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
2x^{2} + mx  12 = 0 (1)
has a common root with the quadratic equation
x^{2}  (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 x^{2} 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
m^{3}  10 m^{2}  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 relative 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((ba)^{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((ba)^{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((ba)^{2} + (b' a')^{2})
2 sqrt((ba)^{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 relative 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
9x^{2} + mx + 4
f(x) =  with m as real parameter
2x  7
Calculate m such that G and the xaxis 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 yaxis.
Calculate
/
 f(x) dx
/
The graph and the xaxis have just one common point
<=>
9x^{2} + 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
(2x7)(18x+m) (9x^{2} + mx + 4).(2)
f'(x) = 
(2x7)^{2}
The slope is f'(0) = (7m  8)/49
/ 9x^{2} + 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.
9x^{2} + 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)x^{2} + (m/2 + 63/4)x + (7m/2 + 457/4)(1/2)ln2x  7 + C
Given : x and y in ]0, pi/2[
Show that :
1 + 2cos(4y)  (sin(6x)/sin(2x)) = 16.sin(xy)sin(x+y)cos(xy)cos(x+y)
Right side = 4.2.sin(xy)cos(xy).2.sin(x+y)cos(x+y)
= 4.sin 2(x+y).sin 2(xy)
= 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 rightangled triangle if
sin(2 b) sin(2 c) = 2  2 sin^{2}(b) cos^{2}(c)  2 cos^{2}(b) sin^{2}(c)
a, b, c form arithmetic sequence and x, y, z form a geometric sequence.
Prove that
x^{b}.y^{c}.z^{a} = x^{c}.y^{a}.z^{b}
a, b, c form arithmetic sequence => b = (a+c)/2
x, y, z form a geometric sequence => y.y = x.z
x^{b}.y^{c}.z^{a} = x^{(a+c)/2} .(xz)^{c/2} . z^{a}
= x^{(a/2+c)} . z^{(c/2+a)}
x^{c}.y^{a}.z^{b}= x^{c}.(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(2m1) , 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)cos^{2}(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)cos^{2}(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(2x^{2} + k)
The relative maximum or minimum occurs if
2x  sqrt(2x^{2} + 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.n^{2} + 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(2n1) = ... = 4a n  a + b
t'(n)  t'(n1) = ... = 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 relative 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}  x^{5}  y^{5}
From all this we know that
z = (x + y)^{5}  x^{5}  y^{5}= x.y.(x + y).(5x^{2}+ axy + 5y^{2})
For x = 1 and y = 1 we have
32 2 = 2.(5 + a + 5) => a = 5
Hence,
z = (x + y)^{5}  x^{5}  y^{5}= 5x.y.(x + y).(x^{2}+ xy + y^{2})
Given : The function
mx^{2}  7x + 5
f(x) = 
5x^{2}  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.
mx^{2}  7x + 5
 = b
5x^{2}  7x + m
<=>
The following equation has a solution for all fixed real numbers b.
(m  5b)x^{2}  7(1  b)x + (5  mb) = 0
<=>
The previous equation has a positive discriminant for all b
<=>
(49  20m)b^{2} + (4m^{2} + 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 m1
Row 1  Row 4 gives
0 0 0 2m
1 1 m 1  = 0
3 0 1 2 
2 m 1 m1
<=>
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 x^{2} + 3 x  4
f : x > 
m + 3 x  4 x^{2}
is equal to R.
For each real y there must be an x such that
m x^{2} + 3 x  4
y = 
m + 3 x  4 x^{2}
<=> y.(m + 3 x  4 x^{2}) = m x^{2} + 3 x  4
<=> (4y + m)x^{2} + (3y  3)x + (my + 4) = 0
For each y the D must be not negative
<=> (16 m + 9) y^{2}+ (4 m^{2} +46)y + (16m +9) >= 0
<=> 16 m + 9 > 0 and (4 m^{2} +46)^{2} 4(16 m + 9)(16 m + 9) <= 0
<=> 16 m + 9 > 0 and 16m^{4}  656 m^{2}  1152m + 1792 <= 0
<=> 16 m + 9 > 0 and 16(m7)(m1)(m+4)^{2}<= 0
<=> m > 9/16 and m in [1,7]
<=> m in [1,7]
For m = 1 or m = 7
m x^{2} + 3 x  4

m + 3 x  4 x^{2}
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
/
 e^{2x} + cos^{2}(y) = 1


 e^{3x} cos(y) = cos(3 y)
\
<=>
/
 e^{2x} = sin^{2}(y)


 e^{3x} cos(y) = cos(3 y)
\
<=>
/
 e^{x} = sin (y)
(*) 

 sin^{3} (y) cos(y) = 4 cos^{3} (y)  3 cos(y)
\
OR
/
 e^{x} =  sin (y)

