ESAT Paper 1 Advanced Mathematics Sample Questions

Correct Answer: A. 0.344

Step-by-Step Breakdown:

1. Multiply above and below by the conjugate of the denominator

The denominator is $\sqrt{21}+\sqrt{5}$, whose conjugate is $\sqrt{21}-\sqrt{5}$. Multiplying numerator and denominator by the same quantity leaves the value unchanged:
$${\displaystyle \frac{\sqrt{21}-\sqrt{5}}{\sqrt{21}+\sqrt{5}}}$$$$=\frac{(\sqrt{21}-\sqrt{5}{)}^2}{(\sqrt{21}{)}^2-(\sqrt{5}{)}^2}$$$$=\frac{(\sqrt{21}-\sqrt{5}{)}^2}{21-5}$$
The denominator is now the rational number $16$.

2. Expand the numerator, then divide every term of it

$$(\sqrt{21}-\sqrt{5}{)}^2=21-2\sqrt{21}\sqrt{5}+5$$$$=26-2\sqrt{105}$$
so the quotient is ${\displaystyle \frac{26-2\sqrt{105}}{16}}$. Both $21$ and $5$ are odd, so $26$ and $16$ are both even and the factor $2$ comes out of the whole numerator and out of the denominator together:
$$\frac{26-2\sqrt{105}}{16}$$$$=\frac{2(13-\sqrt{105})}{2\times 8}=\frac{13-\sqrt{105}}{8}$$
The factor cancels out of the whole bracket, not out of one term of it.

3. Put a number to the exact form

$\sqrt{105}$ is close to $10.2$, and it is smaller than $13$, since $13$ is the arithmetic mean of $21$ and $5$ while $\sqrt{105}$ is their geometric mean. The numerator is therefore positive, and to 3 significant figures the fraction ${\displaystyle \frac{13-\sqrt{105}}{8}}$ is $0.344$.

Sanity check

Take $\sqrt{21}\approx 4.58$ and $\sqrt{5}\approx 2.24$. The original quotient is then roughly $4.58-2.24$ over $4.58+2.24$, which is close to $0.344$. The value also has to be positive and below $1$, because $\sqrt{21}-\sqrt{5}$ is the smaller of two positive quantities and $\sqrt{21}+\sqrt{5}$ is the larger, and it is.

The key is $0.344$.

Why the Other Options Are Wrong (❌):

