PARTIAL FRACTIONS

FIND THE PARTIAL FRACTION OF (x+14)/(x-4)(x+2)

(x+14)/(x-4)(x+2)=a/(x-4)+b/(x+2).......1.

(x+14)/(x-4)(x+2)=⟦a(x+2)+b(x-4)⟧(x-4)(x+2)

or

(x+14)=⟦a(x+2)+b(x-4)⟧

put 

x=-2

-2+14=⟦a(-2+2)+b(-2-4)⟧

12=b(-6)

b=-2

and

put

x=4

4+14=⟦a(4+2)+b(4-4)⟧

18=6a

a=3

hence

from 1.

(x+14)/(x-4)(x+2)=3/(x-4)-2/(x+2) 

Simplify (1/16x)+(1/8)) / ((1/4)+(1/8x))?

((1/16x)+(1/8)) / ((1/4)+(1/8x)) =y assume y=N/D

N=numerator=((1/16x)+(1/8))

and

D=(1/4)+(1/8x)

N=((1/16x)+(1/8))=

Now solve this

N=(1+2x)/16x

and

D=(1/4)+(1/8x)

D=(2x+1)/8x

Now y=N/D=(1+2x)/16x/(2x+1)/8x=8x/16x=1/2

SPEED, TIME AND DISTANCE

The time taken by a train running at 30kmph to pass a man walking in the same direction at a speed of 6 kmph is 3/7 th of the time taken by the train to cross a platform. If the length of the train is 96 mts. What is the length of the platform?

let the time taken for train to cross the man is t1

then 

relative speed=30-6=24kmph=6.66mps

t1=96/6.66

and

t2=time taken for train to cross the platform is t2

speed of train=30kmph=8.33mps

t2=(96+l)

l=length of train

Now according to given condition

t1=(3/7)t2

96/6.66=(3/7)(96+l)/8.33

solve for l

l=184m

A boat takes twice as much time going upstream as it takes to go downstream. While covering the same distance. If the speed of the boat in still water is 6 km/hr. What is the speed of the current?

———→ downstream 

Assume speed of current is V.

Relative velocity= V+6 and time= T1

and 

←———upstream

Relative velocity=6-V  and time T2

Note: Relative velocity=6-V not V-6 becuse if V>6 i.e. velocity of current is greater then velocity of boat then boat will not go to upstream.

Now,

i.e.T2=2T1

d/(6-V)=2d/(V+6)

V+6=2(6-V)

V+6=12-2V

3V=6

V=2kmph

The speed of the boat in still water is 22 kmph. It takes the same time for the boat to travel 1.2 km upstream as it does to travel 32km downstream. What is the speed of the current?

———→ downstream

Assume speed of current is V.

Relative velocity= V+22 and time= T1 

Distance=32km

and 

←———upstream

Relative velocity=22-V  and time T2

Distance=1.2km

Note: Relative velocity=22-V not V-22 becuse if V>22 i.e. velocity of current is greater then velocity of boat then boat will not go to upstream.

Now,

i.e.T2=T1

1.2/(22-V)=32/(V+22)

(V+22)1.2=32(22-V)

1.2V+22×1.2=32×22-32V

V=20.4kmph

SURDS AND INDICES

How do I solve 4x−2x−2=0?

4x−2x−2=0

2^2X-2^X-2=0

Assume 2^x=t

then

above equation become

t²-t-2=0

t²-2t+t-2=0

from this

t=2 and t=-1

but 2^x=t

2^x=2

x=1 

and

2^x=-1

for above condition x value not possible

hence only solution x=1

How can I simplify ab√√ab√3 with exponent properties?

y=√ab/∛a∛b

y=√a√b/∛a∛b     because √ab=√a√b

y=a^(1/2-1/3)×b^(1/2-1/3)  because a^x/a^y=a^x-y

y=a^(5/6)×b^(5/6)      because a^x×b^x=(ab)^x

y=(ab)^(5/6)

If the value of (a + 1/a) is equal to 3, then what will be the value of a2 + 1/a2?

a+1/a=3

squering both sides

a²+1/a²+2×a×(1/a)=9

a²+1/a²+2=9  use formula (x+y)²=x²+y²+2xy

a²+1/a²=7

If (8x²)^8-r(2x)^-r=2^ax^{b   then find a and b in terms of r ?}
(8x²)^8-r(2x)^-r=2^ax^b

(2^³x²)^8-r(2x)^-r=2^ax^b

2^3(8-r)x^2(8-r)=2^ax^b

2^(24-3r)x^(16-2r)=2^ax^b

^{now compare the power of 2 and x}

a=(24-3r)

and b=16-2r

IF (8-√18)/√2=a+b√2 then find a and b ?
we have to eliminate √2 denominator.

for this multiply and divide by √2

i.e.

