..READ THIS
FIRST..Miscellaneous
solved problems
READ THIS
FIRST
If a problem is solved. It is not 'the' answer.No attempt is made to search for the most elegant answer.
I highly recommend that you at least try to solve the problem before you read the solution.
Miscellaneous solved problems
About roots of a quadratic equation
Given is the equation
(m2 - 5m + 6)x2 + (4 - m2)x + 20 = 0
Say the roots are x' and x", with x' < x" or x'=x".Calculate the values of the parameter m such that x' < 1 < x". |
Let f(x) = ax2+ bx + c .
If a > 0 then, f(1) < 0 <=> x' < 1 < x"
If a < 0 then, f(1) > 0 <=> x' < 1 < x"
Conclusion : a.f(1) < 0 <=> x' < 1 < x"
Thus, in this exercise we have :
x' < 1 < x"
<=>
(m2 - 5m + 6).( m2 - 5m + 6 + 4 - m2 + 20 ) < 0
<=>
(m2 - 5m + 6).(30 - 5m) < 0
<=>
2 < m < 3 or m > 6
Calculate m such that in the equation
x2 + mx - 24 = 0 (1)
a root is twice a root of the equation
x2 - (m+1)x + m = 0 (2)
|
Denote x1 and x2 the roots of (1).
First, we write a quadratic equation with roots x1/2 and x2/2.
x2 + (m/2) x - 6 = 0
<=> 2x2 + mx - 12 = 0 (3)
The problem is now, find m such that (2) and (3) have a common root.
Take the matrix of the coefficients
[1 -m-1 m ]
[2 m -12]
A = 12 m + 12 - m2 B = 2 m + 12 C = 3 m + 2
(1) and (3) have a common root if and only if
B.B = A.C
<=> (2 m + 12) (2 m + 12) = (12 m - m2 + 12) (3 m + 2)
<=> 3 m3 - 30 m2 - 12 m + 120
<=> 3 (m + 2) (m - 2) (m - 10) = 0
<=> m = -2 or m = 2 or m = 10
Formula about the area of a triangle ABC
| Given : the coordinates of A,B and C with respect to an orthonormal
coordinate system in a plane. A(a,a') , B(b,b') , C(c,c') Prove that the area of the triangle is the absolute value of
1 |a a' 1|
--- .|b b' 1|
2 |c c' 1|
|
|AB| = sqrt((b-a)2+ (b'- a')2)
The equation of the line AB is
|x y 1|
|a a' 1| = 0
|b b' 1|
<=>
(a' - b')x - (a - b)y +(ab'-ba') = 0
The normal equation of this line is
(a' - b')x - (a - b)y +(ab'-ba')
--------------------------------- = 0
sqrt((a' - b')2+ (a - b )2)
<=>
|x y 1|
|a a' 1|
|b b' 1|
--------------------------- = 0
sqrt((a' - b')2+ (a - b )2)
The distance from C(c,c') to AB is the absolute value of
|c c' 1|
|a a' 1|
|b b' 1|
-------------------------
sqrt((b-a)2 + (b'- a')2)
The area of the triangle is then the absolute value of
|c c' 1|
|a a' 1|
1 |b b' 1|
--- . ------------------------- .qrt((b-a)2 + (b'- a')2)
2 sqrt((b-a)2 + (b'- a')2)
1 |c c' 1|
= --- . |a a' 1|
2 |b b' 1|
1 |a a' 1|
= --- . |b b' 1|
2 |c c' 1|
Conclusion:Given : the coordinates of A,B and C with respect to an orthonormal coordinate system in a plane.