• B. 2.91 · Cross-term sign slip
  The cross term was expanded with the wrong sign, so $(\sqrt{21}-\sqrt{5}{)}^2$ was written as $21+2\sqrt{105}+5=26+2\sqrt{105}$. Dividing by the correct denominator $16$ then gives about $2.91$, which is greater than $1$, although the quotient in the stem has the smaller of the two positive quantities on top and so cannot exceed $1$. Squaring a difference subtracts the cross term.
• C. 1.97 · Partial cancellation
  From the correct ${\displaystyle \frac{26-2\sqrt{105}}{16}}$ the factor $2$ was cancelled against $16$ but applied to the surd term only, leaving ${\displaystyle \frac{26-\sqrt{105}}{8}}$, which is about $1.97$. The whole numerator is being divided, so $26$ has to be halved to $13$ as well.
• D. 0.212 · Denominator added instead of differenced
  The denominator was multiplied out as $\sqrt{21}\times \sqrt{21}+\sqrt{5}\times \sqrt{5}=21+5=26$ rather than as the difference of two squares $21-5=16$. That gives ${\displaystyle \frac{26-2\sqrt{105}}{26}}\approx 0.212$, which is smaller than the true value of about $0.344$. A conjugate pair multiplies to a difference of squares, never to a sum.
• E. -0.281 · Second square subtracted in the expansion
  The second square was subtracted rather than added while expanding the numerator: $21-2\sqrt{105}-5=16-2\sqrt{105}$. Dividing by $16$ gives about $-0.281$, a negative value, although $\sqrt{21}-\sqrt{5}$ and $\sqrt{21}+\sqrt{5}$ are both positive and so their quotient cannot be negative. Both squares are added when a bracket is squared; only the cross term changes sign.
• F. 0.147 · Conjugate multiplied into the denominator only
  The conjugate was multiplied into the bottom of the fraction but not into the top, so ${\displaystyle \frac{\sqrt{21}-\sqrt{5}}{\sqrt{21}+\sqrt{5}}}$ was recorded as ${\displaystyle \frac{\sqrt{21}-\sqrt{5}}{16}}$. Multiplying the denominator alone changes the value of the quotient, so the same factor has to go into the numerator, where it squares $\sqrt{21}-\sqrt{5}$ and produces $26-2\sqrt{105}$. The quotient is therefore ${\displaystyle \frac{26-2\sqrt{105}}{16}}\approx 0.344$.
• G. 0.985 · Cross term counted once
  Squaring the numerator produced a single copy of the cross term, so $(\sqrt{21}-\sqrt{5}{)}^2$ was written as $26-\sqrt{105}$ and then divided by $16$. Expanding $(\sqrt{21}-\sqrt{5})(\sqrt{21}-\sqrt{5})$ gives $-\sqrt{105}$ twice, once from each bracket, so the numerator is $26-2\sqrt{105}$ and the value is $0.344$.
• H. 0.784 · Surds collected under one root
  Top and bottom were each collapsed into a single surd, as though $\sqrt{21}-\sqrt{5}=\sqrt{16}$ and $\sqrt{21}+\sqrt{5}=\sqrt{26}$, which leaves $\sqrt{{\displaystyle \frac{16}{26}}}$. A root does not spread across a sum or a difference: $\sqrt{21}-\sqrt{5}$ squares to $26-2\sqrt{105}$ and not to $16$. Clearing the denominator with the conjugate leaves ${\displaystyle \frac{26-2\sqrt{105}}{16}}\approx 0.344$.

Common Mistake (⚠️):
Dividing only the surd term by the rationalised denominator. Once the quotient reads ${\displaystyle \frac{26-2\sqrt{105}}{16}}$, the whole numerator is being divided, so $26$ has to be reduced to $13$ at the same time as the $2$ in front of the surd is taken away. Cancelling into one term only leaves a value roughly $1.97$, which is above $1$ and therefore impossible here.

Takeaway (📌):
The conjugate is the whole method. It turns $\sqrt{21}+\sqrt{5}$ into $21-5=16$ and turns the numerator into $(\sqrt{21}-\sqrt{5}{)}^2=26-2\sqrt{105}$. Expect the exact answer in the form ${\displaystyle \frac{13-\sqrt{105}}{8}}$, and divide every term of the numerator, never the surd alone.

Correct Answer: A. -13

Step-by-Step Breakdown:

1. Complete the square

Halve the coefficient of $x$: half of $-6$ is $-3$, so the bracket is $(x-3{)}^2$. Expanding that bracket introduces an extra $9$ that the original expression does not have, so the same amount has to come back off:

$$y=(x-3{)}^2-9-4$$

The two constants collect to $-9-4=-13$. A square is never negative and $(x-3{)}^2$ is $0$ at $x=3$, so the lowest point of the curve is $(3,-13)$ and its least $y$-value is $-13$.

2. Set the discriminant to zero

The line $y=k$ meets the curve wherever $x^2-6x-4=k$. In standard form that is

$$x^2-6x-4-k=0$$

so, with $a=1$, $b=-6$ and $c=-4-k$, the discriminant is

$$D=(-6{)}^2-4(1)(-4-k)$$$$=36-4(-4-k)$$

One point of contact is a repeated root, which needs $D=0$:

$$36-4(-4-k)=0⟹-4-k=\frac{36}{4}$$$$=9⟹k=-4-9=-13$$

3. Check the value

The two routes agree, which is the check worth doing: the completed square puts the least value of the curve at $-13$, and the discriminant puts the repeated root at the same height. Substituting $x=3$ into $x^2-6x-4$ gives $-13$ directly, so the line $y=-13$ touches the curve at $(3,-13)$. Any $k$ above $-13$ gives $D>0$ and two crossings; any $k$ below $-13$ gives $D<0$ and no crossing at all.

