Past Exam Questions

Answer: \(-1\lt k\lt 3\).

For the quadratic to have no real roots, its discriminant must be negative.

Here \(a=4\), \(b=-4k\), and \(c=2k+3\). Therefore

\(b^2-4ac\lt 0.\)

So

\((-4k)^2-4(4)(2k+3)\lt 0.\)

This gives

\(16k^2-32k-48\lt 0.\)

Divide by \(16\):

\(k^2-2k-3\lt 0.\)

Factorise:

\((k-3)(k+1)\lt 0.\)

The expression is negative between the roots, so

\(\boxed{-1\lt k\lt 3}.\)

Answer: \(k\lt -4\) or \(k\gt 4\).

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

\(2x^2+k+4=(k+4)x.\)

Rearranging gives the quadratic equation

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

For the curve and the line to intersect at two distinct points, this quadratic must have two distinct real roots. Therefore its discriminant must be positive:

\([-(k+4)]^2-4(2)(k+4)\gt 0.\)

So

\((k+4)^2-8(k+4)\gt 0.\)

Factorising,

\((k+4)(k-4)\gt 0.\)

This product is positive when both factors have the same sign, so

\(k\lt -4\quad\text{or}\quad k\gt 4.\)

Answer: \(k\lt-15\) or \(k\gt-3\).

For the quadratic to have two distinct real roots, its discriminant must be positive.

Here

\(a=1,\qquad b=k+9,\qquad c=9.\)

So

\((k+9)^2-4(1)(9)\gt0.\)

This gives

\((k+9)^2-36\gt0.\)

Factorise as a difference of two squares:

\((k+9-6)(k+9+6)\gt0.\)

Therefore

\((k+3)(k+15)\gt0.\)

The product is positive outside the two critical values, so

\(k\lt-15\quad\text{or}\quad k\gt-3.\)

Answer: \(x=2+\sqrt7\) or \(x=\dfrac23+\dfrac13\sqrt7\).

Use the quadratic formula with

\(a=\sqrt7-2,\qquad b=-4,\qquad c=\sqrt7+2.\)

The discriminant is

\(b^2-4ac=(-4)^2-4(\sqrt7-2)(\sqrt7+2).\)

Now

\((\sqrt7-2)(\sqrt7+2)=7-4=3.\)

So

\(b^2-4ac=16-4(3)=4.\)

Hence

\(x=\frac{4\pm\sqrt4}{2(\sqrt7-2)}.\)

So

\(x=\frac{4\pm2}{2(\sqrt7-2)}.\)

The first root is

\(x=\frac6{2(\sqrt7-2)} =\frac3{\sqrt7-2}.\)

Rationalise the denominator:

\(\frac3{\sqrt7-2}\cdot\frac{\sqrt7+2}{\sqrt7+2} =\frac{3(\sqrt7+2)}{7-4} =\sqrt7+2.\)

The second root is

\(x=\frac2{2(\sqrt7-2)} =\frac1{\sqrt7-2}.\)

Rationalise the denominator:

\(\frac1{\sqrt7-2}\cdot\frac{\sqrt7+2}{\sqrt7+2} =\frac{\sqrt7+2}{3}.\)

Therefore

\(x=2+\sqrt7\quad\text{or}\quad x=\frac23+\frac13\sqrt7.\)

Answer: The discriminant is \((m-5)^2\), so it is always non-negative.

At an intersection,

\(mx+4=x^2+3x+m.\)

Rearrange to form a quadratic in \(x\):

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

For the line to touch or intersect the curve, this quadratic must have at least one real root. Its discriminant is

\((3-m)^2-4(1)(m-4).\)

Now simplify:

\((3-m)^2-4(m-4) =m^2-6m+9-4m+16 =m^2-10m+25 =(m-5)^2.\)

Since \((m-5)^2\geq0\) for every real value of \(m\), the quadratic always has real root(s). Hence the line will always touch or intersect the curve.

Answer: \(k\lt 0\) or \(k\gt \dfrac45\).

For real and distinct roots, the discriminant must be positive:

\(b^2-4ac\gt 0.\)

Here \(a=k-1\), \(b=k\), and \(c=-k\). Hence

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

Simplifying,

\(k^2+4k(k-1)\gt 0\)

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

So

\(k(5k-4)\gt 0.\)

The critical values are \(k=0\) and \(k=\frac45\). The product is positive outside these values, so

\(\boxed{k\lt 0\quad\text{or}\quad k\gt \frac45}.\)

Answer: \(-1\lt k\lt 15\).

