ARITHMETIC OPERATIONS WITH POLYNOMIAL
The polynomials with same power of variables can be added and subtracted.
X + X = 2X
As X and Y are different
but
Now let's come to multiplication
The power of variable gets added
Similarly
CREATING QUADRATIC EQUATIONS
EX) A car started moving from an initial velocity 5M/S and has an acceleration of 3M/S2 if in t sec it travelled 200M then find t.
Given the Equation of motion is S = ut + 1/2at2
ANS) Initial velocity = u = 5M/S
Displacement = S = 200M
time = t
acceleration = a = 3M/S2
equation of motion is S = ut + 1/2at2
200 = 5t + 3/2t2
SUM OF SOLUTIONS
You can be asked to find the sum of the solution of an equation. in it you just have to add all the solutions of an equation.
EX) Find the sum of solutions of X2 + 5X + 6 = 0
In this question we will do splitting the middle terms
X2 + 2X + 3X + 6 = 0
X(X + 2) + 3(X + 2) = 0
(X + 3)(X + 2) = 0
X = -3, -2
So sum of solution is X = (-3) + (-2) = -5
Here in the question of sum of solutions we cannot plug in the options as they will be sum of the 2 roots, not the roots.
CREATING EXPONENTIAL EQUATION FROM STATMENTS
EX) A bacteria grows at the rate of 30% per hour. The researcher wants to estimate that if at the beginning of the experiment there were 20 bacterias then how many bacteria will be there after t hours.
ANS) At the beginning there are 20 bacteria
there is a growth rate of 30% in 1 hour
so after 1 hour bacteria will be
= 20(1 + 30/100)
= 20(1.3)
After 2 hours there will be
= 20(1.3)(1 + 30/100)
= 20(1.3)(1.3)
= 20(1.3)2
After 3 hours there will be
= 20(1.3)2(1 + 30/100)
= 20(1.3)3
So if you can notice
1 hour = 20(1.3)1
3 hour = 20(1.3)3
similarly for t hour = 20(1.3)t
Thus the general equation will be
bacteria after t hour = 20(1.3)t
Now i don't want you to do these questions in this way in exam. I want you to directly understand this in this way-
Bacteria grows with time as = 20(1.3)(1.3)(1.3)(1.3).............
For t hour = 20(1.3)t
Exponenential function like Y = a(1 + r)t is for increasing
while Y = a(1 - r)t is for decreasing
EX) Solve the algebraic expression
(X3/2/X)1/2.(X-5/4/X3).(X-1/2)2.Y2
ANS) (X3/2/X)1/2.(X-5/4/X3).(X-1/2)2.Y2
This may seem difficult to you but notice what i had done i kept X and Y separate then i followed these simple steps
And with help of them keep simplifying. While simplifying always remember that the power of bracket is to be multiplies with both numerator and denominator.
# SOLVING EQUATIONS WITH VARIABLE ON DENOMINATOR
They are of the form
Now let's work on it with an example
Eg -
Multiplying both sides by X+3 and x+5
(X + 5) + (X + 3)(X + 2) = 1(X + 3)(X + 5)
X + 5 + X2 + 5X + 6 = X2+ 8X + 15
6X + 11 = 8X + 14
-4 = 2X
X = -2
Here the final equation came to be linear thus we could easily solve it but in SAT it may had been quadratic so be ready to use quadratic formula.
I would always recommend you to plug in the answers in it.
SOLVING SYSTEM OF 1 LINEAR AND 1 QUADRATIC EQUATION
Now there is a question that can come in SAT which is quite tough is solving a system of equations with 1 linear and 1 quadratic equation.
Let's take an example
EX) Solve the system of equations
x = y -1
y2+ 2y - 2x = 11
From equation 1 x = y -1
substituting in equation 2
y2+ 2y - 2y + 2 = 11
y2 = 9
y = +3,-3
Thus solutions are (3,2) and (-3,-4)
If needed, we can also use the quadratic formula if there was a quadratic equation at the end.
