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Mth101 Assignment No. 1 Solution

Thursday, October 20, 2011 Posted In Edit This
Assignment 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.                                                            

(Note: In order to get full marks, do all necessary steps)



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


:::::::::::::::::::::::::::::::::::::::::::::



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.

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