TRIGONOMETRY

How do I prove that tan70×tan20=1?
tan90=tan(70+20)=(tan70+tan20)/(1-tan70tan20)

but tan90=∞=1/0

or

1/0=tan90=tan(70+20)=(tan70+tan20)/(1-tan70tan20)

solving

0=(1-tan70tan20)

tan70tan20=1 

hence proved

What is the value of Sin 15+Cos 15?
sin15+cos15 =I

squring both sides

sin²15+cos²15+2sin15cos15=I²

1+sin30=I²

1+(1/2)=I²

3/2=I²

I=√(3/2)

FUNCTIONS

• State the domain and range of the following relation. Is the relation a function?

{(2, –3), (4, 6), (3, –1), (6, 6), (2, 3)}

The domain is all x -values.

and

Range is all y-values.

Hence in the above sets

Domain {2,4,3,6,2}

and

Range {-3,6,-1,6,3}

for  function {(2,1),(10,5)}

hence 
Domain {2,10}
and
Range {1,5}

Function and Relations

Determine the Domain of the function:

y=1/(x²+3x+2)

y is defined for all value of x except

where (x²+3x+2)=0

x²+2x+x+2=0

from this x=-2 and x=-1

hence the range is all real value of x except -1 and -2

Determine the Domain of the function:

y=x/√(x²-3x+2)

y is defined for
where (x²-3x+2)>0  here we take greater than zero(>)                    because the square root is in the denominator. If equal to zero then the (x²-3x+2)=0 and y=∞, which is not possible.                                               

x²-2x-x+2>0

x(x-2)-1(x-2)>0

(x-1)(x-2)>0

from this we conclude that

x<1 and x>2

hence the range of the function is x<1 and x>2

Determine the Domain of the function:

y=√(x²-5x+6)

y is defined for
where (x²-5x+6)≥0       

here we take greater than equal to zero(≥)                                           because the square root is in the numerator                                                         . 

x²-3x-2x+6≥0

x(x-3)-2(x-3)≥0

(x-3)(x-2)≥0

from this we conclude that

x≤2 and x≥3

Determine the range of the function:

y=x²+4

Range is the all y value.

To find the range write x in term of x

i.e.

y-4=x²

or

x=√(y-4)

for x to be defined (y-4)≥0

or

y≥0

hence the range of the function is y≥0

Determine the range of the function:

y-4=4/x²

Range is the all y value.

for range:

write the x in terms of y

i.e.

x²=4/(y-4)

or

x=2/√(y-4)

for x to be defined (y-4)>0

or

y>4

hence the range of function is y>4

Determine the domain and range of the y=9/(√(x²-4))

for domain

for y to be defined x²-4>0

from this

(x-2)(x+2)>0

or

x>2 and x<-2

hence the domain of the function x>2 and x<-2

for range

for x to be defined

write x in terms of y

i.e. 

x²-4=(81/y²)

or

x²=(81/y²)+4

or

x=√((81/y²)+4)

x is defined for all value of y except when y=0

hence range is all real value of y except when y=0

What is the domain and range of the function f(x)=1/x+3
Given function

f(x)=1/x+3

Domain is the all x value for which f(x) is defined, hence the f(x) is defined for all real number except at x=0

Domain= for all real number except x=0

Range

All value of y for which x is defined

To find range write x in terms of y

y=1/x+3

y-3=1/x

x=1/(y-3)

hence x is defined for all real number except when y-3=0 i.e. y=3

Range: All real number except y=3

If f(x)=2x+2 , then what is f(f(3)) equal to?

f(x)=2x+3

now

f(f(x))=2f(x)+3

Now

f(f(3))=2f(3)+3

but 

f(3)=2×3+2=8

f(f(x))=2f(x)+3=2×8+3

f(f(x))=19

What is the domain and range of √9+x^2?

y=√9+x²

Domain is all value of x for y is defined.