A(a,a') , B(b,b') , C(c,c')
The area of the triangle is THE ABSOLUTE VALUE OF
1 |a a' 1|
= --- . |b b' 1|
2 |c c' 1|
Equation with integral
Calculate all positive x-values such that
/x dt
| --- = 2
/1 t
|
/x dt 2
| --- = 2 <=> ln(x) - ln(1) = 2 <=> ln(x) = 2 <=> x = e
/1 t
Asymptotes, tangent line, integration
Let G be the graph of
9x2 + mx + 4
f(x) = ---------------- with m as real parameter
2x - 7
|
-
The graph and the x-axis have just one common point <=> 9x2 + mx + 4 = 0 has just one root (not 7/2) <=> m = 12 or m = -12 -
Each oblique asymptote D has an equation y = ax + b. According to the givens a = 9/2 and b = 3/2. f(x) lim ---- = 9/2 (does not depend on m) infty x b = lim (f(x) - (9/2)x ) = ... = m/2 - 63/4 infty So, m/2 - 63/4 = 3 <=> m = 57/2 -
(2x+7)(18x+m)- (9x2 + mx + 4).(2) f'(x) = ------------------------------------------- (2x+7)(2x+7) The slope is f'(0) = (7m - 8)/49 -
/ 9x2 + mx + 4 | --------------- dx / 2x - 7 We have to integrate an improper fraction. It can be written as the sum of a polynomial and a proper fraction. 9x2 + mx + 4 -------------- 2x - 7 7m/2 + 457/4 = (9/2)x + (m/2 + 63/4) + --------------- 2x + 7 Integration of the right side gives (9/4)x2 + (m/2 + 63/4)x + (7m/2 + 457/4)(1/2)ln|2x + 7| + C
Integral equation
Calculate all positive x-values such that
/e dt /x dt
| --- = | --
/1/e t /e t
|
<=> ln(e) - ln(1/e) = ln(x) - ln(e) <=> ln(x) = 3.ln(e) <=> ln(x) = ln(e3) <=> x = e3
Trigonometry - formulas
Given : x and y in ]0, pi/2[ Show that : 1 + 2cos(4y) - (sin(6x)/sin(2x)) = 16.sin(x-y)sin(x+y)cos(x-y)cos(x+y) |
Right side = 4.2.sin(x-y)cos(x-y).2.sin(x+y)cos(x+y)
= 4.sin 2(x+y).sin 2(x-y)
= 2 (cos(4y) - cos(4x))
Left side = 1 - (sin(6x)/sin(2x)) + 2cos(4y)
sin(2x) - sin(6x)
= ------------------------ + 2cos(4y)
sin(2x)
2.cos(4x).sin(-2x)
= ------------------------ + 2cos(4y)
sin(2x)
= -2.cos(4x) + 2.cos(4y)
Solve
_______________
| 2 x
| 2 cos --- - 1 > 2 sin(x) - 3
\| 2
|
2 x
Since 1 + cos(x) = 2 cos --- we have to solve
2
________
V cos(x) > 2 sin(x) - 3
The right side is strictly < 0.
The left side is positive if it exists.
The set of solutions is the set of all x values so that
cos(x) > 0 or cos(x) = 0
<=> -pi/2 + 2.k.pi =< x =< pi/2 + 2.k.pi
a,b and c are the angles of a triangle.
Show that we have a right-angled triangle if
sin(2 b) sin(2 c) = 2 - 2 sin2(b) cos2(c) - 2 cos2(b) sin2(c)
|
sin(2 b) sin(2 c) = 2 - 2 sin2(b) cos2(c) - 2 cos2(b) sin2(c)
<=> 4sin(b) cos(b) sin(c) cos(c) + 2 sin2(b) cos2(c) + 2 cos2(b) sin2(c) =2
<=> 2(sin(b) cos(c) + cos(b) sin(c))2 = 2
<=> sin2(b+c) = 1
<=> sin2(a) = 1
<=> sin (a) = 1
<=> a = pi/2
Arithmetic and geometric sequences
a, b, c form arithmetic sequence and x, y, z form a geometric sequence.
