Past Exam Questions

Answer: \(x=1,\ -\frac32,\ \frac43\).

Answer: \(x=1,\ -\frac32,\ \frac43\).

At the points of intersection, the two expressions for \(y\) are equal:

\(7x^3-7x^2-17x-4=x^3-2x^2-4x-16.\)

Rearrange:

\(6x^3-5x^2-13x+12=0.\)

Testing simple roots, \(x=1\) is a root, so \((x-1)\) is a factor. Dividing gives

\(6x^3-5x^2-13x+12=(x-1)(6x^2+x-12).\)

Now factorise the quadratic:

\(6x^2+x-12=(2x+3)(3x-4).\)

Therefore

\((x-1)(2x+3)(3x-4)=0.\)

Hence

\(x=1, \quad x=-\frac32, \quad x=\frac43.\)

Answer: \(k=3\), and the remaining quadratic factor has no real roots.

Answer: \(k=3\), and the line intersects the curve at only one point.

At an intersection,

\(kx+6=x^3-4x^2+3kx+2.\)

Substitute \(x=2\):

\(2k+6=8-16+6k+2.\)

So

\(2k+6=6k-6,\)

which gives

\(k=3.\)

Now substitute \(k=3\) into the intersection equation:

\(3x+6=x^3-4x^2+9x+2.\)

Hence

\(x^3-4x^2+6x-4=0.\)

Since \(x=2\) is a root, factorise:

\(x^3-4x^2+6x-4=(x-2)(x^2-2x+2).\)

The quadratic factor has discriminant

\((-2)^2-4(1)(2)=4-8=-4\lt 0.\)

So it has no real roots. Therefore the only real intersection occurs at \(x=2\), and the line intersects the curve at only one point.

Answer: \(\left(\frac12,\frac14\right)\) and \((2,1)\).

From

\(x^2=5y-1,\)

we get

\(y=\frac{x^2+1}{5}.\)

At an intersection point,

\(\frac{x^2+1}{5}=x^2-2x+1.\)

Multiplying by 5 gives

\(x^2+1=5x^2-10x+5.\)

So

\(4x^2-10x+4=0,\)

or

\(2x^2-5x+2=0.\)

Factorise:

\((2x-1)(x-2)=0.\)

Thus

\(x=\frac12\quad\text{or}\quad x=2.\)

When \(x=\frac12\),

\(y=\frac{(\frac12)^2+1}{5}=\frac14.\)

When \(x=2\),

\(y=\frac{2^2+1}{5}=1.\)

Therefore the points of intersection are

\(\boxed{\left(\frac12,\frac14\right)\text{ and }(2,1)}.\)

Answer: \(k\lt \frac19\) or \(k\gt 1\).

At an intersection point,

\(x-3=k^2x^2+5kx+1.\)

Bring all terms to one side:

\(k^2x^2+(5k-1)x+4=0.\)

For two distinct intersections, this quadratic in \(x\) must have two distinct roots, so its discriminant must be positive:

\((5k-1)^2-4(k^2)(4)\gt 0.\)

Hence

\(25k^2-10k+1-16k^2\gt 0,\)

so

\(9k^2-10k+1\gt 0.\)

Factorise:

\((9k-1)(k-1)\gt 0.\)

The critical values are \(k=\frac19\) and \(k=1\). Since the quadratic in \(k\) is positive outside these roots,

\(\boxed{k\lt \frac19\text{ or }k\gt 1}.\)

Answer: \((3,0)\) and \(\left(-\frac95,-\frac{16}{5}\right)\).

