SOLUTION: 1) Find the exact solution of the equation:
32/e^x - e^2x = 4
2) A curve y = 3x^2 - 7x +5. The point P on the curve has x coordinate 2. The tangent and the norm
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Question 182880: 1) Find the exact solution of the equation:
32/e^x - e^2x = 4
2) A curve y = 3x^2 - 7x +5. The point P on the curve has x coordinate 2. The tangent and the normal to the curve at P meet the x axis at A and B respectively. Find the area of the triangle APB.
3) a Show that ∑_(r=1)^n▒(r=n(n+1) /2
b Hence, or otherwise, find the sum of all the integers from 1 to 99 inclusive which are not multiples of 5.
4) The line L passes through the points with coordinates (5,5) and (11,10).
a) Show that the equation for L is 6y = 5x + 5.
The curve C with equation xy = 5 intersects the line L in two points.
b) Find the coordinates of these two points.
The line L passes through (1,0) and is perpendicular to L.
c) Find an equation for L.
d) Show, using algebra, that L never meets C.
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
1) Find the exact solution of the equation:
)
)
)
=============================
2) A curve 
. The point P on the curve has x coordinate 2. The tangent and the normal to the curve at P meet the x-axis at A and B respectively. Find the area of triangle 
.
y-coordinate of P:
 = 3x^2 - 7x + 5)
 = 3(2)^2 - 7(2) + 5 = 3)
Slope of line tangent to
at )
:
 = 6x - 7)
 = 6(2) - 7 = 5)
Equation of tangent line (Equation of line containing segment 
):
)
x-intercept of above line, hence x-coordinate of A:
Hence, )
:
Slope of tangent is 5, so slope of normal must be 
. Equation of normal (Equation of line containing segment 
):
)
x-intercept of above line, hence x-coordinate of B:
Hence, Hence, )
Length of segment 
^2 + (0 - 0)^2} = 17 - \frac{7}{5} = \frac{85}{5} - \frac{7}{5} = \frac{78}{5})
Length of altitude equals y coordinate of point 
Triangle area:
\left(3\right)\left(\frac{1}{2}\right) = \frac{117}{5})
========================================
3)a) Show that
}{2})
Show that it is true for n = 1:
}{2}= \frac{2}{2} = 1)
If is true for some n, is it true for n + 1?
Assuming:
Then:
(n+2)}{2})
But
 = \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 n, it is true for n + 1. But we proved it true for n = 1, therefore it is true for n = 2. True for 2, must be true for 3, and so on, ad infinitum.
==============================================
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:
}{2})
where a is the first term, l is the last term, and n is the number of terms:
99 numbers from 1 to 99, less 19 that are divisible by 5 so n = 80:
}{2} = 4000)
====================================================
4) The line L passes through the points with coordinates (5,5) and (11,10).
a) Show that the equation for L is 
.
)
You can do your own arithmetic.
===================================================
The curve C with equation 
intersects the line L in two points.
b) Find the coordinates of these two points.
(x - 2) = 0)
so
and the first point is:
 )
so
and the second point is:
 )
==================================================
The line 
passes through (1,0) and is perpendicular to 
.
c) Find an equation for .
Slope of 
, so slope of perpendicular 
.
)
.
Again, you can do your own arithmetic.
=================================================
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 
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
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