Prove that
xb.yc.za = xc.ya.zb
|
a, b, c form arithmetic sequence => b = (a+c)/2
x, y, z form a geometric sequence => y.y = x.z
xb.yc.za = x(a+c)/2 .(xz)c/2 . za
= x(a/2+c) . z(c/2+a)
xc.ya.zb= xc.(xz)(a/2). z(a+c)/2
= x(a/2+c) . z(c/2+a)
Simplify a trigonometric expression
Simplify
(1 + tan(a).tan(a/2))-1
|
1 1
------------------- = ------------------------------
1 + tan(a).tan(a/2) 2 tan(a/2).tan(a/2)
1 + ----------------------
1 - tan(a/2).tan(a/2)
1 - tan(a/2).tan(a/2)
= ------------------------ = cos(a)
1 + tan(a/2).tan(a/2)
Trigonometric equation
If m is a real parameter, then solve :
m.cos2(x) + (2m2 - m + 1)sin(x) -3m + 1 = 0
|
m.cos2(x) + (2m2 - m + 1)sin(x) -3m + 1 = 0
<=> m.(1 - sin2(x))+ (2m2 - m + 1)sin(x) -3m + 1 = 0
<=> -m.sin2(x) + (2m2 - m + 1)sin(x) -2m + 1 = 0
Discriminant = ... = ( 2m2 - m - 1)2
<=> sin(x) = ... = 1/m or sin(x) = ... = 2m - 1
-
sin(x) = 1/m has roots if and only if m is not in ]-1,1[. Let b = arcsin(1/m) , then sin(x) = sin(b) The soltions are x = b + 2.k.pi and x = pi - b + 2.k.pi
-
sin(x) = 2m - 1 has roots if and only if m is in [0,1]. Let c = arcsin(2m-1) , then sin(x) = sin(c) The soltions are x = c + 2.k.pi and x = pi - c + 2.k.pi
Relative maxima and minima and trigonometry
Given:
f(x) = (m - 1)cos2(x) -3mcos(x) + 2m
with m a real parameter ( m is not 1).
Determine all m such that there are four different x values in
[0, 2.pi[ so that the image is a relative maximum or minimum.
|
f(x) = (m - 1)cos2(x) -3mcos(x) + 2m
f'(x) = -2(m - 1)cos(x).sin(x) + 3m.sin(x)
f'(x) = -sin(x) .(2(m - 1)cos(x) - 3m)
From this we see that there are maxima or minima for x = 0 and x = pi.
Other maxima or minima occur for
3m
cos(x) = -------------
2(m - 1)
This equation has two solutions in ]0, 2.pi[ \ {pi}
if and only if
3m
-1 < --------- < 1
2(m - 1)
...
-2 < m < 2/5
Hence, if -2 < m < 2/5 there are four different x values in
[0, 2.pi[ so that the image is a relative maximum or minimum.
Relative maximum and minimum
| Given : f(x) = sqrt(2x.x + k) - x/3 - 2 Calculate k such that the x value, corresponding with the relative maximum or minimum of f(x), is equal to that reative maximum or minimum. |
2x
f'(x) = --------------- - 1/3
sqrt(2x2 + k)
The relative maximum or minimum occurs if
2x - sqrt(2x2 + k)/3 = 0
<=> ... <=> x = sqrt(k/34)
We have to calculate k such that for x = sqrt(k/34)
f(x) = x
<=> ... <=> k = 306/49 = 6.2449
Sequences and sums
| Given : Two constant values a and b. A sequence {t(n)}. The sum of the first n terms is S(n)=a.n2 + b.n With the sequence we construct a new sequence {t'(n)} such that t'(n) = t(2n). Calculate the sum S'(n) of the first n terms of {t'(n)}. |
t'(n) = t(2n) = S(2n) - S(2n-1) = ... = 4a n - a - b
t'(n) - t'(n-1) = ... = 4a = constant
So, {t'(n)} is an arithmetic sequence.
t'(1) = 3a - b
S'(n) = n. (t'(1) + t'(n))/2 = ... = n.(a - b + 2an)
Elimination and trigonometry
| The numbers a,b,c are constant. Eliminate x out of sin(a + x) = 2b ; sin(a - x) = 2c. |
sin(a + x) + sin(a - x) = 2sin(a)cos(x) = 2b + 2c
sin(a + x) - sin(a - x) = 2cos(a)sin(x) = 2b - 2c
cos(x) = (b + c)/sin(a)
sin(x) = (b - c)/cos(a)
Hence, the value x exists if and only if
2 2
(b + c) (b - c)
-------- + -------- = 1
2 2
sin (a) cos (a)
Factoring
Factor
z = (x + y)5 - x5 - y5
|
First we see that
-
z is symmetric and homogenious with respect to x and y.
