Question
A curve has equation \(y=x^{2}-x+3\) and a line has equation y = 3x + a, where a is a constant.
(i) Show that the x-coordinates of the points of intersection of the line and the curve are given by the equation \(x^{2}-4x+(3+a)=0\).
(ii) For the case where the line intersects the curve at two points, it is given that the x-coordinate of one of the points of intersection is −1. Find the x-coordinate of the other point of intersection.
(iii) For the case where the line is a tangent to the curve at a point P, find the value of a and the coordinates of P.
▶️Answer/Explanation
(i) To find the x-coordinates of the points of intersection between the line \(y = 3x + a\) and the curve \(y = x^{2} – x + 3\), we need to set the two equations equal to each other and solve for x:
\(x^{2} – x + 3 = 3x +a\)
\(x^{2} – 4x + (3 -a) = 0\)
Hence, the x-coordinates of the points of intersection are given by the equation \(x^{2} – 4x + (3-a) = 0\).
(ii) Given that one of the points of intersection has an x-coordinate of -1, we can substitute this value into the equation \(x^{2} – 4x + (3 + a) = 0\) and solve for a:
\((-1)^{2} – 4(-1) + (3 -a) = 0\)
\(1 + 4 + (3 -a) = 0\)
\(8 – a = 0\)
Solving for a, we find \(a = 8\).
To find the x-coordinate of the other point of intersection, we can substitute \(a = 8\) back into the equation \(x^{2} – 4x + (3- a) = 0\) and solve for x:
\(x^{2} – 4x + (3 – 8) = 0\)
\(x^{2} – 4x – 5 = 0\)
\((x – 5)(x + 1) = 0\)
\(x – 5 = 0 \quad \text{or} \quad x + 1 = 0\)
Solving for x, we find \(x = 5\) or \(x = -1\).
Since we are given that one of the points of intersection has an x-coordinate of -1, the x-coordinate of the other point of intersection is \(x = 5\).
(iii) If the line is tangent to the curve at a point P, it means that the line intersects the curve at only one point. In this case, the discriminant of the quadratic equation \(x^{2} – 4x + (3 – a) = 0\) must be zero.
The discriminant is given by \(b^{2} – 4ac\), where \(a = 1\), \(b = -4\), and \(c = 3 – a\):
\((-4)^{2} – 4(1)(3 – a) = 0\)
\(16 – 4(3 – a) = 0\)
\(16 – 12 + 4a = 0\)
\(4 + 4a = 0\)
Solving for a, we find \(a = -1\).
Substituting this value of a back into the equation \(y = 3x + a\), we get \(y = 3x – 1\).
To find the coordinates of point P, we substitute \(y = 3x + 1\) into the equation of the curve:
\(3x – 1 = x^{2} – x + 3\)
\(x^{2} – 4x + 4 = 0\)
\left ( x-2 \right )^{2}=0
\(x=2,2\)
So, the x-coordinates of the point P are \(x=2\) and \(x=2.\)
To find the corresponding y-coordinates, we substitute these values of x into the equation \(y=3x-1\)
\(y=3\times 2-1\)
\(y=5\)
Therefore, the coordinates of the point P is\( (2,5).\)
Question.
(a) Find the values of the constant m for which the line y = mx is a tangent to the curve \(y = 2x^2 – 4x + 8\).
(b) The function f is defined for x \in R by \(f(x) = x^2 + ax + b\), where a and b are constants. The
solutions of the equation f(x) = 0 are x = 1 and x = 9.
Find
(i) the values of a and b,
(ii) the coordinates of the vertex of the curve y = f(x).
▶️Answer/Explanation
(a)(i) To find the values of the constant m for which the line y = mx is a tangent to the curve \(y = 2x^2 – 4x + 8\), we need to equate the slopes of the line and the curve at their point of intersection.
The equation of the line is y = mx, where the slope is m.
The equation of the curve is \(y = 2x^2 – 4x + 8\), where the slope can be found by taking the derivative of the curve with respect to x,
\(y’ = \frac{d}{dx}(2x^2 – 4x + 8)\)
= \(4x – 4\)
Now,for finding point of intersection ,we need to equate equation f ;ine equals equation of curve,
\(mx = 2x^2 – 4x + 8\)
\(2x^2 – (4 + m)x + 8 = 0\)
Since the line is a tangent, it intersects the curve at only one point. This means the quadratic equation above should have exactly one solution. For a quadratic equation to have one solution, the discriminant \((b^2 – 4ac)\) should be equal to zero.
\((4 + m)^2 – 4(2)(8) = 0\)
\(16 + 8m + m^2 – 64 = 0\)
\(m^2 + 8m – 48 = 0\)
\((m + 12)(m – 4) = 0\)
Setting each factor equal to zero and solving for m:
\(m + 12 = 0\Rightarrow m = -12\)
\(m – 4 = 0\Rightarrow m = 4\)
So the values of the constant m for which the line y = mx is a tangent to the curve \(y = 2x^2 – 4x + 8\) are m = -12 and m = 4.