FUNCTION
Without going too much in depth a function just provides an output for every input
Examples of functions
ALGEBRAIC FUNCTION - y = bxa + c; Where X and Y are variable
Here let a = 3, b = 1, and c = 2
So, y = x3 + 2
For X = 0, Y = 2
X = 1, Y = 3
X = 2, Y = 10
Here 0,1 and 2 which are the values of X are called as DOMAIN. Domain are those inputs which we discussed about in function definition
While the values 2,3 and 10 are called as RANGE. Range is the output we get from a function.
So basically a function does arithmetic operations with domain and provides us with the desired value called range.
If you still couldn't understand it then let's discuss it with another example.
EXPONENTIAL FUNCTION -
Y = A(B)X + C
Where X is a variable abd a.b.c are constant. So let's give them values a = 1, b = 2 and c = 3
Y = 2X + 3
Here I can put any value of X thus domain ∈ R. That is domain belongs to Real numbers which mean domain consists of all the real numbers.
Now let's think about the range
can it be 0? 2X No is always going to be positive
and thus can be - 3.
Range does not belongs to R, but whatever values you get on putting domain, it is going to be range. Let's think about the maximum and minimum values it can have
at X = infinity(∞)
Y = 2∞ + 3
➝ Y = ∞
Since nothing can be bigger than infinity thus it is the maximum value.
Now let's think about it's minimum value for Y = 2X + 3
#MASTERING POLYNOMIALS
You need to be comfortable with polynomials. These formulas will help you a lot while working with them -
(a + b)2
= a2 + b2 + 2ab
(a - b)2
= a2 + b2 - 2ab
a2
- b2 = (a + b)(a – b) EX) (X3 + 3X2 + B)(X + 2) = X4 + 5X3 + 6X2 +6X + 12
In the above equation, B is a constant. If the equation is true for all value of X find B.
ANS 1) CLASSICAL METHOD
Solve LHS
(X3 + 3X2 + B)(X + 2)
= X4 + 2X3+ 3X3
+ 6X2 +BX + 2B
= X4 + 5X3 + 6X2
+BX + 2B
Equating it with RHS
X4 + 5X3 + 6X2 +BX + 2B = X4 + 5X3 + 6X2 +6X + 12
BX = 6X and 2B = 12
→ B = 6
ANS 2) THE METHOD TO DO
Let X = 0(we can put any value to get our answer)
LHS = (X3 + 3X2 + B)(X + 2)
= (0+0+B)(0+2) = 2B
RHS = X4 + 5X3 + 6X2 + 6X + 12
= 0 + 0 + 0 + 0 + 12
= 12
→ 2B = 12
→ B = 6
Here putting X = 0 was most favorable but if it wasn't so, you could have assigned it any other value since B is a constant and X is a variable the value of B won't change for any value of X.
Now questions could come which would involve the formulas listed above
EX) Find (7X2 – 3Y) (7X2 + 3Y)
ANS) it is (7X2)2 – (3Y)2
which is equal to 49X4 – 9Y2
The reverse of which could have also been there. Remember time is your greatest enemy in this exam.
ARITHMETIC OPERATIONS WITH POLYNOMIAL
The polynomials with same power of variables can be added and subtracted.
X + X = 2X
2X2 + 5X2 = 7X2
X2 + 3Y2 = X2 + 3Y2
As X and Y are different
but
2XY2 + 3XY2 = 5XY2
Because variable in both of them is XY2 Now let's come to multiplication
The power of variable gets added
X2
× X = X2 + 1 = X3
X2
× Y2 = X2Y2
If you
couldn’t understand this then let me help you
X2.Y0
× X0.Y2 = X2+0.Y2+0 = X2Y2
As X0 = Y0 = 1
Similarly
X2Y
× XY = X2+1.Y1+1
= X3 Y2
CREATING QUADRATIC EQUATIONS
EX) A car started moving from an initial velocity 5M/S and has an acceleration of 3M/S2 if in t sec it travelled 200M then find t.