-
z = 0 for x = 0 => z contains a factor x. z = 0 for y = 0 => z contains a factor y. z = 0 for x = -y => z contains a factor (x + y)
-
z = (x + y)5 - x5 - y5
Range or image of a function
Given : The function
mx2 - 7x + 5
f(x) = ---------------
5x2 - 7x + m
The set of all the images is called the range or image of the function.
Determine m such that the range of f(x) is the set of all real numbers.
|
The range of f(x) is the set of all real numbers.
<=>
The following equation has a solution for all fixed real numbers b.
mx2 - 7x + 5
----------- = b
5x2 - 7x + m
<=>
The following equation has a solution for all fixed real numbers b.
(m - 5b)x2 - 7(1 - b)x + (5 - mb) = 0
<=>
The previous equation has a positive discriminant for all b
<=>
(49 - 20m)b2 + (4m2 + 2)b + (49 - 20m) is positive for all b
<=>
(49 - 20m)>0 and
previous expression has a strictly negative discriminant
<=>
(49 - 20m)>0 and
(m - 5)2.(m + 12)(m - 2) < 0
<=> -12 < m < 2
Lines in space and linear systems
Given : Two lines in space with equations
/ 2x + my + z = 1 / 3x + z = 2
| and |
\ x - y + mz = 1 \ 2x + my + z = m - 1
Determine the values of m such that the two lines have
an intersection point.
|
The intersection point of the two lines is a solution of the system
/ 2x + my + z = 1
| x - y + mz = 1
| 3x + z = 2
\ 2x + my + z = m - 1
The matrix of coefficients is
[2 m 1]
[1 -1 m]
[3 0 1]
[2 m 1]
The determinant formed by the first three rows is
|2 m 1|
|1 -1 m| = 3m.m - m + 1 and this is never zero !
|3 0 1|
Hence, the system has a solution if and only if the
characteristic determinant of the last equation is zero.
This condition is :
|2 m 1 1 |
|1 -1 m 1 | = 0
|3 0 1 2 |
|2 m 1 m-1|
Row 1 - Row 4 gives
|0 0 0 2-m|
|1 -1 m 1 | = 0
|3 0 1 2 |
|2 m 1 m-1|
<=>
|1 -1 m|
(2 - m).|3 0 1| = 0
|2 m 1|
<=>
(2 - m).(3m.m - m + 1) = 0
<=> m = 2
The lines are intersecting only for m = 2.
Maybe this is a predictable result, but the method is instructive.
Lines and eigen values
In space we take a transformation T , defined by
x' = 2x + y + z
y' = x + 2y + z
z' = x + y + 2z
By this transformation, a point P(x,y,z) is transformed in
another point P'(x',y',z').
A line b is transformed in another line b'.
Calculate all lines b, such that b' is parallel to b.
|
Let (u,v,w) = the direction numbers of such line b, then
we search for (u,v,w) such that (u'= ru ; v'= rv ; w'= rw)
In other words :
2u + v + w = ru
u + 2v + w = rv
u + v + 2w = rw
<=>
[2 1 1][u] [u]
[1 2 1][v] = r. [v]
[1 1 2][w] [w]
We see that (u,v,w) is a characteristic vector of the transformation.
To calculate (u,v,w), we calculate the eigenvalues first.
The eigenvalues are the numbers r such that
|2-r 1 1|
|1 2-r 1| = 0
|1 1 2-r|
The roots are r = 1 and r = 4 .
First take r = 4 .
The characteristic vectors are the solutions of
[2 1 1][u] [u]
[1 2 1][v] = 4. [v]
[1 1 2][w] [w]
We find u = v = w
Hence, the line with the direction numbers (1,1,1) it is transformed in
a parallel line.