Substitute

\(y=\frac23x-2\)

into

\(x^2+xy=9.\)

This gives

\(x^2+x\left(\frac23x-2\right)=9.\)

Multiplying by \(3\),

\(3x^2+2x^2-6x=27.\)

So

\(5x^2-6x-27=0.\)

Factorising,

\((x-3)(5x+9)=0.\)

Hence

\(x=3\quad\text{or}\quad x=-\frac95.\)

When \(x=3\),

\(y=\frac23(3)-2=0.\)

When \(x=-\frac95\),

\(y=\frac23\left(-\frac95\right)-2=-\frac65-2=-\frac{16}{5}.\)

The points of intersection are

\((3,0)\quad\text{and}\quad \left(-\frac95,-\frac{16}{5}\right).\)

Answer: \(\left(-\frac32,0\right)\), \((-2,-3)\), and \(\left(\frac32,18\right)\).

At the intersections, the two expressions for \(y\) are equal:

\((2x+3)^2(x-1)=3(2x+3).\)

Move all terms to one side and factorise:

\((2x+3)^2(x-1)-3(2x+3)=0,\)

\((2x+3)\big((2x+3)(x-1)-3\big)=0.\)

So either \(2x+3=0\), giving \(x=-\frac32\), or

\((2x+3)(x-1)-3=0.\)

Expanding the quadratic gives

\(2x^2+x-6=0=(2x-3)(x+2).\)

Hence \(x=\frac32\) or \(x=-2\).

Substitute into \(y=3(2x+3)\):

\(x=-\frac32 \Rightarrow y=0,\qquad x=-2 \Rightarrow y=-3,\qquad x=\frac32 \Rightarrow y=18.\)

The coordinates are

\(\boxed{\left(-\frac32,0\right),\ (-2,-3),\ \left(\frac32,18\right)}.\)

Answer: \(\left(-\dfrac1{20},-\dfrac{31}{5}\right)\).

At the intersections,

\(10x^3-19x^2-x=4x-6.\)

Bring all terms to one side:

\(10x^3-19x^2-5x+6=0.\)

Since \(C=(2,2)\), \(x=2\) is one root. Therefore \(x-2\) is a factor.

Dividing gives

\(10x^3-19x^2-5x+6=(x-2)(10x^2+x-3).\)

Now factorise the quadratic:

\(10x^2+x-3=(5x+3)(2x-1).\)

So the other two \(x\)-coordinates are

\(x=-\frac35,\qquad x=\frac12.\)

These correspond to \(A\) and \(B\). Use \(y=4x-6\) to find their \(y\)-coordinates:

\(x=-\frac35\quad\Rightarrow\quad y=4\left(-\frac35\right)-6=-\frac{42}{5},\)

and

\(x=\frac12\quad\Rightarrow\quad y=4\left(\frac12\right)-6=-4.\)

The midpoint of \(AB\) is therefore

\(\left(\frac{-\frac35+\frac12}{2},\frac{-\frac{42}{5}-4}{2}\right).\)

This simplifies to

\(\boxed{\left(-\frac1{20},-\frac{31}{5}\right)}.\)

Answer: \(C=\left(\frac12,\frac{21}{2}\right)\), so \(x+y=11\).

Substitute \(y=x+10\) into \(xy=11x+5\):

\(x(x+10)=11x+5.\)

So

\(x^2-x-5=0.\)

Using the quadratic formula,

\(x=\frac{1\pm\sqrt{21}}{2}.\)

Since \(y=x+10\), the corresponding \(y\)-coordinates are

\(y=\frac{21\pm\sqrt{21}}{2}.\)

The midpoint \(C\) has coordinates

\(\left(\frac{\frac{1+\sqrt{21}}2+\frac{1-\sqrt{21}}2}{2},\frac{\frac{21+\sqrt{21}}2+\frac{21-\sqrt{21}}2}{2}\right).\)

This simplifies to

\(C=\left(\frac12,\frac{21}{2}\right).\)

Now

\(\frac12+\frac{21}{2}=\frac{22}{2}=11.\)

Therefore \(C\) lies on the line

\(\boxed{x+y=11}.\)

Answer: \(-1\lt k\lt 11\).