Given the Equation of motion is S = ut + 1/2at2
ANS) Initial velocity = u = 5M/S
Displacement = S = 200M
time = t
acceleration = a = 3M/S2
equation of motion is S = ut + 1/2at2
200 = 5t + 3/2t2
3t2
+ 10t – 400 = 0
Since time cannot be negative so -80/6 is rejected
and thus time = t = 10S
SUM OF SOLUTIONS
You can be asked to find the sum of the solution of an equation. in it you just have to add all the solutions of an equation.
EX) Find the sum of solutions of X2 + 5X + 6 = 0
In this question we will do splitting the middle terms
X2 + 2X + 3X + 6 = 0
X(X + 2) + 3(X + 2) = 0
(X + 3)(X + 2) = 0
X = -3, -2
So sum of solution is X = (-3) + (-2) = -5
Here in the question of sum of solutions we cannot plug in the options as they will be sum of the 2 roots, not the roots.
CREATING EXPONENTIAL EQUATION FROM STATMENTS
EX) A bacteria grows at the rate of 30% per hour. The researcher wants to estimate that if at the beginning of the experiment there were 20 bacterias then how many bacteria will be there after t hours.
ANS) At the beginning there are 20 bacteria
there is a growth rate of 30% in 1 hour
so after 1 hour bacteria will be
= 20(1 + 30/100)
= 20(1.3)
After 2 hours there will be
= 20(1.3)(1 + 30/100)
= 20(1.3)(1.3)
= 20(1.3)2
After 3 hours there will be
= 20(1.3)2(1 + 30/100)
= 20(1.3)3
So if you can notice
1 hour = 20(1.3)1
3 hour = 20(1.3)3
similarly for t hour = 20(1.3)t
Thus the general equation will be
bacteria after t hour = 20(1.3)t
Now i don't want you to do these questions in this way in exam. I want you to directly understand this in this way-
Bacteria grows with time as = 20(1.3)(1.3)(1.3)(1.3).............
For t hour = 20(1.3)t
Exponenential function like Y = a(1 + r)t is for increasing
while Y = a(1 - r)t is for decreasing
EX) Solve the algebraic expression
(X3/2/X)1/2.(X-5/4/X3).(X-1/2)2.Y2
ANS) (X3/2/X)1/2.(X-5/4/X3).(X-1/2)2.Y2
= (X1/2)1/2.(1
/(X3. X5/4)).(X-1).Y2
= (1 /(X3.
X)).(1/X).Y2
= Y2/X5
= X-5Y2
This may seem difficult to you but notice what i had done i kept X and Y separate then i followed these simple steps
(XA)B
= XAB
XA.XB
= XA+B
XA/XB
= XA-B
XA.YB
= XA.YB
And with help of them keep simplifying. While simplifying always remember that the power of bracket is to be multiplies with both numerator and denominator.
# SOLVING EQUATIONS WITH VARIABLE ON DENOMINATOR
They are of the form
Eg -
(X + 5) + (X + 3)(X + 2) = 1(X + 3)(X + 5)
X + 5 + X2 + 5X + 6 = X2+ 8X + 15
6X + 11 = 8X + 14
-4 = 2X
X = -2
Here the final equation came to be linear thus we could easily solve it but in SAT it may had been quadratic so be ready to use quadratic formula.
I would always recommend you to plug in the answers in it.
SOLVING SYSTEM OF 1 LINEAR AND 1 QUADRATIC EQUATION
Now there is a question that can come in SAT which is quite tough is solving a system of equations with 1 linear and 1 quadratic equation.
Let's take an example
EX) Solve the system of equations
x = y -1
y2+ 2y - 2x = 11
From equation 1 x = y -1
substituting in equation 2
y2+ 2y - 2y + 2 = 11
y2 = 9
y = +3,-3
Thus solutions are (3,2) and (-3,-4)
If needed, we can also use the quadratic formula if there was a quadratic equation at the end.