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

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

Rearrange this into a quadratic equation in \(x\):

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

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

\((3-k)^2-4(2)(k+3)\lt 0.\)

Expanding and simplifying gives

\(k^2-14k-15\lt 0.\)

Factorise:

\((k+1)(k-15)\lt 0.\)

The quadratic is negative between its two roots, so

\(\boxed{-1\lt k\lt 15}.\)

Answer: \(-\dfrac12\lt k\lt \dfrac32\).

For the quadratic to have no real roots, its discriminant must be negative.

Here

\(a=3-2k,\quad b=2k-3,\quad c=1.\)

So the discriminant is

\((2k-3)^2-4(3-2k)(1).\)

Expand and simplify:

\((2k-3)^2-4(3-2k)=4k^2-12k+9-12+8k=4k^2-4k-3.\)

Factorise:

\(4k^2-4k-3=(2k-3)(2k+1).\)

We need

\((2k-3)(2k+1)\lt 0.\)

The critical values are \(k=-\dfrac12\) and \(k=\dfrac32\). Since the quadratic expression is negative between its roots,

\(\boxed{-\dfrac12\lt k\lt \dfrac32}.\)

Answer: The discriminant is \((p+q)^2\ge 0\), so the roots are real.

For the quadratic

\(px^2+(p-q)x-q=0,\)

the discriminant is

\(\Delta=(p-q)^2-4(p)(-q).\)

Simplify:

\(\Delta=p^2-2pq+q^2+4pq=p^2+2pq+q^2.\)

So

\(\Delta=(p+q)^2.\)

A square is always non-negative for all real \(p\) and \(q\), so

\(\Delta\ge 0.\)

Therefore the equation has real roots for all real values of \(p\) and \(q\).

Answer: \(k\lt -\dfrac12\) or \(k\gt \dfrac92\).

For a quadratic equation to have no real roots, its discriminant must be negative.

Here

\(a=k,\quad b=3,\quad c=k-4.\)

So

\(b^2-4ac\lt 0.\)

Hence

\(9-4k(k-4)\lt 0.\)

Expand:

\(9-4k^2+16k\lt 0.\)

Multiply by \(-1\), reversing the inequality:

\(4k^2-16k-9\gt 0.\)

Factor:

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

This product is positive outside the two critical values, so

\(\boxed{k\lt -\dfrac12\ \text{or}\ k\gt \dfrac92}.\)

Answer: \(2x+6y+1=0\).

At the points where the line and curve meet, their \(y\)-values are equal:

\(3x+4=2x^2+8x+1\).

Rearrange to form a quadratic:

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

Factorising,

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

So

\(x=\frac12\quad\text{or}\quad x=-3\).

Use the straight line \(y=3x+4\) to find the corresponding \(y\)-coordinates:

when \(x=\frac12\), \(y=3\left(\frac12\right)+4=\frac{11}{2}\);

when \(x=-3\), \(y=3(-3)+4=-5\).

Thus the two points are

\(A\left(\frac12,\frac{11}{2}\right)\quad\text{and}\quad B(-3,-5)\).

The midpoint of \(AB\) is

\(\left(\dfrac{\frac12+(-3)}{2},\dfrac{\frac{11}{2}+(-5)}{2}\right)=\left(-\frac54,\frac14\right)\).

The gradient of \(AB\) is the same as the gradient of the line \(y=3x+4\), so it is \(3\).

Therefore the perpendicular bisector has gradient

\(-\frac13\).

Using the midpoint \(\left(-\frac54,\frac14\right)\), the perpendicular bisector is

\(y-\frac14=-\frac13\left(x+\frac54\right)\).

Multiply by \(12\):

\(12y-3=-4x-5\).

Rearranging,

\(4x+12y+2=0\).

Dividing by \(2\),

\(2x+6y+1=0\).

Answer: \(A=\left(-\frac32,-9\right)\), \(B=(2,5)\), and the area of \(\triangle POQ\) is \(\dfrac{961}{128}\).

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

\(4x-3=3+5x-2x^2\).

Rearranging gives

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

Factorising,

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

Hence

\(x=-\frac32\quad\text{or}\quad x=2\).

