Friday, May 8, 2020

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

Thursday, May 7, 2020

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


Wednesday, May 6, 2020

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

Tuesday, May 5, 2020

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 ?
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

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