Past Exam Questions

Answer: (a) \(x\lt 1\) or \(x\gt 4\); (b)(i) \(3(x-2)^2+4\); (b)(ii) \(y=4\).

For a quadratic expression, first find its roots or complete the square. The sign of the quadratic then follows from its shape and the position of the roots.

(a) Start with

\(3x^2-12x+16\gt 3x+4.\)

Bring all terms to one side:

\(3x^2-15x+12\gt 0.\)

Factorise:

\(3(x^2-5x+4)\gt 0,\)

so

\(3(x-1)(x-4)\gt 0.\)

The quadratic is positive outside its roots, so

\(x\lt 1\quad\text{or}\quad x\gt 4.\)

(b)(i) Complete the square:

\(3x^2-12x+16=3(x^2-4x)+16.\)

Since

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

we get

\(3x^2-12x+16=3\bigl((x-2)^2-4\bigr)+16.\)

Hence

\(3x^2-12x+16=3(x-2)^2+4.\)

(b)(ii) The minimum point is \((2,4)\). At the minimum point, the tangent is horizontal, so its equation is

\(y=4.\)

Answer: \(-4\lt x\lt -2\).

Answer: \(-4\lt x\lt -2\).

Expand both sides:

\((2x+3)(x-4)=2x^2-5x-12,\)

and

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

So

\(2x^2-5x-12\gt 3x^2+x-4.\)

Bring all terms to one side:

\(0\gt x^2+6x+8.\)

Hence

\(x^2+6x+8\lt 0.\)

Factorise:

\((x+2)(x+4)\lt 0.\)

The product is negative between the two roots, so

\(-4\lt x\lt -2.\)

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

Expand the left side:

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

So

\(x^2+3x-10\gt 3x+6.\)

Subtracting \(3x+6\) from both sides gives

\(x^2-16\gt 0.\)

Factorise:

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

The product is positive outside the two roots, so

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

Answer: \(k\lt 2-2\sqrt3\) or \(k\gt 2+2\sqrt3\).

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

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

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

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

The line meets the curve at two distinct points exactly when this quadratic has two distinct real roots. Therefore its discriminant must be positive:

\((2-k)^2-4(1)(3)\gt0.\)

So

\((2-k)^2\gt12.\)

Taking square roots gives

\(|2-k|\gt2\sqrt3.\)

This means either

\(2-k\gt2\sqrt3\quad\text{or}\quad 2-k\lt-2\sqrt3.\)

Hence

\(k\lt2-2\sqrt3\quad\text{or}\quad k\gt2+2\sqrt3.\)

Answer: \(k=-\frac34\) or \(k=1\).

For a quadratic equation to have two equal roots, its discriminant must be zero.

The equation is

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

Here

\(a=1,\quad b=4k,\quad c=k+3.\)

So the discriminant condition is

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

Simplify:

\(16k^2-4k-12=0.\)

Divide by \(4\):

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

Factorise:

\((4k+3)(k-1)=0.\)

Therefore

\(k=-\frac34\quad\text{or}\quad k=1.\)

Answer: \(k\lt \frac{2}{9}\) or \(k\gt 2\).

For a quadratic equation to have real and distinct roots, its discriminant must be positive.

The equation is

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

Here

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

So

\((3k-2)^2-4(2)(k)\gt 0.\)

Expand and simplify:

\(9k^2-12k+4-8k\gt 0.\)

Therefore

\(9k^2-20k+4\gt 0.\)

Factorise:

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

The critical values are

\(k=\frac29\quad\text{and}\quad k=2.\)

The quadratic expression in \(k\) has positive leading coefficient, so it is positive outside the two roots.

Hence the possible values of \(k\) are

\(k\lt \frac29\quad\text{or}\quad k\gt 2.\)

Answer: (a)(i) \(c=12\), (a)(ii) \(c=19\). (b) \(c\gt16\).

Use differentiation to find the gradient information needed. Stationary points occur when the derivative is zero, and tangents or normals are found from the gradient at the given point.

For part (a)(i), the curve crosses the \(x\)-axis at \(x=2\), so \(y=0\) when \(x=2\):

\(0=2^2-8(2)+c=4-16+c.\)

Thus \(c=12\).