Using the straight line \(y=4x-3\), the corresponding coordinates are

\(x=-\frac32:\quad y=4\left(-\frac32\right)-3=-9\),

and

\(x=2:\quad y=4(2)-3=5\).

So the two points are

\(A=\left(-\frac32,-9\right)\), \(B=(2,5)\).

The gradient of \(AB\) is

\(\dfrac{5-(-9)}{2-\left(-\frac32\right)}=\dfrac{14}{\frac72}=4\).

Therefore the gradient of the perpendicular bisector is \(-\dfrac14\).

The midpoint of \(AB\) is

\(\left(\dfrac{-\frac32+2}{2},\dfrac{-9+5}{2}\right)=\left(\frac14,-2\right)\).

So the perpendicular bisector is

\(y+2=-\dfrac14\left(x-\frac14\right)\).

Simplifying,

\(y=-\dfrac14x-\dfrac{31}{16}\).

When \(y=0\), \(x=-\dfrac{31}{4}\), so \(P=\left(-\dfrac{31}{4},0\right)\).

When \(x=0\), \(y=-\dfrac{31}{16}\), so \(Q=\left(0,-\dfrac{31}{16}\right)\).

The triangle \(POQ\) is right-angled at \(O\), hence

\(\text{area}=\dfrac12\left|OP\right|\left|OQ\right|=\dfrac12\cdot\dfrac{31}{4}\cdot\dfrac{31}{16}=\dfrac{961}{128}\).

Answer: \(k=-\frac45\). The possible coordinates of \(D\) are \(\left(\frac92,-2\right)\) and \(\left(-\frac{51}{10},\frac{14}{5}\right)\).

Work through the given conditions step by step, keeping exact values until the final answer is required.

Substitute \(y=2x+1\) into \(y+xy+3x^2=15\):

\(2x+1+x(2x+1)+3x^2=15.\)

This simplifies to

\(5x^2+3x-14=0.\)

Solving gives \(x=\frac75\) or \(x=-2\). The corresponding \(y\)-values are \(\frac{19}{5}\) and \(-3\).

The midpoint of \(AB\) is

\(\left(\frac{\frac75-2}{2},\frac{\frac{19}{5}-3}{2}\right)=\left(-\frac{3}{10},\frac25\right).\)

The line \(AB\) has gradient \(2\), so the perpendicular bisector has gradient \(-\frac12\). Its equation is

\(y-\frac25=-\frac12\left(x+\frac{3}{10}\right).\)

Since \(C=\left(\frac{21}{10},k\right)\) lies on this line,

\(k-\frac25=-\frac12\left(\frac{21}{10}+\frac{3}{10}\right)=-\frac65.\)

Thus \(k=-\frac45\).

The vector from the midpoint to \(C\) is

\(\left(\frac{21}{10},-\frac45\right)-\left(-\frac{3}{10},\frac25\right)=\left(\frac{12}{5},-\frac65\right).\)

A point twice as far from \(AB\) on the same perpendicular line can lie on either side of the midpoint. Hence

\(D=\left(-\frac{3}{10},\frac25\right)+2\left(\frac{12}{5},-\frac65\right)=\left(\frac92,-2\right)\),

or

\(D=\left(-\frac{3}{10},\frac25\right)-2\left(\frac{12}{5},-\frac65\right)=\left(-\frac{51}{10},\frac{14}{5}\right).\)

This gives the required answer: \(k=-\frac45\). The possible coordinates of \(D\) are \(\left(\frac92,-2\right)\) and \(\left(-\frac{51}{10},\frac{14}{5}\right)\).

Answer: \(P=(\sqrt6,\frac{\sqrt6}{3})\), \(Q=(-\sqrt6,-\frac{\sqrt6}{3})\); \(PQ=\frac{4\sqrt{15}}{3}\).

We start with the main method. Substitute the line into the curve; a tangent corresponds to a repeated root, so the discriminant is zero.