Now, take r = 1 .
The characteristic vectors are the solutions of
[2 1 1][u] [u]
[1 2 1][v] = 1. [v]
[1 1 2][w] [w]
This system is equivalent to
u + v + w = 0
Hence, each line with direction numbers such that u + v + w = 0, is
transformed in a parallel line.
calculation about the range of a function
Calculate m such that the range of the function
m x2 + 3 x - 4
f : x -> -----------------
m + 3 x - 4 x2
is equal to R.
|
For each real y there must be an x such that
m x2 + 3 x - 4
y = ----------------
m + 3 x - 4 x2
<=> y.(m + 3 x - 4 x2) = m x2 + 3 x - 4
<=> -(4y + m)x2 + (3y - 3)x + (my + 4) = 0
For each y the D must be not negative
<=> (16 m + 9) y2+ (4 m2 +46)y + (16m +9) >= 0
<=> 16 m + 9 > 0 and (4 m2 +46)2- 4(16 m + 9)(16 m + 9) <= 0
<=> 16 m + 9 > 0 and 16m4 - 656 m2 - 1152m + 1792 <= 0
<=> 16 m + 9 > 0 and 16(m-7)(m-1)(m+4)2<= 0
<=> m > -9/16 and m in [1,7]
<=> m in [1,7]
For m = 1 or m = 7
m x2 + 3 x - 4
----------------
m + 3 x - 4 x2
can be simplified and then the range is not R because of a horizontal
asymptote.
So, the range is R if and only if m in ]1,7[.
Solving a system with exponential and trigonometric functions
Solve the system
/
| e2x + cos2(y) = 1
|
|
| e3x cos(y) = cos(3 y)
\
|
<=>
/
| e2x = sin2(y)
|
|
| e3x cos(y) = cos(3 y)
\
<=>
/
| ex = sin (y)
(*) |
|
| sin3 (y) cos(y) = 4 cos3 (y) - 3 cos(y)
\
OR
/
| ex = - sin (y)
|
(**) |
| - sin3 (y) cos(y) = 4 cos3 (y) - 3 cos(y)
\
First consider the case cos(y) = 0 , THEN
x x
sin(y) = 1 or -1 and e = 1 since e = - 1 is impossible.
Then x = 0.
This gives the set of solutions
x = 0 and y = pi/2 + k. pi (k is an integer)
Now consider the first system (*) and cos(y) not 0, THEN
(*)
<=>
/
| ex = sin (y)
|
|
| sin3 (y) = 4 cos2 (y) - 3
\
<=>
/
| ex = sin (y)
|
|
| sin3 (y) = 4 (1 -sin2 (y)) - 3
\
Let t = sin(y) ; then the last equation becomes
t3 + 4 t2 - 1 = 0
Since sin(y) = ex > 0, we are looking for the positive root.
Studying this curve we find that there is only one positive root, and this
root is in [0,1].
With the Newton approximation method we find t = 0.472833908996.
So,
sin(y) = sin(0.492504155559)
y = 0.492504155559 + k.2.pi or y = pi - 0.492504155559 + k.2.pi
x
e = 0.472833908996 => x = -0.749011095927
The solutions of the first system (*) with cos(y) not 0 are
(x = -0.749011095927 ; y = 0.492504155559 + k.2.pi)
and (x = -0.749011095927 ; y = 2.64908849803 + k.2.pi)
Similarly you can find the solutions of the system (**) with cos(y) not 0.
Furthermore it is easy to see that
(xo,yo) is a solution of (*)
<=>
(xo,-yo) is a solution of (**)
So, the solutions of the system (**) with cos(y) not 0 are
(x = -0.749011095927 ; y = - 0.492504155559 + k.2.pi)
and (x = -0.749011095927 ; y = - 2.64908849803 + k.2.pi)
Trigonometry and inequalities.
