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.
For which values of m the quadratic equation
2x2 + mx - 12 = 0 (1)
has a common root with the quadratic equation
x2 - (m+1)x + m = 0 (2)
The question is equivalent to:
Find the necessary and sufficient condition so that the system formed by the two
quadratic equations has at least one solution for x.
( see page elimination )
We have to eliminate x from te system formed by (1) and (2).
First we make sure that x2 occurs in only 1 of the equations.
(1) - 2.(2) gives
(3m+2)x - 2m - 12 = 0. (3)
From this we calculate x and we put this value in (1). This gives rise to the coexistence
condition. We find
m3 - 10 m2 - 4 m + 40 = 0
After factoring : (m + 2) (m - 2) (m - 10) = 0
Conclusion: For m = -2 or m = 2 or m = 10 the two equations (1) and (2) have a common root.
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
Let G be the graph of
9x2 + mx + 4
f(x) = ---------------- with m as real parameter
2x - 7
Calculate m such that G and the x-axis have just one point in common.
Calculate m such that the line 9x - 2y + 3 = 0 is asymptote of G.
Calculate the slope of the tangent line in the intersection point of G
and the y-axis.
Calculate
/
| f(x) dx
/
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 = 3/2 <=> m = -57/2
(2x-7)(18x+m)- (9x2 + mx + 4).(2)
f'(x) = -------------------------------------------
(2x-7)2
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
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)
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)
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 solutions 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 solutions are x = c + 2.k.pi and x = pi - c + 2.k.pi
Given:
f(x) = (m - 1)cos2(x) -3m cos(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) -3m cos(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.
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 relative 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
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)
z is symmetric and homogeneous 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
From all this we know that
z = (x + y)5 - x5 - y5= x.y.(x + y).(5x2+ axy + 5y2)
For x = 1 and y = 1 we have
32 -2 = 2.(5 + a + 5) => a = 5
Hence,
z = (x + y)5 - x5 - y5= 5x.y.(x + y).(x2+ xy + y2)
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
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.
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[.
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)
Send all suggestions, remarks and reports on errors to Johan.Claeys@ping.be
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