(8-√18)/√2=(8-√18)×√2/(√2×√2) 

                   =(8-√18)×√2/2

                   =(8√2-√18√2)/2

                   =(8√2-√36)/2

                   =(8√2-2√3)/2

                    =4√2-√9

                      =-3+4√2

a+b√2=-3+4√2

a=-3 and b=4
What is the value of 22014−22013

2^2014 - 2^2013

= 2^(2013+1) - 2^2013

= 2^2013 (2^1- 1)

= 2^ 2013 * (2–1)

= 2^ 2013 * 1

= 2^2013

CIRCLE

Circle

Curve that is radius away from centre point.
or
All points are same distance from centre. 



Radius: Distance from centre to outwards only one side.
Diameter: Distance from centre to outwards on both sides.
Circumference: It is a distance once around the circle.

The ratio of circumference to the diameter is called Π.
important relation:
circumference

         ↑ ×Π

     Diameter

         ↑×2

      Radius

Area of circle: It is the Π times the radius square.

i,e,

Area of circle=Πr²

Unit of Area: If the unit of the radius is meter(m) then the unit of area is meter square (m²).

Example: Find the area of circle whose radius is 3m?

Area of circle: Πr²

A=Π×r×r

A=Π×3×3

A=3.14×3×3

A=28.26m²

Lines

Chord: Chord is the line between any two points on the circumference.

Diameter: Diameter is the longest chord of the circle.

Tangent: When the line touches at only one point on the circumference, it is called tangent.

General Equation of circle

General equation of circle if (a,b) and r is the radius is given by:

(x-a)²+(y-b)²=r²

How do I find an equation of a circle whose centre is at point (4,5) and touches the x axis? How do I find the coordinates of points at which the circle cuts the y axis?

Centre (4,5)

Radius=5 since it is touching the x axis hence y coordinate of center will be radius.

General equation of circle if (a,b) and r is the radius:

(x-a)²+(y-b)²=r²

(x-4)²+(y-5)²=5²

(x-4)²+(y-5)²=25

when it cut the y-axis where x=0

i.e.

(0-4)²+(y-5)²=25

16+(y-5)²=25

(y-5)²=25-16

(y-5)²=9

y-5=±3

from this

y=5+3 and 5-3

y=8 and 2

hence the circle cut the y axis at 2 and 8

The equation for a circle is x^2 + y^2 + 10y - 7 = 0. What is the centre of the circle?

Given

x²+y²+10y-7=0

General equation of circle if (a,b) and r is the radius:

(x-a)²+(y-b)²=r²
x²+y²+10y-7=0
x²+y²+10y+25=25+7

x²+(y+5)²=(√32)²

(x-a)²+(y-b)²=r²

comparing both
Centre =(0,-5)
and
Radius =√32

In a circle with a radius of 5 cm, what is the length of the chord which is at a distance of 3 cm from its center?
Given radius (r)=5cm

and let BAC is the chord. Midpoint is B

perpendicular from centre to chord (OA)=3cm

then BA=(√(5²-3²)=√16=4cm

length of chord (BAC)=2BA because BA=AC

length of chord (BAC)=2*4=8cm

DEFINITE INTEGRAL

DEFINITE INTEGRAL

Definite integral has starting and End limits.

i.e.

It is bounded by limits, boundaries.

Definite integration directly gives the area under the curve within the given limits.

i.e.

if 
y=∫f(x)dx

is an indefinite integral 

and 

if the lower and upper limits of integration are given then it is a definite integral.

Important properties to find integral: 

What is the area bounded by the curve y=logx, x-axis and the ordinates x=1 and x=e?
y=∫logx dx

y=logx.x-∫(1/x)xdx    integration by parts

y=logx.x-x⟧₁^e

^{because the limit is from 1 to e}

^{y=loge.e-e-(log1.1-1)}

^y=1