For part (a)(ii), complete the square:

\(y=x^2-8x+c=(x-4)^2-16+c.\)

The minimum value is therefore \(c-16\). Given that the minimum value is \(3\),

\(c-16=3\), so \(c=19\).

For \(y\) to be always greater than \(0\), the minimum value must be greater than \(0\):

\(c-16\gt0.\)

Therefore \(c\gt16\).

The result has been obtained using the exact condition from the question, so no additional solutions are introduced.

Answer: no real solutions, since the discriminant is \(-120\).

We start with the main method. Use the discriminant to decide how many real roots the quadratic equation has.

Expand and collect terms:

\(5x(x-3)=5x-26\)

\(5x^2-15x=5x-26.\)

So

\(5x^2-20x+26=0.\)

For this quadratic,

\(a=5,\quad b=-20,\quad c=26.\)

The discriminant is

\(b^2-4ac=(-20)^2-4(5)(26).\)

Thus

\(b^2-4ac=400-520=-120.\)

Since the discriminant is negative, the equation has no real solutions.

This completes the solution and gives the required result.

Answer: \(6\lt k\lt10\).

We start with the main method. Use the discriminant to decide how many real roots the quadratic equation has.

For the curve

\(y=x^2+kx+4k-15\)

to be completely above the \(x\)-axis, the quadratic must have no real roots and open upwards.

Since the coefficient of \(x^2\) is \(1\), it opens upwards.

So we need the discriminant to be negative:

\(k^2-4(1)(4k-15)\lt0.\)

This gives

\(k^2-16k+60\lt0.\)

Factorising,

\((k-6)(k-10)\lt0.\)

Therefore \(k\) lies between the two roots:

\(6\lt k\lt10.\)

This completes the solution and gives the required result.

Answer: \(k\gt \dfrac18\).

We start with the main method. Use the discriminant to decide how many real roots the quadratic equation has.

For the quadratic \(kx^2+(2k-1)x+k+1=0\) to have no real roots, its discriminant must be negative.

Here \(a=k\), \(b=2k-1\), and \(c=k+1\). So

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

Simplifying,

\(\Delta=4k^2-4k+1-4k^2-4k=1-8k.\)

Now require

\(1-8k\lt 0.\)

Thus \(8k\gt 1\), so

\(k\gt \frac18.\)

This completes the solution and gives the required result.

Answer: \(x=-6+3\sqrt5\) or \(x=-2+\sqrt5\).

Rewrite the equation as a standard quadratic in the chosen variable, then solve and reject any value that does not satisfy the original form.

Rewrite the equation as

\((2+\sqrt5)x^2-4x-3(2-\sqrt5)=0.\)

Using the quadratic formula,

\(x=\frac{4\pm\sqrt{16+12(2+\sqrt5)(2-\sqrt5)}}{2(2+\sqrt5)}.\)

Since

\((2+\sqrt5)(2-\sqrt5)=4-5=-1,\)

this becomes

\(x=\frac{4\pm\sqrt{16-12}}{2(2+\sqrt5)} =\frac{4\pm2}{2(2+\sqrt5)}.\)

So

Rationalising,

\(\frac{3}{2+\sqrt5} =\frac{3(2-\sqrt5)}{4-5} =-6+3\sqrt5,\)

and

\(\frac{1}{2+\sqrt5} =\frac{2-\sqrt5}{4-5} =-2+\sqrt5.\)

Therefore the result matches the required answer.

Answer: \(\frac13\lt k\lt 3\).

Answer: \(\frac13\lt k\lt 3\).

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

\(9kx+1=kx^2+3x(2k+1)+4\).

Rearrange into a quadratic in \(x\):

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

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

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

Simplifying,

\(9(1-k)^2-12k\lt 0\),

\(9k^2-30k+9\lt 0\),

\(3k^2-10k+3\lt 0\).

Factorise:

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

Therefore \(k\) lies between the two critical values:

\(\frac13\lt k\lt 3\).

Answer: \(k\le-25\) or \(k\ge-1\).

Answer: \(k\le-25\) or \(k\ge-1\).

For the equation to have real roots, the discriminant must be non-negative:

\(b^2-4ac\ge0\).

Here \(a=k\), \(b=k+5\) and \(c=-4\). Therefore

\((k+5)^2-4(k)(-4)\ge0\).

Expanding gives

\(k^2+10k+25+16k\ge0\),

so

\(k^2+26k+25\ge0\).

