TRIGONOMETRY

If tan(x+y)=tanx+tany , what is tanxtany ?

tan(x+y)=(tanx+tany)/(1-tanxtany)

but given tan(x+y)=tanx+tany

so tan(x+y)=tan(x+y)/(1-tanxtany)

or (1-tanxtany)=tan(x+y)/tan(x+y)

or (1-tanxtany)=1

or -tanxtany=0

or tanxtany=0

MENSURATION

If four interior angles of a quadrilateral are (x+15), (x+10), (2x-45) and (x-30), what is the value of x?

sum of interior angle of quadrilateral is 360

so, (x+15)+(x+10)+(2x-45)+(x-30)=360

5x+25-75=360

5x=360+50

5x=410

x=82

LIMITS

Why is the limit as x approaches positive or negative infinity of the function ƒ(x) = (x+1) /(x-2) equal to 1?

f(x)=(x+1)/(x-2) when x→∞

if we put x=∞ then it is ∞/∞ form

or if let x=1/t then t→0

f(1/t)=(1/t+1)/(1/t-2) when t→0

or f(1/t)=(1+t)/(1-2t) when t→0

or t=0, f(1/t)=(1+0)/(1-2×0)=1

hence proved

QUESTIONS ON GRAPH

What is the solution of sinx=3?

No Real solution

Because if we draw the graph of y=sinx and y=3 then it will not intersect each other.

as the maximum value of sinx is 1 when x=90° and y=3 not cut the graph of sinx.

GENERAL QUESTIONS

A cylinder of diameter 14 and height 7 is converted into a cone of radius 6. Now, what could be the percentage height of the new shape?

volume of cylinder=Π×r²×h

=Π×7²×7

volume of cone=Π×R²×H/3=Π×6²×H/3

volume of cylinder and cone will be same

Π×7²×7=Π×6²×H/3

H=28.58

% of height of cone=(28.58-7/7)×100=308.28

If the volume of a cylinder is 300 m3 and the height to diameter ratio is 3:1, what is the height and diameter of the cylinder?

Given h/d=3/1 or h/2r=3/1 or h/r=6 or h=6r

volume of cyinder=Π×r²×h

300=Π×r²×(6r)

300=Π×6×r³

300/6=Π×r³

50/Π=r³

solve for r=2.51

and h=6×2.51=15.06

DOMAIN AND RANGE

What is the range of the real function f defined by f(x) =√(x-1)?     

Given Y=√(x-1)

Range is the value of y for the function to be defined.

convert x is in the form of y

i.e.

y²=x-1

or x=1+y²

hence the x is always defined for all real value of y

so the range of function is y∈R i.e. all real value

What is the domain function f, where f(x) = √(1-x) (x-2)?

Given y=√(1-x) (x-2)

for function to be defined i.e. y to defined domain is value of x for which y is defined

y is defined when value under squre root is greater than and equal to zero

i.e.

(1-x)(x-2)≥0

or (x-1)(x-2)≤0

or 1≤x≤2

What will be the range of the function f(x) =1/ (√ (x^2-1))?

Range is the value of the y for which x is defined

rewritting as y=1/ (√ (x^2-1))

or y²=1/(x^2-1)

or x²-1=1/y²

or x²=1/y²+1

or x=√((1/y²)+1

hence the x is defined for all value of y except when y is zero

range y∈R-(1)

What are the domain and range of the function y=-(x-2)^2 +2?

Domain is the value of x for which function is defined, here for all value of x i.e. all real value of x.

Range is the value of y for which function is defined.

(y-2)=-(x-2)²

or (2-y)=(x-2)²

or (x-2)=√2-y

for this function to defined (2-y)≥0

or (y-2)≤0

or y≤2

What is the range of f(x)=(sinx)sinx+1?

range is the value of y for which function to be defined 

min value of sinx=0 when x=0

hence min value when x=0 is 0⁰+1

which is not defined hence solve by limit

limx→0 (sinx)^sinx 

^{or taking log on both sides}

^{lny=limx→0ln}(sinx)^sinx 

^{or lny=}limx→0sinxln(sinx)

or lny=limx→0ln(sinx)/coescx

solve using L hospital rule

lny=limx→0cosx/(-cosecx.cotx)(sinx)

lny=-1

or y=1/e

hence min value of y is (1/e)Λ(1/e)+1

and max value when x=90°

ie. y(max)=1+1=2

range is ((1/e)Λ(1/e)+1),2)

DEFINITE INTEGRAL

Definite Integration

The definite integral has limit fixed and hence the value obtained of integral is without constant(C).