At an intersection,

\(kx+3=x^2+5x+12.\)

Rearranging gives

\(x^2+(5-k)x+9=0.\)

For the line not to meet the curve, this quadratic must have no real roots. Therefore its discriminant must be negative:

\((5-k)^2-4(1)(9)\lt0.\)

So

\((5-k)^2\lt36.\)

This gives

\(-6\lt5-k\lt6.\)

Hence

\(\boxed{-1\lt k\lt 11}.\)

Answer: \(-13\lt k\lt11\).

At a point of intersection, the two expressions for \(y\) are equal:

\(1-2kx=9x^2-(3k+1)x+5.\)

Bring all terms to one side:

\(9x^2-(3k+1)x+5-1+2kx=0.\)

So

\(9x^2-(k+1)x+4=0.\)

For the line not to meet the curve, this quadratic must have no real roots. Therefore its discriminant must be negative:

\((k+1)^2-4(9)(4)\lt0.\)

Thus

\((k+1)^2\lt144.\)

Taking square roots gives

\(-12\lt k+1\lt12.\)

Therefore

\(-13\lt k\lt11.\)

Answer: \(\left(\dfrac5{12},\dfrac5{12}\right)\).

At the intersections, substitute \(y=2x+5\) into \(y+xy=5\):

\((2x+5)+x(2x+5)=5.\)

This gives

\(2x^2+7x=0.\)

So

\(x(2x+7)=0,\)

and hence \(x=0\) or \(x=-\frac72\).

Using \(y=2x+5\), the intersection points are

\((0,5)\quad\text{and}\quad\left(-\frac72,-2\right).\)

The midpoint of \(AB\) is

\(\left(-\frac74,\frac32\right).\)

The gradient of \(AB\) is

\(\frac{5-(-2)}{0-(-7/2)}=2.\)

Therefore the gradient of the perpendicular bisector is \(-\frac12\). Its equation is

\(y-\frac32=-\frac12\left(x+\frac74\right).\)

So

\(y=-\frac12x+\frac58.\)

At the intersection with \(y=x\),

\(x=-\frac12x+\frac58.\)

Hence

\(\frac32x=\frac58,\qquad x=\frac5{12}.\)

Since \(y=x\), the required point is

\(\boxed{\left(\frac5{12},\frac5{12}\right)}.\)

Answer: \(k=-10\) or \(k=2\); \(A=(-4,20)\), \(B=(2,8)\); perpendicular bisector \(y=\dfrac12x+14.5\).

(i) At a tangent point, the line and curve meet with one repeated solution. Equate the line and the curve:

\(12-2x=k+6+kx-x^2.\)

Bring all terms to one side:

\(x^2-(k+2)x+6-k=0.\)

For tangency, the quadratic has equal roots, so its discriminant is zero:

\((k+2)^2-4(6-k)=0.\)

Expand and simplify:

\(k^2+4k+4-24+4k=0,\)

so

\(k^2+8k-20=0.\)

Factorise:

\((k+10)(k-2)=0.\)

Therefore

\(k=-10\quad\text{or}\quad k=2.\)

(ii) If \(k=-10\), the quadratic becomes

\(x^2+8x+16=0,\)

so \(x=-4\). On the line \(y=12-2x\),

\(y=12-2(-4)=20.\)

So one tangent point is \((-4,20)\).

If \(k=2\), the quadratic becomes

\(x^2-4x+4=0,\)

so \(x=2\). Then

\(y=12-2(2)=8.\)

So the other tangent point is \((2,8)\).

(iii) Let

\(A=(-4,20),\qquad B=(2,8).\)

The gradient of \(AB\) is

\(\frac{8-20}{2-(-4)}=\frac{-12}{6}=-2.\)

So the perpendicular gradient is \(\dfrac12\).

The midpoint of \(AB\) is

\(\left(\frac{-4+2}{2},\frac{20+8}{2}\right)=(-1,14).\)

The perpendicular bisector is therefore

\(y-14=\frac12(x+1).\)

Hence

\(\boxed{y=\frac12x+14.5}.\)

Answer: (i) \(k=6\) or \(k=10\). (ii) \((2,1)\).

