Question 182880
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1)  Find the exact solution of the equation:


*[tex \LARGE \text{          }\math \frac{32}{e^x} - e^{2x} = 4]


*[tex \LARGE \text{          }\math \frac{32}{e^x} = e^{2x} + 4]


*[tex \LARGE \text{          }\math 32 = e^{3x} + 4e^x]
                

*[tex \LARGE \text{          }\math 32 = e^x(e^{3} + 4)]


*[tex \LARGE \text{          }\math   e^x = \frac{32}{e^{3} + 4}]


*[tex \LARGE \text{          }\math   \ln{e^x} = \ln\left(\frac{32}{e^{3} + 4}\right)]


*[tex \LARGE \text{          }\math   x = \ln{32}-\ln\left({e^{3} + 4}\right)]


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2) A curve *[tex \Large y = 3x^2 - 7x +5]. The point <i>P</i> on the curve has <i>x</i> coordinate 2. The tangent and the normal to the curve at <i>P</i> meet the <i>x</i>-axis at <i>A</i> and <i>B</i> respectively. Find the area of triangle *[tex \Large APB].


<i>y</i>-coordinate of <i>P</i>:


*[tex \LARGE \text{          }\math f(x) = 3x^2 - 7x + 5]


*[tex \LARGE \text{          }\math f(2) = 3(2)^2 - 7(2) + 5 = 3]


Slope of line tangent to


*[tex \LARGE \text{          }\math f(x) = 3x^2 - 7x + 5]


at *[tex \Large (2, 3)]:



*[tex \LARGE \text{          }\math f'(x) = 6x - 7]


*[tex \LARGE \text{          }\math f'(2) = 6(2) - 7 = 5]


Equation of tangent line (Equation of line containing segment *[tex \Large \overline{PA}]):


*[tex \LARGE \text{          }\math y - 3 = 5(x - 2)]


*[tex \LARGE \text{          }\math y = 5x - 7]


<i>x</i>-intercept of above line, hence <i>x</i>-coordinate of <i>A</i>:


*[tex \LARGE \text{          }\math 0 = 5x - 7 \ \ \Rightarrow\ \ x = \frac{7}{5}]


Hence, *[tex \Large A = (\frac{7}{5}, 0)]:


Slope of tangent is 5, so slope of normal must be *[tex \Large -\frac{1}{5}].  Equation of normal (Equation of line containing segment *[tex \Large \overline{PB}]):


*[tex \LARGE \text{          }\math y - 3 = -\frac{1}{5}(x - 2)]


*[tex \LARGE \text{          }\math y = -\frac{1}{5}x + \frac{17}{5}]


<i>x</i>-intercept of above line, hence <i>x</i>-coordinate of <i>B</i>:


*[tex \LARGE \text{          }\math 0 = -\frac{1}{5}x + \frac{17}{5}]


*[tex \LARGE \text{          }\math x = 17]


Hence, Hence, *[tex \Large B = (17, 0)]


Length of segment *[tex \Large \overline{AB}]


*[tex \LARGE \text{          }\math d = sqrt{( 17 - \frac{7}{5} )^2 + (0 - 0)^2} = 17 - \frac{7}{5} = \frac{85}{5} - \frac{7}{5} = \frac{78}{5}]


Length of altitude equals <i>y</i> coordinate of point *[tex \Large P = 3]


Triangle area:


*[tex \LARGE \text{          }\math A = \frac{bh}{2} = \left(\frac{78}{5}\right)\left(3\right)\left(\frac{1}{2}\right) = \frac{117}{5}]


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3)a) Show that 


*[tex \LARGE \text{          }\math \sum_{r=1}^n\,{r} = \frac{n(n+1)}{2}]


Show that it is true for <i>n</i> = 1:


*[tex \LARGE \text{          }\math \sum_{r=1}^1\,{r} =^? \frac{1(1+1)}{2}= \frac{2}{2} = 1]


If *[tex \LARGE \text{          }\math \sum_{r=1}^n\,{r} = \frac{n(n+1)}{2}] is true for some <i>n</i>, is it true for <i>n</i> + 1?