(a) Substitute \(y=\frac2x\) into

\(4x^2-3y^2+xy=24.\)

This gives

\(4x^2-3\left(\frac2x\right)^2+x\left(\frac2x\right)=24.\)

So

\(4x^2-\frac{12}{x^2}+2=24.\)

Multiply by \(x^2\):

\(4x^4-12+2x^2=24x^2.\)

Hence

\(4x^4-22x^2-12=0,\)

or

\(2x^4-11x^2-6=0.\)

Let \(u=x^2\). Then

\(2u^2-11u-6=0.\)

Factorise:

\((2u+1)(u-6)=0.\)

Since \(u=x^2\geq0\), we take \(u=6\). Thus

\(x=\pm\sqrt6.\)

Using \(y=\frac2x\), the points are

\(\left(\sqrt6,\frac{\sqrt6}{3}\right)\quad\text{and}\quad \left(-\sqrt6,-\frac{\sqrt6}{3}\right).\)

(b) The distance between these two points is

\(PQ=\sqrt{(2\sqrt6)^2+\left(\frac{2\sqrt6}{3}\right)^2}.\)

So

\(PQ=\sqrt{24+\frac83}=\sqrt{\frac{80}{3}} =\frac{4\sqrt{15}}{3}.\)

This completes the solution and gives the required result.

Answer: \(k=-\dfrac{19}{8}\), and \(D=\left(\dfrac{29}{8},-\dfrac98\right)\).

We start with the main method. Substitute the line into the curve; a tangent corresponds to a repeated root, so the discriminant is zero.

Substitute \(y=3x-2\) into

\(2x^2-xy+y^2=2.\)

This gives

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

Expanding and simplifying,

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

Factorise:

\((4x-1)(x-1)=0.\)

So

\(x=\frac14\quad\text{or}\quad x=1.\)

Using \(y=3x-2\), the intersection points are

\(\left(\frac14,-\frac54\right)\quad\text{and}\quad(1,1).\)

The midpoint of \(AB\) is

\(\left(\frac{\frac14+1}{2},\frac{-\frac54+1}{2}\right) =\left(\frac58,-\frac18\right).\)

The gradient of \(AB\) is \(3\), because \(AB\) lies on \(y=3x-2\).

Therefore the perpendicular bisector has gradient \(-\frac13\).

Its equation is

\(y+\frac18=-\frac13\left(x-\frac58\right).\)

The point \(C=\left(k,\frac78\right)\) lies on this line, so substitute \(y=\frac78\):

\(\frac78+\frac18=-\frac13\left(k-\frac58\right).\)

So

\(1=-\frac13\left(k-\frac58\right).\)

Hence

\(k-\frac58=-3\), and

\(k=-\frac{19}{8}.\)

For part (b), \(D\) is the reflection of \(C\) in the line \(AB\).

The perpendicular bisector meets \(AB\) at the midpoint of \(CD\), which is also the midpoint of \(AB\):

\(M=\left(\frac58,-\frac18\right).\)

Since \(M\) is the midpoint of \(C\) and \(D\),

\(D=2M-C.\)

Thus

\(D=\left(2\cdot\frac58-\left(-\frac{19}{8}\right),\ 2\cdot\left(-\frac18\right)-\frac78\right).\)

Therefore

\(D=\left(\frac{29}{8},-\frac98\right).\)

This completes the solution and gives the required result.

Answer: (a) \(3\sqrt3\); (b) \(\left(-10+\sqrt{17},-\dfrac{10+\sqrt{17}}{83}\right)\) and \(\left(-10-\sqrt{17},-\dfrac{10-\sqrt{17}}{83}\right)\).

We start with the main method. Substitute the line into the curve; a tangent corresponds to a repeated root, so the discriminant is zero.

For part (a), the curve cuts the \(y\)-axis where \(x=0\).

Substitute \(x=0\) into

\(9(x-1)^2+4(y-3)^2=36.\)

This gives

\(9+4(y-3)^2=36.\)

So

\(4(y-3)^2=27\), and hence

\(y=3\pm\sqrt{\frac{27}{4}}=3\pm\frac{3\sqrt3}{2}.\)

The distance between the two points on the \(y\)-axis is therefore

\(\left(3+\frac{3\sqrt3}{2}\right)-\left(3-\frac{3\sqrt3}{2}\right)=3\sqrt3.\)

For part (b), substitute \(y=\frac1x\) into the other curve:

\(2x^2+83x\left(\frac1x\right)=x^3\left(\frac1x\right)-20x.\)

So

\(2x^2+83=x^2-20x.\)

Hence

\(x^2+20x+83=0.\)

Using the quadratic formula,

\(x=\frac{-20\pm\sqrt{20^2-4(1)(83)}}{2}.\)