(i) At a point of contact, the tangent and curve have exactly one common point. Equate the line and curve:

\(2k+1-kx=12-4x-x^2.\)

Bring all terms to one side:

\(x^2+(4-k)x+(2k-11)=0.\)

For tangency, this quadratic has a repeated root, so its discriminant is zero:

\((4-k)^2-4(2k-11)=0.\)

Expand and simplify:

\(k^2-8k+16-8k+44=0,\)

so

\(k^2-16k+60=0.\)

Factorise:

\((k-6)(k-10)=0.\)

Thus

\(k=6\quad\text{or}\quad k=10.\)

(ii) If \(k=6\), the tangent is

\(y=13-6x.\)

If \(k=10\), the tangent is

\(y=21-10x.\)

At their intersection,

\(13-6x=21-10x.\)

So

\(4x=8, \qquad x=2.\)

Substitute into \(y=13-6x\):

\(y=13-12=1.\)

Therefore the point of intersection is

\(\boxed{(2,1)}.\)

Answer: (i) \(k=11\) or \(k=3\). (ii) For \(k=11\): tangent \(y=-5x+16\), curve \(y=7-11x-x^2\), point \((-3,31)\). For \(k=3\): tangent \(y=-5x+8\), curve \(y=7-3x-x^2\), point \((1,3)\). (iii) Distance \(=\sqrt{800}=20\sqrt2\).

(i) At a point of contact, the line and curve meet in one repeated root. Equate the two expressions for \(y\):

\(-5x+k+5=7-kx-x^2.\)

Bring all terms to one side:

\(x^2+(k-5)x+(k-2)=0.\)

For tangency, the discriminant is zero:

\((k-5)^2-4(k-2)=0.\)

Expand and simplify:

\(k^2-10k+25-4k+8=0.\)

So

\(k^2-14k+33=0.\)

Factorise:

\((k-11)(k-3)=0.\)

Hence

\(k=11 \quad\text{or}\quad k=3.\)

(ii) If \(k=11\), the tangent is

\(y=-5x+16\)

and the curve is

\(y=7-11x-x^2.\)

Equate them:

\(-5x+16=7-11x-x^2.\)

This gives

\(x^2+6x+9=0,\)

so

\((x+3)^2=0, \qquad x=-3.\)

Then

\(y=-5(-3)+16=31.\)

So the point of contact is \((-3,31)\).

If \(k=3\), the tangent is

\(y=-5x+8\)

and the curve is

\(y=7-3x-x^2.\)

Equate them:

\(-5x+8=7-3x-x^2.\)

This gives

\(x^2-2x+1=0.\)

So

\((x-1)^2=0, \qquad x=1.\)

Then

\(y=-5(1)+8=3.\)

So the point of contact is \((1,3)\).

(iii) The two points are \((-3,31)\) and \((1,3)\). Their distance is

\(\sqrt{(1-(-3))^2+(3-31)^2}.\)

So

\(d=\sqrt{4^2+(-28)^2} = \sqrt{16+784} = \sqrt{800} = 20\sqrt2.\)

Answer: \(A\left(-\dfrac13,\dfrac83\right)\), \(B(1,0)\); perpendicular bisector \(3x-6y=-7\).