Factorising,

\((k+1)(k+25)\ge0\).

The product is non-negative outside the two roots \(-25\) and \(-1\). Hence

\(k\le-25\quad\text{or}\quad k\ge-1\).

Answer: (a) \(x\lt \frac15\) or \(x\gt 6\). (b) The discriminant is always positive.

Answer: (a) \(x\lt \frac15\) or \(x\gt 6\). (b) The discriminant is always positive.

(a) The critical values are found from

\(5x-1=0\quad\text{and}\quad 6-x=0\),

so \(x=\frac15\) and \(x=6\).

The product \((5x-1)(6-x)\) is negative outside these two critical values, so

\(x\lt \frac15\quad\text{or}\quad x\gt 6\).

(b) For the quadratic equation

\((2k+1)x^2-4kx+2k-1=0\),

the discriminant is

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

Since

\((2k+1)(2k-1)=4k^2-1\),

we get

\(\Delta=16k^2-4(4k^2-1)=4\).

Thus \(\Delta\gt 0\), so the equation has distinct real roots. Also, \(k\ne-\frac12\) ensures that the coefficient of \(x^2\) is not zero, so the equation is genuinely quadratic.

Answer: \(k=9\) or \(k=\dfrac19\).

For equal roots, the discriminant is zero.

Here

\(a=k,\qquad b=-3(k+1),\qquad c=25.\)

So

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

Hence

\([-3(k+1)]^2-4(k)(25)=0.\)

This gives

\(9(k+1)^2-100k=0.\)

Expanding,

\(9k^2+18k+9-100k=0,\)

so

\(9k^2-82k+9=0.\)

Factorising,

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

Therefore

\(k=\frac19\quad\text{or}\quad k=9.\)

Answer: \(k\lt 0\) or \(k\gt 1\).

The key point is to use the discriminant condition for the required number of real roots.

For the equation to have two different real roots, the discriminant must be positive.

Here

\(a=k,\qquad b=4k,\qquad c=3k+1.\)

So the discriminant is

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

Simplifying,

\(16k^2-12k^2-4k=4k^2-4k.\)

Therefore

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

Factorise:

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

This product is positive when both factors are positive or both are negative. Hence

\(k\gt 1\quad\text{or}\quad k\lt 0.\)

Answer: \(k=10\pm\sqrt{48}\).

The key point is to differentiate first, then substitute the required value or solve the resulting equation.

At an intersection of the line and the curve,

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

Rearrange:

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

For the line to be a tangent, this quadratic must have equal roots. Therefore its discriminant is zero:

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

Expand and simplify:

\(k^2-4k+4-16k+48=0.\)

So

\(k^2-20k+52=0.\)

Using the quadratic formula,

\(k=\frac{20\pm\sqrt{(-20)^2-4(1)(52)}}{2}.\)

Thus

\(k=\frac{20\pm\sqrt{192}}{2} =10\pm\sqrt{48}.\)

Answer: \(-\frac52\leq k\leq2\).

The key point is to use the discriminant condition for the required number of real roots.

For real roots, the discriminant must be non-negative:

\(b^2-4ac\geq0.\)

Here

\(a=2k-1,\qquad b=6,\qquad c=k+1.\)

So

\(6^2-4(2k-1)(k+1)\geq0.\)

Expand:

\(36-4(2k^2+k-1)\geq0.\)

Therefore

\(40-4k-8k^2\geq0.\)

Divide by \(-4\), reversing the inequality:

\(2k^2+k-10\leq0.\)

Factorise:

\((2k+5)(k-2)\leq0.\)

Hence

\(-\frac52\leq k\leq2.\)

Answer: \(2\lt m\lt 26\).

At an intersection,

\(mx-2=(m+1)x^2+8x+1.\)

Rearrange into a quadratic in \(x\):

\((m+1)x^2+(8-m)x+3=0.\)

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

\((8-m)^2-4(m+1)(3)\lt 0.\)

Simplify:

\((8-m)^2-12(m+1)\lt 0,\)

\(m^2-16m+64-12m-12\lt 0,\)

\(m^2-28m+52\lt 0.\)

Factorise:

\((m-2)(m-26)\lt 0.\)

The product is negative between the roots, so

\(2\lt m\lt 26.\)

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 \(2\),

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