Mth101 Assignment No. 1 Solution
Thursday, October 20, 2011 Posted In Mth Edit ThisAssignment No. 1
MTH 101 (fall 2011)
Maximum Marks: 20
Due Date: 26th Oct. 2011
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Question # 01: 10
Find equation of the line which is perpendicular to the line
and has y-intercept = 6. Question # 02: 10
Find the vertex, x and y intercepts of the parabola given by the equation
Question # 03:
Find the centre and radius of the circle given by the equation
. Question # 04:
Given that 
Find
. Also describe the domain of composite function.
Find. Also describe the domain of composite function. (Note: In order to get full marks, do all necessary steps)
Solution:
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MTH101 ASSIGNMENT SOLUTION
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Q1 Solution
y=5x+9
From the given equation slop of line is 5
Since Slop of perpendicular line =m=-1/5
and y intercept is= 6
By slop intercept form
y=mx+b
y=-1/5x+6
Find equation of the line which is perpendicular to the line
y= 5x+9 and has y-intercept = 6.
I think correct ansewr is
y=5x+9 is darivative w.r.t x
dy/dx=5+0
m=5
then slope of normal is m= -1/5
for normal equation is (y-y1)m =(x-x1)
(y-6)-1/5=(x-0)
-y1/5+6=x
-y+30=5x
y+5x-30=0
this is normal equation
for parpandicular is
y=mx+b
y=-1/5x+6
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Question # 01:
Find equation of the line which is perpendicular to the line and has y-intercept = 6.
Solution:
Y = 5x + 9
From the given equation slop of line is 5
Since slop of perpendicular line = m = - 1/5
And Y intercept is = 6
By slop intercept form:
Y = mx + b
Y = -1/5x + 6
For the equation of the line which is perpendicular to the line:
Y = 5x + 9 and has Y-intercept = 6
Y = 5x + 9 is derivative w.r.t x
Dy/dx = 5 + 0
M=5
Then slope of normal is m= -1/5
for normal equation is (Y-Y1)m = (X-X1)
(Y-6)-1/5 =( x - 0)
5y – 30 -1 = 5x
5y- 31 = 5x
5y = 5x+31
y = (5x+31)/5
y = x+(31/5)
This is normal equation
for perpendicular is
Y = mx + b
Y=x+(31/5)
___________________________________________
Question # 02:
Find the vertex, x and y intercepts of the parabola given by the equation
Solution:
y = 3x2 - 2x + 1
In this problem: a = 3, b = -2 , and c = 1
Since "a" is positive we'll have a parabola that opens upward (is U shaped).
To find the x-intercepts we plug in 0 for y:
once it's found that the discriminant is 4-12=-8, the solutions must therefore be imaginary numbers (complex conjugates, that is), so there are no x- intercepts.