(i) Substitute \(y=2(1-x)=2-2x\) into the curve:

\(3x^2+x(2-2x)-(2-2x)^2+4(2-2x)-3=0.\)

Expand and simplify:

\(3x^2+2x-2x^2-(4-8x+4x^2)+8-8x-3=0,\)

\(-3x^2+2x+1=0.\)

Equivalently,

\(3x^2-2x-1=0.\)

Factorise:

\((3x+1)(x-1)=0.\)

So

\(x=-\frac13 \quad\text{or}\quad x=1.\)

Find the corresponding \(y\)-values from \(y=2-2x\):

If \(x=-\dfrac13\), then

\(y=2-2\left(-\frac13\right)=2+\frac23=\frac83.\)

If \(x=1\), then

\(y=2-2(1)=0.\)

Hence the intersection points are

\(\boxed{\left(-\frac13,\frac83\right)\text{ and }(1,0)}.\)

(ii) The midpoint of \(AB\) is

\(\left(\frac{-1/3+1}{2},\frac{8/3+0}{2}\right) =\left(\frac13,\frac43\right).\)

The gradient of \(AB\) is

\(\frac{0-\frac83}{1-(-1/3)}=\frac{-8/3}{4/3}=-2.\)

So the perpendicular bisector has gradient

\(\frac12.\)

Its equation through \(\left(\frac13,\frac43\right)\) is

\(y-\frac43=\frac12\left(x-\frac13\right).\)

Multiply through by \(6\):

\(6y-8=3x-1.\)

Hence

\(3x-6y=-7.\)

Therefore the perpendicular bisector is

\(\boxed{3x-6y=-7}.\)

(i) To find the value of h, we use the fact that ∠ABC = 90° and lines BC and AD are parallel. The gradient of AB is \(-\frac{1}{2}\) and the gradient of BC is \(\frac{3h - 2}{h}\). Since BC is perpendicular to AB, the product of their gradients is \(-1\):

\(\left(-\frac{1}{2}\right) \times \frac{3h - 2}{h} = -1\)

Solving for h, we get:

\(\frac{3h - 2}{h} = 2\)

\(3h - 2 = 2h\)

\(h = 2\)

(ii) With h = 2, the coordinates of C are (2, 6). Since CD is parallel to the x-axis, the y-coordinate of D is the same as C, which is 6. The x-coordinate of D can be found using the fact that AD is parallel to BC, so the gradient of AD is also 2:

\(\frac{6 - 0}{x - 4} = 2\)

\(6 = 2(x - 4)\)

\(6 = 2x - 8\)

\(2x = 14\)

\(x = 7\)

Thus, the coordinates of D are (7, 6).

(i) To find the equation of the perpendicular bisector of AC:

The midpoint of AC is \(\left( \frac{-1 + 5}{2}, \frac{2 + 4}{2} \right) = (2, 3)\).

The gradient of AC is \(\frac{4 - 2}{5 + 1} = \frac{1}{3}\).

The gradient of the perpendicular bisector is the negative reciprocal, \(-3\).

Using the point-slope form, the equation is \(y - 3 = -3(x - 2)\).

(ii) To find the coordinates of B and D:

Since B lies on the y-axis, its x-coordinate is 0. Using the perpendicular bisector equation, \(y - 3 = -3(0 - 2)\), we find \(y = 9\). Thus, \(B(0, 9)\).

To find D, use the vector move from A to C, which is \((5 - (-1), 4 - 2) = (6, 2)\).

Since D is opposite B, apply the vector move to B: \((0 + 4, 9 - 12) = (4, -3)\). Thus, \(D(4, -3)\).

(iii) To find the area of the rhombus:

The length of AC is \(\sqrt{(5 - (-1))^2 + (4 - 2)^2} = \sqrt{40}\).

The length of BD is \(\sqrt{(4 - 0)^2 + (-3 - 9)^2} = \sqrt{160}\).

The area of the rhombus is \(\frac{1}{2} \times \sqrt{40} \times \sqrt{160} = 40\).

The gradient of OB is calculated as follows:

Gradient of OB = \(\frac{1 - 0}{3 - 0} = \frac{1}{3}\).

Since L₁ is parallel to OB, it has the same gradient: \(\frac{1}{3}\).

The equation of L₁ passing through A (-1, 3) is:

\(y - 3 = \frac{1}{3}(x + 1)\).