FUNCTION
Without going too much in depth a function just provides an output for every input
Examples of functions
ALGEBRAIC FUNCTION - y = bxa + c; Where X and Y are variable
Here let a = 3, b = 1, and c = 2
So, y = x3 + 2
For X = 0, Y = 2
X = 1, Y = 3
X = 2, Y = 10
Here 0,1 and 2 which are the values of X are called as DOMAIN. Domain are those inputs which we discussed about in function definition
While the values 2,3 and 10 are called as RANGE. Range is the output we get from a function.
So basically a function does arithmetic operations with domain and provides us with the desired value called range.
If you still couldn't understand it then let's discuss it with another example.
EXPONENTIAL FUNCTION -
Y = A(B)X + C
Where X is a variable abd a.b.c are constant. So let's give them values a = 1, b = 2 and c = 3
Y = 2X + 3
Here I can put any value of X thus domain ∈ R. That is domain belongs to Real numbers which mean domain consists of all the real numbers.
Now let's think about the range
can it be 0? 2X No is always going to be positive
and thus can be - 3.
Range does not belongs to R, but whatever values you get on putting domain, it is going to be range. Let's think about the maximum and minimum values it can have
at X = infinity(∞)
Y = 2∞ + 3
➝ Y = ∞
Since nothing can be bigger than infinity thus it is the maximum value.
Now let's think about it's minimum value for Y = 2X + 3
2X is a variable, so if x is negative it's value will decrease
at x → -∞ it will be 1/2∞= 1/∞ = 0. thus it has minimum value of 3.
Now the maximum value of a function is called as MAXIMA while the minimum value is called MINIMA.
INTERCEPT - The point where the function touches axes.
X - INTERCEPT - The point where the function touches X - axis.
Y - INTERCEPT - The point where the function touches Y - axis.
Now let's think how can you find an intercept without seeing it's graph.
EX) Find X - intercept of function Y = -2x2 + 8
ANS) So at X - intercept the function will touch X - Axis thus Y coordinate is 0. So
Y = -2x2 + 8
0 = -2x2 + 8
X = ± 2
This means there will be 2 X - intercepts at (2,0) and (-2,0).
If you are thinking why Y is 0 at X - intercept then it's because X and Y are perpendicular and meet each other at 0. Thus if it touches one the other by default is 0.
ZEROES AND FACTORS
The concept about factors remains same as we have learned in our elementary school but then the difference is now it's with variables.
Factors are expressions from which our function is divisible
Like for the function Y = x2 - 4
X -2 ans X + 2 are the factors.
Now coming to zeroes, they are values of X for which the value of function is 0. It can be pretty much understood by it's name.
The easy way to do is put the factors = 0. like in above function
X - 2 = 0 and X + 2 = 0
and you will get X = +2,-2
You can verify at these values of X the Y will be 0.
END BEHAVIOR
You must be knowing that the number line begins from -∞ and goes to +∞. These are in some ways, the end of the number line. The behavior of a function while approaching - ∞ and + ∞ is termed as End Behavior.
VARIABLE AND CONSTANT SEGREGATION
Sometimes, specially in Algebraic function you will be required to segregate constants and variable to find maximum and minimum value.
EX) Find the minimum value of x2 + 4x + 100 = 0
#Fog(X)
In this, you will be asked questions from a function whose domain will be the range of another function.
Let, g(X) = x2
and f(X) = 3x2 + 1
So fog(X) = 3(g(X))2 + 1
Substituting value of g(X)
fog(X) = 3(x2)2 + 1
It may seem tricky in the beginning but it's quite easy let's take another example.
g(X) = 3x + 1
f(X) = 5x + 2
Substituting value of g(X)
f(X) = 5(3x + 1) + 1
Thus you can see it's very simple.