(**) 
  sin^{3} (y) cos(y) = 4 cos^{3} (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
(*)
<=>
/
 e^{x} = sin (y)


 sin^{3} (y) = 4 cos^{2} (y)  3
\
<=>
/
 e^{x} = sin (y)


 sin^{3} (y) = 4 (1 sin^{2} (y))  3
\
Let t = sin(y) ; then the last equation becomes
t^{3} + 4 t^{2}  1 = 0
Since sin(y) = e^{x} > 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)
The the tangent line in a variable point P of a parabola y = ax^{2} intersects the xaxis and the yaxis
in the points A and B. Find the locus of the center of the circle through A, B and O (0,0).
The slope of the tangent line to the parabola in point P(x,y) is 2ax.
Using a parameter t, we let slide point P along the parabola.
So, P(t,at^{2}) is a variable point of the parabola and the slope of the tangent line is 2at.
The tangent line is y  at^{2} = 2at (xt) <=> y = 2atx  at^{2}.
The intersection points A and B are A(t/2,0) and B(0,at^{2}).
Since the triangle OAB is a right triangle, AB a diameter of the circle through O, A and B.
The center of [AB] is M(t/4,at^{2}/2). M is also the center of the circle.
M is the intersection point af the associated lines
x = t/4
y = at^{2}/2
The equation of the locus arises by eliminating t.
We find the parabola y = 8x^{2} as the requested locus.
Find the locus of the points S such that the tangent lines from S, to the parabola
y=x^{2}, are orthogonal.
Direct method
Say S(a,b) is a point of the locus.
A line through S has equation y = m(xa) + b
The intersection points of the line and the parabola are the solutions of the system
/ y = x^{2}
\ y = m(xa) + b
So, the xvalues of the intersection points the solutions of
x^{2}  m(xa)  b = 0
<=> x^{2}  m x + m a  b = 0
The line is tangent to the parabola
<=> The intersection points of the line and the parabola coincide
<=> The previous quadratic equation has two equal roots
<=> m^{2}  4(m a  b) = 0
<=> m^{2}  4 a m + 4 b = 0
The roots of the last quadratic equation are the slopes
of the two tangent lines from S, to the parabola.
The two tangent lines are orthogonal
<=> the product of the two slopes is 1
<=> 4b = 1
<=> b = 1/4
Conclusion:
The two tangent lines from S(a,b) to the parabola y = x^{2} are orthogonal
<=> b = 1/4 and a is arbitrary
The locus is described by S(a,1/4), with a is arbitrary.
The equation of the locus is y = 1/4
Method with associated lines and a binding equation
Take a variable point P(a, a^{2}) of the parabola.
The equation of the tangent line in P to the parabola is
y = 2 a x  a^{2} (1)
Take a variable point Q(b, b^{2}) of the parabola with Q different from P.
a  b is not 0
The equation of the tangent line in Q to the parabola is
y = 2 b x  b^{2} (2)
The tangent lines in P and Q are orthogonal if and only if
4 a b = 1 (3)
The two tangent lines (1) and (2) are the associated lines
and the parameters a and b are connected with the binding equation (3)
The equation of the locus arises by eliminating a and b from
/ y = 2 a x  a^{2}
 y = 2 b x  b^{2}
\ 4 a b = 1
It is not simple to eliminate a and b.
By carefully treating the system we are able to establish the elimination.
/ y = 2 a x  a^{2} (1)
 y = 2 b x  b^{2} (2)
\ 4 a b = 1 (3)
we replace (2) by (1)  (2)
<=>
/ y = 2 a x  a^{2}
 0 = 2(ab)x  (a^{2}  b^{2})
\ 4 a b = 1
<=>
/ y = 2 a x  a^{2}
 0 = 2(ab)x  (ab)(a+b)
\ 4 a b = 1
but a  b is not 0
<=>
/ y = 2 a x  a^{2}
 0 = 2x  (a+b)
\ 4 a b = 1
<=>
/ y = 2 a x  a^{2} (1)
 a + b = 2x
\ a b = 1/4
From the last two equations, we see that a and b are two numbers with
a sum equal to 2x and their product is 1/4.
a and b are the solutions of the quadratic equation
X^{2}  (2x) X  1/4 = 0
So, for a we have
a^{2}  (2x) a  1/4 = 0
<=>
2 a x  a^{2} = 1/4
We substitute this result in (1) and a and b are eliminated.
We find : y = 1/4
This is the equation of the locus.
Send all suggestions, remarks and reports on errors to Johan.Claeys@ping.be
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