The gradient of AB is:

Gradient of AB = \(\frac{1 - 3}{3 + 1} = -\frac{1}{2}\).

The gradient of L₂, being perpendicular to AB, is the negative reciprocal:

\(Gradient of L₂ = 2.\)

The equation of L₂ passing through B (3, 1) is:

\(y - 1 = 2(x - 3)\).

Solving the simultaneous equations:

1. \(y = \frac{1}{3}x + \frac{10}{3}\)

2. \(y = 2x - 5\)

Equating the two expressions for \(y\):

\(\frac{1}{3}x + \frac{10}{3} = 2x - 5\)

Multiply through by 3 to clear the fraction:

\(x + 10 = 6x - 15\)

Rearrange to solve for \(x\):

\(5x = 25\)

\(x = 5\)

Substitute \(x = 5\) back into one of the equations to find \(y\):

\(y = 2(5) - 5 = 5\)

Thus, the coordinates of C are (5, 5).

(i) The gradient of AB is calculated as:

\(\frac{22 - (-2)}{15 - 3} = \frac{24}{12} = 2\)

Since the gradient of AB is given as 2m, equating gives:

\(2m = 2\)

Thus, \(m = 1\).

(ii) The equation of line AC is:

\(y + 2 = -2(x - 3)\)

\(y = -2x + 6 - 2\)

\(y = -2x + 4\)

The equation of line BC is:

\(y - 22 = 1(x - 15)\)

\(y = x + 7\)

Solving the simultaneous equations:

\(y = -2x + 4\)

\(y = x + 7\)

Equating gives:

\(-2x + 4 = x + 7\)

\(-3x = 3\)

\(x = -1\)

Substitute \(x = -1\) into \(y = x + 7\):

\(y = -1 + 7 = 6\)

Thus, \(C = (-1, 6)\).

(iii) The midpoint M of AB is:

\(M = \left( \frac{3 + 15}{2}, \frac{-2 + 22}{2} \right) = (9, 10)\)

The gradient of the perpendicular bisector is:

\(-\frac{1}{2}\)

The equation of the perpendicular bisector is:

\(y - 10 = -\frac{1}{2}(x - 9)\)

\(2y - 20 = -x + 9\)

\(2y + x = 29\)

Solving the equations \(y = x + 7\) and \(2y + x = 29\):

Substitute \(y = x + 7\) into \(2y + x = 29\):

\(2(x + 7) + x = 29\)

\(2x + 14 + x = 29\)

\(3x = 15\)

\(x = 5\)

Substitute \(x = 5\) into \(y = x + 7\):

\(y = 5 + 7 = 12\)

Thus, \(D = (5, 12)\).

(i) The y-coordinate of D is the same as the y-coordinate of the midpoint of AC. The midpoint of AC is \(\left( \frac{0+12}{2}, \frac{-2+14}{2} \right) = (6, 6)\). Therefore, the y-coordinate of D is 6.

(ii) The gradient of AD is calculated as \(\frac{y_2 - y_1}{x_2 - x_1} = \frac{6 - (-2)}{h - 0} = \frac{8}{h}\) or \(\frac{h-12}{8}\). The gradient of CD is \(\frac{6 - 14}{h - 12} = \frac{-8}{h-12}\) or \(\frac{-h}{8}\).

(iii) Since BD is parallel to the x-axis, the product of the gradients of AD and CD is -1. Solving \(\frac{8}{h} \times \frac{-8}{h-12} = -1\) gives \(h^2 - 12h - 64 = 0\). Solving this quadratic equation, we find \(h = 16\) or \(h = -4\). Thus, \(x_D = 16\) and \(x_B = -4\).

(iv) The area of rectangle ABCD can be calculated using the lengths of the sides. The length of AC is \(\sqrt{(12-0)^2 + (14-(-2))^2} = 20\). The length of BD is \(\sqrt{(16-(-4))^2 + (6-6)^2} = 20\). The area is \(20 \times 8 = 160\).