INCREASING FUNCTION - The function whose Y keeps on increasing with the value of X.
For EX : X, 3x,X3
DECREASING FUNCTION - The function whose Y keeps on decreasing with the value of X.
For EX : -5x , 2-x ,-X
SAT HOMEPAGE
SAT EXAM FORMAT
SAT READING
SAT WRITING
SAT MATH
SAT MATH: HEART OF ALGEBRA
SAT MATH: PROBLEM SOLVING AND DATA ANALYSIS
SAT MATH: ADDITIONAL TOPIC IN MATH
SAT MATH: PASSPORT TO ADVANCE MATH
SAT MATH:GRAPH
SAT ESSAY
at x → -∞ it will be 1/2∞= 1/∞ = 0. thus it has minimum value of 3.
Now the maximum value of a function is called as MAXIMA while the minimum value is called MINIMA.
INTERCEPT - The point where the function touches axes.
X - INTERCEPT - The point where the function touches X - axis.
Y - INTERCEPT - The point where the function touches Y - axis.
Now let's think how can you find an intercept without seeing it's graph.
EX) Find X - intercept of function Y = -2x2 + 8
ANS) So at X - intercept the function will touch X - Axis thus Y coordinate is 0. So
Y = -2x2 + 8
0 = -2x2 + 8
2x2
= 8
x2
= 4
X = ± 2
This means there will be 2 X - intercepts at (2,0) and (-2,0).
If you are thinking why Y is 0 at X - intercept then it's because X and Y are perpendicular and meet each other at 0. Thus if it touches one the other by default is 0.
ZEROES AND FACTORS
The concept about factors remains same as we have learned in our elementary school but then the difference is now it's with variables.
Factors are expressions from which our function is divisible
Like for the function Y = x2 - 4
X -2 ans X + 2 are the factors.
Now coming to zeroes, they are values of X for which the value of function is 0. It can be pretty much understood by it's name.
The easy way to do is put the factors = 0. like in above function
X - 2 = 0 and X + 2 = 0
and you will get X = +2,-2
You can verify at these values of X the Y will be 0.
END BEHAVIOR
You must be knowing that the number line begins from -∞ and goes to +∞. These are in some ways, the end of the number line. The behavior of a function while approaching - ∞ and + ∞ is termed as End Behavior.
VARIABLE AND CONSTANT SEGREGATION
Sometimes, specially in Algebraic function you will be required to segregate constants and variable to find maximum and minimum value.
EX) Find the minimum value of x2 + 4x + 100 = 0
ANS) By completing the square we get (x + 2)2 + 96 = 0
Here you can see that minimum value is 96 at X = -2 But seeing is possible only when you will be able to segregate constants and variables.
#Fog(X)
In this, you will be asked questions from a function whose domain will be the range of another function.
Let, g(X) = x2
and f(X) = 3x2 + 1
So fog(X) = 3(g(X))2 + 1
Substituting value of g(X)
fog(X) = 3(x2)2 + 1
It may seem tricky in the beginning but it's quite easy let's take another example.
g(X) = 3x + 1
f(X) = 5x + 2
Substituting value of g(X)
f(X) = 5(3x + 1) + 1
Thus you can see it's very simple.
INCREASING FUNCTION - The function whose Y keeps on increasing with the value of X.
For EX : X, 3x,X3
DECREASING FUNCTION - The function whose Y keeps on decreasing with the value of X.
For EX : -5x , 2-x ,-X
BONNE CHANCE !
SAT HOMEPAGE
SAT EXAM FORMAT
SAT READING
SAT WRITING
SAT MATH
SAT MATH: HEART OF ALGEBRA
SAT MATH: PROBLEM SOLVING AND DATA ANALYSIS
SAT MATH: ADDITIONAL TOPIC IN MATH
SAT MATH: PASSPORT TO ADVANCE MATH
SAT MATH:GRAPH
SAT ESSAY
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