Draw, in an orthonormal coordinate system, the area covered by the
points P(x,y) defined by the requirements
sin(pi.x/2) < cos(pi.y/2)
x2 + y2 < 4
|
The second requirement means that the points are inside the circle c with origin o(0,0) and radius = 2.
As to the first requirement, we first calculate the points such that
sin(pi.x/2) = cos(pi.y/2)
<=> cos(pi/2 - pi.x/2) = cos(pi.y/2)
<=> pi/2 - pi.x/2 = pi.y/2 + 2.k.pi or pi/2 - pi.x/2 = - pi.y/2 + 2.k.pi
<=> 1 - x = y + 4.k or 1 - x = - y + 4.k
<=> y = - x + 1 - 4.k or y = x - 1 + 4.k
Here, k is an integer and we get a set of lines. Since x and y must be in
the circle c, it is sufficient to take only these lines that are in or near this
circle. The lines divide the circle in 4 regions.
The function sin(pi.x/2) - cos(pi.y/2) only changes sign, if we cross a blue
line.
Therefore the sign of this function is fixed for all the points in the
same region.
In region 1 we have sin(pi.x/2) > cos(pi.y/2)
In region 2 we have
sin(pi.x/2) < cos(pi.y/2)
In region 3 we have sin(pi.x/2) >
cos(pi.y/2)
In region 4 we have sin(pi.x/2) < cos(pi.y/2)
The points P(x,y) defined by the two requirements are the points inside the circle c and in the regions 2 or 4.
A circle, projections, distance, area.
| Take a circle C with two fixed diameters AA' and BB' enclosing a sharp
angle s and point P is rotating on C. Consider the orthogonal projection Q and R of point P on AA' and BB'. We'll study the properties of this moving figure. |
Coordinate system
Choose an orthonormal coordinate system with the midpoint of the circle as origin and the x-axis on AA'. Choose the radius of the circle as unity.
Then we can write the coordinates of variable point P as (cos(t),sin(t)). Then we have Q(cos(t),0);
To calculate the coordinates of R, we use
R = (P.B')B' = cos(t-s) . B' Then R(cos(t-s).cos(s),cos(t-s).sin(s))
The distance |QR| using coordinates
|QR|2 = (cos(t-s).cos(s) - cos(t))2 + (cos(t-s).sin(s)))2
= ...
= sin2(s)
=> |QR| = sin(s) is independent of the position of point P.
The distance |QR| geometrical
The circle K through PRO has central line PO.
The circle K' through PQO
has central line PO.
Thus circle K = K' and contains points P,R,Q,O and has
radius 1/2.
The sine rule in triangle QRO gives |QR| = sin(s) and is
independent of the position of point P.
The area QOR
let f = angle OQR = angle OPR = pi/2 - angle POR = pi/2 - (t-s)
Thus f = pi/2 - (t-s)
The area QOR = (1/2).|OQ|.|QR|.sin(f)
= (1/2).| cos(t).sin(s). sin(f) |
= (1/2).| cos(t).sin(s).cos(t-s) |
= (1/2).sin(s). | cos(t).cos(t-s) |
But cos(t).cos(t-s) = (1/2).(cos(s) + cos(2t-s))
So, the area QOR = (1/4).sin(s).|cos(s) + cos(2t-s)|
If t changes there is a maximum area for
cos(2t-s) = 1
<=> 2t - s = 2.k. pi
<=> t = s/2 + k.pi
The maximum area is then (1/4).sin(s).(cos(s) + 1)
= sin(s/2).cos3(s/2)
If t changes there is a minimum area for
cos(2t-s) = -1
<=> 2t - s = (2k+1). pi
<=> t = s/2 + k.pi + pi/2
The minimum area is then (1/4).sin(s).(1 - cos(s))
= cos(s/2).sin3(s/2)
Movement of QR
Since f = pi/2 - (t-s) If P moves one time around on the circle, t grows from
0 to 2.pi.
Then f varies from (pi/2 + s) to (pi/2 + s - 2 pi).
Thus, with
respect to point Q, point R describes a circle with radius sin(s).
Information Provided by Johan.Claeys@ping.be