9+x²≥0 which is true for all value of x.

hence domain is all real value.

and 

Range

y²=9+x²

or x=√y²-9

range is all value of y for which x is defined

y²-9≥0

or (y-3)(y+3)≥0

or

range is⟦-∞,-3⟧U⟦3,∞⟧

What is the maximum value of f(x) =√3 sin x + cos x?
Max value of asinx+bcosx is √(a²+b²)

so max value of √3sinx+cosx is √(√3)²+1²=√(3+1)=√4=2

•   

$\mathbf{\color{green}{\mathit{y} = \dfrac{\mathit{x}^2 + \mathit{x} - 2}{\mathit{x}^2 - \mathit{x} - 2}}}$

GEOMETRIC PROGRESSION

GEOMETRIC PROGRESSION (GP)

Geometric Progression(GP) or Geometric Sequence is sequence of non-zero numbers in which the ratio of any term and its preceding term is always constant.

For example:

3,6,12,24......is an GP because the common ratio (6/3=12/6=24/12=2) is constant.

General representation of GP:

a,ar²,ar³.....

here a is the first term and r is common ratio of the GP

• nth term of a GP = a rn-1
• Geometric Mean = nth root of product of n terms in the GP
• Sum of ‘n’ terms of a GP (r < 1) = [a (1 – rn)] / [1 – r]
• Sum of ‘n’ terms of a GP (r > 1) = [a (rn – 1)] / [r – 1]
• Sum of infinite terms of a GP (r < 1) = (a) / (1 – r)

Given x=1+a+a2+a3+⋯ and y=1+b+b2+b3+⋯, where a and b are proper fractions, what does 1+ab+a2b2+a3b3+⋯ equal to?

Sum of first infinite series

x=1/(1-a)..............1.

where 1 is the first term and a is the common ratio.

Sum of second infinite series

y=1/(1-b)..............2.

where 1 is the first term and b is the common ratio.

Now the sum of the third infinite series(s)

s=1/(1-ab)

where 1 is the first term and ab is the common ratio.

from 1.

1-a=(1/x)

a=(x-1)/x

and 

b=(y-1)/y

s=1/(1-ab)

s=1/(1-((x-1)(y-1))/xy

or

s=xy/(xy-(xy-x-y+1)

s=xy/(xy-xy+x+y-1)

s=xy/(x+y-1)

ARITHMETIC PROGRESSION

ARITHMETIC PROGRESSION (AP)

Arithmetic progression in which the difference between the consecutive term is constant.

In other words the next term is calculated by adding a fixed number in the previous term. This fixed term is called common difference.

For example:

3,6,9,12.....is an arithmetic progression because the common difference is same i.e. ⟦ 6-3=9-6=12-9=3⟧ 

General representation of AP:

a,a+d,a+2d,a+3d.........

Here a is called the first term and d is the common difference.

nth term= a+(n-1)d

Arithmetic mean= sum of terms of AP/No of terms of AP

Sum of AP (S)=n/2⟦2a+(n-1)⟧d

If the a is first term and T is the last term then

 Sum of AP (S)=n/2⟦a+T⟧

Example:

1. Find the nth term of 1,3,5,7.....
    nth term=a+(n-1)d
                  =1+(n-1)2

                   =1+2n-2
     nth term=2n-1

 2. Find the 10th term of 2,4,6,8,10...
     for this sequence a=2
       and common difference=2
      nth term=a+(n-1)d
      10th term=2+(10-1)2
                      =2+9*2
       10th term =20

3.Find the no of term in the 8,12,16..........72
  for this sequence a=8
    and common difference=4
  No. of term=(l-a)/d+1
                     =(72-8)/4+1
                      =64/4+1
                       =16+1
                       =17
No of term=17

4. Find 3+7+11+15.....+20 terms
  For this a=3
and common difference=4
Sum= n/2⟦2a+(n-1)⟧d
        =20/2[2*3+(20-1)]4
         =10[6+19]4
          =40*25
    Sum   =1000


   

LINEAR EQUATIONS WITH TWO VARIABLES

THE EQUATIONS WITH TWO VARIABLES x AND y CALLED LINEAR EQUATIONS WITH TWO VARIABLES.

