# CIE A level -Pure Mathematics 1 : Topic : 1.1 Quadratics: discriminant of a quadratic polynomial : Exam Questions Paper 1

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.

(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).

(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.

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:
$$k ∈ (-∞,0) ∪ (\frac{8}{9},+∞)$$

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