Thus

\(x=\frac{-20\pm\sqrt{68}}{2}=-10\pm\sqrt{17}.\)

Now \(y=\frac1x\). For \(x=-10+\sqrt{17}\),

\(y=\frac{1}{-10+\sqrt{17}}\cdot\frac{-10-\sqrt{17}}{-10-\sqrt{17}} =\frac{-10-\sqrt{17}}{100-17} =-\frac{10+\sqrt{17}}{83}.\)

For \(x=-10-\sqrt{17}\),

\(y=\frac{1}{-10-\sqrt{17}}\cdot\frac{-10+\sqrt{17}}{-10+\sqrt{17}} =\frac{-10+\sqrt{17}}{100-17} =-\frac{10-\sqrt{17}}{83}.\)

Therefore the two points are

\(\left(-10+\sqrt{17},-\frac{10+\sqrt{17}}{83}\right)\)

and

\(\left(-10-\sqrt{17},-\frac{10-\sqrt{17}}{83}\right).\)

This completes the solution and gives the required result.

Answer: \(a=3\).

We start with the main method. Substitute the line into the curve; a tangent corresponds to a repeated root, so the discriminant is zero.

At the point of contact, the line and curve have equal \(y\)-values:

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

Bring all terms to one side:

\(ax^2-(2a+6)x+12=0.\)

For the line to be a tangent, this quadratic in \(x\) must have one repeated root.

So its discriminant is zero:

\(\big(-(2a+6)\big)^2-4(a)(12)=0.\)

Hence

\((2a+6)^2-48a=0.\)

Expanding gives

\(4a^2+24a+36-48a=0.\)

So

\(4a^2-24a+36=0.\)

Divide by \(4\):

\(a^2-6a+9=0.\)

Therefore

\((a-3)^2=0,\)

so

\(a=3.\)

This completes the solution and gives the required result.

Answer: \(k=4\).

We start with the main method. Substitute the line into the curve; a tangent corresponds to a repeated root, so the discriminant is zero.

At points of intersection,

\(x(x+2k)=-2x-6k-1.\)

So

\(x^2+2kx+2x+6k+1=0,\)

or

\(x^2+(2k+2)x+6k+1=0.\)

For the line to be a tangent to the curve, this quadratic in \(x\) must have exactly one repeated root.

Therefore its discriminant is zero:

\((2k+2)^2-4(6k+1)=0.\)

Expanding gives

\(4k^2+8k+4-24k-4=0.\)

So

\(4k^2-16k=0.\)

Factorising,

\(4k(k-4)=0.\)

The question asks for the non-zero value, so

\(k=4.\)

This completes the solution and gives the required result.

Answer: \(\displaystyle k=\frac{1+\sqrt{13}}{6}\) or \(\displaystyle k=\frac{1-\sqrt{13}}{6}\).

At a tangent point, the line and the curve meet at one repeated point. After equating the two expressions for \(y\), use the discriminant condition \(b^2-4ac=0\).

At a point of intersection, the line and the curve have the same \(y\)-value:

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

Rearranging gives

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

For the line to be a tangent to the curve, this quadratic in \(x\) must have exactly one repeated root. Therefore its discriminant is zero:

\(2^2-4(k)(3k-1)=0.\)

So

\(4=4k(3k-1).\)

Dividing by 4,

\(1=3k^2-k.\)

Hence

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

Using the quadratic formula,

\(k=\frac{1\pm\sqrt{(-1)^2-4(3)(-1)}}{2(3)} =\frac{1\pm\sqrt{13}}{6}.\)

Answer: \(x=-5,\ \frac32,\ 2\).

Answer: \(x=-5,\ \frac32,\ 2\).

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

\(2x^3+3x^2-26x+22=3x-8.\)

Rearranging gives

\(2x^3+3x^2-29x+30=0.\)

Testing simple integer roots gives \(x=2\):

\(2(2)^3+3(2)^2-29(2)+30=0.\)

So \((x-2)\) is a factor. Dividing gives

\(2x^3+3x^2-29x+30=(x-2)(2x^2+7x-15).\)

Now factorise the quadratic:

\(2x^2+7x-15=(2x-3)(x+5).\)

Therefore

\((x-2)(2x-3)(x+5)=0.\)

The required \(x\)-coordinates are

\(x=2,\quad x=\frac32,\quad x=-5.\)