(b) Given that the solutions of the equation f(x) = 0 are x = 1 and x = 9, we can write the quadratic equation in the form:
\(f(x) = (x – 1)(x – 9) = 0\)
\(f(x) = x^2 – 10x + 9 = 0\)
Comparing this equation to the general form \(f(x) = x^2 + ax + b\), we can equate the coefficients:
\(a = -10\)
\(b = 9\)
The values of a and b are \(a = -10\) and \(b = 9.\)
(ii) To find the coordinates of the vertex of the curve y = f(x), we can use the formula \(x=-\frac{b}{2a}\)
Substituting the values of a and b:
\(x = \frac{-(-10)}{2\times 1}\)
\(x = \frac{10}{2}\)
x = 5
To find the y-coordinate of the vertex, we substitute the value of x into the equation
\(y = f(5) = 5^2 + (-10)\times 5 + 9\)
\(y = 25 – 50 + 9\)
\(y = -16\)
Therefore, the coordinates of the vertex of the curve y = f(x) are (5, -16).
Question.
Find the set of values of k for which the equation \(2x^2 + 3kx + k = 0\) has distinct real roots.
▶️Answer/Explanation
For the equation \(2x^2 + 3kx + k = 0\) to have distinct real roots, the discriminant should be greater than zero. The discriminant is given by:
\(\Delta = b^2 – 4ac\)
Comparing the equation \(2x^2 + 3kx + k = 0\) to the general quadratic equation \(ax^2 + bx + c = 0\), we can determine the values of a, b, and c:
\(a = 2\)
\(b = 3k\)
\(c = k\)
\(\Delta = (3k)^2 – 4(2)(k)\)
\(\Delta = 9k^2 – 8k\)
For the equation to have distinct real roots, the discriminant \(\Delta\) should be greater than zero:
\(\Delta > 0\)
\(9k^2 – 8k > 0\)
\(k(9k – 8) > 0\)
Combining both intervals, we find the set of values of k for which the equation \(2x^2 + 3kx + k = 0\) has distinct real roots:
\(k ∈ (-∞,0) ∪ (\frac{8}{9},+∞)\)
Question
The function f is defined by \(f(x)=-2x^{2}+12x-3\) for \(\varepsilon R\)
(i) Express \(-2x^{2}+12x-3 \)in the from \(2(x+a)^{a}+b,\)where a and b are constants.
(ii) State the greatest value of f(x).
The function g is defined by g(x) = 2x + 5 .
(iii) Find the values of x for which gf(x )+1=0.
▶️Answer/Explanation
(i) To express the function \(-2x^2 + 12x – 3\) in the form \(2(x + a)^2 + b\), we need to complete the square. Let’s go through the steps:
\(-2x^2 + 12x – 3 = -2(x^2 – 6x) – 3 = -2(x^2 – 6x + 9 – 9) – 3\)
Expanding and simplifying:
\(-2(x^2 – 6x + 9) + 18 – 3 = -2(x – 3)^2 + 15\)
Therefore, \(-2x^2 + 12x – 3\) can be expressed as \(2(x – 3)^2 + 15\) in the desired form, where \(a = 3\) and \(b = 15\).
(ii) To find the greatest value of \(f(x)\), we observe that the quadratic term \(-2x^2\) has a negative coefficient. This means the graph of \(f(x)\) opens downward and the vertex represents the maximum point. The x-coordinate of the vertex can be found using the formula \(-\frac{b}{2a}\) for a quadratic in the form \(ax^2 + bx + c\).
For \(f(x) = -2x^2 + 12x – 3\), the coefficient of the quadratic term is \(a = -2\) and the coefficient of the linear term is \(b = 12\). Using the formula, we can find:
\(x_{\text{vertex}} = -\frac{12}{2(-2)} = -\frac{12}{-4} = 3\)
Substituting this x-coordinate into the function, we find:
\(f(3) = -2(3)^2 + 12(3) – 3 = -18 + 36 – 3 = 15\)
Therefore, the greatest value of \(f(x)\) is 15.
(iii) To find the values of \(x\) for which \(g(f(x)) + 1\), we need to substitute \(f(x)\) into \(g(x)\) and solve for \(x\):
\(g(f(x)) + 1 = 2f(x) + 5 + 1 = 2(-2x^2 + 12x – 3) + 6 = -4x^2 + 24x – 1+1\)
Now, we set this expression equal to \(x\) and solve for \(x\):
\(-4x^2 + 24x = 0\)
\(4x^{2}-24x=0\)
\(4x(x-6)=0\)
\(x=6\) and\( x=0\)