Assuming:


*[tex \LARGE \text{          }\math \sum_{r=1}^n\,{r} = \frac{n(n+1)}{2}] 


Then:


*[tex \LARGE \text{          }\math \sum_{r=1}^{n+1}\,{r} = \frac{(n +1)(n+2)}{2}] 


But


*[tex \LARGE \text{          }\math \sum_{r=1}^{n+1}\,{r} = \sum_{r=1}^n\,{r} + (n + 1) = \frac{n(n+1)}{2} + (n + 1) = \frac{n(n+1)}{2} + \frac{2(n + 1)}{2} = \frac {n^2 + 3n + 2}{2} = \frac {(n + 1)(n + 2)}{2}]


So now we know that if the relationship is true for some arbitrary <i>n</i>, it is true for <i>n</i> + 1.  But we proved it true for <i>n</i> = 1, therefore it is true for <i>n</i> = 2.  True for 2, must be true for 3, and so on, ad infinitum.


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3) b)

The formula developed in part a) only works for a series of consecutive integers that starts with 1 because the number of terms must equal the value of the last term.  Part b) requires the more general formula for the sum of a series of integers:



*[tex \LARGE \text{          }\math S_n = \frac {n(a + l)}{2}]


where <i>a</i> is the first term, <i>l</i> is the last term, and <i>n</i> is the number of terms:


99 numbers from 1 to 99, less 19 that are divisible by 5 so <i>n</i> = 80:


*[tex \LARGE \text{          }\math S_n = \frac {80(1 + 99)}{2} = 4000]

 
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4) The line L passes through the points with coordinates (5,5) and (11,10).


a) Show that the equation for L is *[tex \Large 6y = 5x + 5].


*[tex \LARGE \text{          }\math y - 5 = \frac{10 - 5}{11 - 5}(x - 5)]


You can do your own arithmetic.


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The curve C with equation *[tex \Large xy = 5] intersects the line L in two points.

 
b) Find the coordinates of these two points. 


*[tex \LARGE \text{          }\math 6y = 5x + 5]


*[tex \LARGE \text{          }\math xy = 5 \ \ \Rightarrow\ \ y = \frac{5}{x} ]


*[tex \LARGE \text{          }\math 6\frac{5}{x} = 5x + 5]


*[tex \LARGE \text{          }\math 5x^2 + 5x - 30 = 0]


*[tex \LARGE \text{          }\math x^2 + x - 6 = 0]


*[tex \LARGE \text{          }\math (x + 3)(x - 2) = 0]


*[tex \LARGE \text{          }\math x = -3] so


*[tex \LARGE \text{          }\math y = -\frac{5}{3} ] and the first point is:


*[tex \LARGE \text{          }\math \left(-3,-\frac{5}{3}\right) ]


*[tex \LARGE \text{          }\math x = 2] so


*[tex \LARGE \text{          }\math y = \frac{5}{2} ] and the second point is:


*[tex \LARGE \text{          }\math \left(2,\frac{5}{2}\right) ]


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The line *[tex \Large L_2] passes through (1,0) and is perpendicular to *[tex \Large L_1]. 
c) Find an equation for *[tex \Large L_2]. 


Slope of *[tex \Large L_1 : m_1 = \frac{5}{6}], so slope of perpendicular *[tex \Large L_2 : m_2 = -\frac{6}{5} ].


*[tex \LARGE \text{          }\math y - 0 = -\frac{6}{5}(x - 1)].


Again, you can do your own arithmetic.


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d) Show, using algebra, that L never meets C.


Using the process shown for part b) of this question, attempt to solve for the coordinates of any points of intersection.  The resulting quadratic will have complex roots indicating that there are no points of intersection in the *[tex \Large R^2] plane.  Hint:  You don't have to actually solve the quadratic, you only need to compute the discriminant and show that it is less than zero.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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