The key is $-13$.

Why the Other Options Are Wrong (❌):

• B. 5 · Sign error completing the square
  A sign slip on the correction term: adding $9$ instead of subtracting it gives $(x-3{)}^2+9-4$, a constant of $5$. Substituting $x=3$ into $x^2-6x-4$ settles it, since the height of the curve there is $-13$.
• C. -4 · Correction term never subtracted
  Writes the completed square as $(x-3{)}^2-4$ and stops, never taking off the $9$ that the bracket introduces. That leaves the constant at $-4$, which is the height of the curve at $x=0$ rather than its least value $-13$.
• D. -40 · Coefficient of $x$ not halved
  Squares the whole coefficient of $x$ instead of half of it: $(x-6{)}^2-36-4$ gives a constant of $-40$. Halving is what makes the bracket reproduce the $-6x$ term, since $(x-6{)}^2$ expands with an $x$ coefficient of $-12$ and not $-6$.
• E. 13 · Sign error rearranging before the discriminant
  Moves $k$ to the wrong side, rearranging to $x^2-6x-4+k=0$ instead of subtracting it. The discriminant then reads $(-6{)}^2-4(-4+k)$, and setting that to zero gives $-4+k=9$, so $k=13$. The curve's least value is $-13$, so a line at height $13$ is not the one that touches it.
• F. 52 · Discriminant taken before $k$ was brought in
  The discriminant was formed from the curve exactly as printed, $(-6{)}^2-4(-4)$, so $k$ never entered the equation at all. The line has to be in it first: $x^2-6x-4=k$ rearranges to $x^2-6x-4-k=0$, whose discriminant is $(-6{)}^2-4(-4-k)$. Setting that to zero gives $-4-k=9$, so $k=-13$, the height of the curve at $x=3$.

Common Mistake (⚠️):
Stopping at the constant term. The completed square form is $(x-3{)}^2-9-4$, and the correction $-9$ has to be carried through: quoting $-4$ gives the height of the curve at $x=0$, not its least value $-13$. Substituting $x=3$ back into $x^2-6x-4$ catches it in one line.

Takeaway (📌):
For a curve $y=x^2+bx+c$ the horizontal line $y=k$ touches it at exactly one point when $k$ is the height of the vertex, and that height is the constant left after completing the square. Halve the coefficient of $x$, square it, subtract: $-4-9=-13$. The discriminant gives the same number and takes longer.

Correct Answer: E. 3.20

Step-by-Step Breakdown:

1. Substitute the line into the curve

Both equations give $y$, so at an intersection

$-x-3=2x^2-8x-2$

Taking every term of the line across, the $x$ term and the constant together, leaves one quadratic:

$2x^2-7x+1=0$

Its $x$ coefficient is $-7$ and its constant is $1$.

2. Evaluate the discriminant

With leading coefficient $2$,

$(-7{)}^2-4(2)(1)=41$

It is positive, which is consistent with the two crossings the question describes.

3. Take the difference of the roots

$x={\displaystyle \frac{-(-7)\pm \sqrt{41}}{2(2)}}$, so the two roots differ by

${\displaystyle \frac{2\sqrt{41}}{2(2)}}={\displaystyle \frac{\sqrt{41}}{2}}=3.20$ to 3 significant figures.

The centre the two roots sit either side of cancels in that subtraction and never has to be worked out.

Sanity check

The roots are near $0.149$ and $3.35$, which are roughly $3.20$ apart. Between them $2x^2-7x+1$ is negative and outside them it is positive, as a quadratic with a positive $x^2$ coefficient and two real roots must be.

The key is $3.20$.