The equation in the form of ax+by+c=0 is called linear equations with two variables.

where a, b and c are constants.

and 

x and y are two variables.

To find two variables i.e. x and y, two equations are required.

Two equations are

ax+by=d

ex+fy=g

Here x and y

and

a,b,c,d,e,f and g are constant.

Solve the following system of linear equations:

{y=2x+4y=3x+2

Now from equation 1 put value of y in equation 2
we get
        y=3x+2
2x+4=3x+2
      4-2=3x-2x
         2=x
Hence x=2
Now putting value of x=2 in equation 1 we get,
y=2x+4
y=2*2+4
y=8

SOLUTION OF THESE TWO EQUATION  x=2 and y=8

 Another example

 x+2y=1      ─⇀ 1

          2x-y = 3        ─⇀ 2

since the coefficient of x and y are both in these two equations

Multiply equation 2 with 2 so that the coefficient of y will be same

i.e.

 x+2y=1            ─⇀ 1

*2(2x-y)=3*2   ─⇀ 2

or

 x+2y=1            ─⇀ 1

4x-2y=6            ─⇀ 2

y=2x+4

y=2x+4

Now adding equation 1 and 2 we get
x+2y+4x-2y=1+6
                5x=7
                 x=7/5
And putting the value of x in equation 1 we get,
      x+2y=1
(7/5)+2y=1
           2y=1-(7/5)
           2y=(5-7)/2
               2y=-2/2
               y=(-1/2)
 Another example

 x+2y=1      ─⇀ 1

          2x-y = 3        ─⇀ 2

since the coefficient of x and y are both in these two equations

Multiply equation 2 with 2 so that the coefficient of y will be same

i.e.

 x+2y=1            ─⇀ 1

*2(2x-y)=3*2   ─⇀ 2

or

 x+2y=1            ─⇀ 1

4x-2y=6            ─⇀ 2

y=2x+4

y=2x+4

Now adding equation 1 and 2 we get
x+2y+4x-2y=1+6
                5x=7
                 x=7/5
And putting the value of x in equation 1 we get,
      x+2y=1
(7/5)+2y=1
           2y=1-(7/5)
           2y=(5-7)/2
               2y=-2/2
               y=(-1/2)

 Another example

 x+2y=1      ─⇀ 1

          2x-y = 3        ─⇀ 2

since the coefficient of x and y are both in these two equations
Multiply equation 2 with 2 so that the coefficient of y will be same
i.e.
 x+2y=1            ─⇀ 1
*2(2x-y)=3*2   ─⇀ 2

or

 x+2y=1            ─⇀ 1

4x-2y=6            ─⇀ 2

y=2x+4

y=2x+4

Now adding equation 1 and 2 we get
x+2y+4x-2y=1+6
                5x=7
                 x=7/5
And putting the value of x in equation 1 we get,
      x+2y=1
(7/5)+2y=1
           2y=1-(7/5)
           2y=(5-7)/2
               2y=-2/2
               y=(-1/2)

What is the value of k for which the system of equations Kx+2y=5 and 3x+y=1 has no solution? Later after solving plz tell me what is no solution?

If two lines are parallel then slope of the line will be same and there will be no solution

kx+2y=5 →1.

3x+y=1  →2.

slope of first line: -k/2

and

slope of second line: -3

For no solution:

-k/2=-3

k=6

No solution means two lines does not intersect each other or two lines are parallel to each other.

METHOD OF SOLVING:

Solve for x and y:

 x+2y=1

3x-2y=7

Solve for x and y:

 x+2y=1

  2x-y=1

For what value of k, system of equations, does x+2y=3&5x+ky=15 have infinite solutions?

given lines x+2y=3 and 5x+ky=15

will have infinite solutions if 

both lines are parallel 

for parallel slope must be equal

-1/2=-5/k

k=10

Work for you:

Find x and y

1.    2x+3y=1

        x+2y=3

2.    x-3y=4

        x+y=2

3.   x+y=2

      x-2y=4