Why the Other Options Are Wrong (❌):

• A. 4.47 · Solved the wrong equation
  The line was left out of the working and the curve's own $x$-intercepts were measured instead. $2x^2-8x-2=0$ has discriminant $(-8{)}^2-4(2)(-2)=80$, so its two roots are ${\displaystyle \frac{\sqrt{80}}{2}}=4.47$ apart to 3 significant figures. Those are the points where the curve crosses the $x$-axis, not the points where it crosses $y=-x-3$.
• B. 4.03 · Constant term of the line left behind
  The $x$ term of the line was taken across but its constant was not, leaving $2x^2-7x-2=0$, whose constant is still $-2$ where the substitution makes it $1$. Its discriminant is $(-7{)}^2-4(2)(-2)=65$, giving ${\displaystyle \frac{\sqrt{65}}{2}}=4.03$ to 3 significant figures. The constant of the line crosses over with its $x$ term, in the same move.
• C. 3.74 · x term of the line left behind
  The constant of the line was taken across but its $x$ term was not, leaving $2x^2-8x+1=0$, whose $x$ coefficient is still $-8$ where the substitution makes it $-7$. Its discriminant is $(-8{)}^2-4(2)(1)=56$, giving ${\displaystyle \frac{\sqrt{56}}{2}}=3.74$ to 3 significant figures. The $x$ term of the line crosses over with its constant, in the same move.
• D. 3.77 · Sign error in the discriminant
  The $4ac$ part of the discriminant was added rather than subtracted: $(-7{)}^2+4(2)(1)=57$ was used in place of $(-7{)}^2-4(2)(1)=41$, which gives ${\displaystyle \frac{\sqrt{57}}{2}}=3.77$ to 3 significant figures. The constant of $2x^2-7x+1=0$ is $1$, and it enters $b^2-4ac$ carrying its own sign.
• F. 1.60 · Only one root's offset from the centre taken
  The quadratic formula puts the roots at ${\displaystyle \frac{-(-7)\pm \sqrt{41}}{2(2)}}$, and the ${\displaystyle \frac{\sqrt{41}}{2(2)}}$ that each root sits away from the centre was reported as the separation. There is one root on each side of that centre, so the distance between them is twice the offset: ${\displaystyle \frac{2\sqrt{41}}{2(2)}}={\displaystyle \frac{\sqrt{41}}{2}}=3.20$ to 3 significant figures. The two roots are near $0.149$ and $3.35$, and the gap between those is the quantity the stem asks for.

Common Mistake (⚠️):
Subtracting only part of the line. Collecting $-x-3=2x^2-8x-2$ means moving the $x$ term and the constant together; stopping after the $x$ term leaves $2x^2-7x-2=0$, whose constant is still $-2$ instead of $1$ and whose discriminant is $65$ instead of $41$. The substitution is one move, not two independent ones.

Takeaway (📌):
Collect the line and the curve into a single quadratic, then the horizontal gap between the crossings is the square root of the discriminant divided by the leading coefficient, here ${\displaystyle \frac{\sqrt{41}}{2}}$. Neither root is ever needed, and the axis of symmetry never has to be found.

Correct Answer: A. 131

Step-by-Step Breakdown:

1. Invert the rule

Make the earlier term the subject:
$$u_{n+1}=\frac{u_n+24}{5}⟹5u_{n+1}$$$$=u_n+24⟹u_n=5u_{n+1}-24$$
This rule takes any term and returns the one before it, so it can be applied as many times as needed.

2. Find the fixed point

A term equal to $L$ would reproduce itself, so $L=\frac{L+24}{5}$, giving $5L=L+24$, then $(5-1)L=24$ and $L=6$. Writing every term as $L$ plus a difference is what turns the chain into one multiplication.

3. Walk back from $u_4$ to $u_1$

The difference between the given term and the fixed point is $u_4-L=7-(6)=1$. Each application of the inverse rule multiplies that difference by $5$, and the number of applications is the difference of the indices, $4-1=3$:
$$u_1=L+5^3\times (1)=6+(125)$$$$=131$$

Check it forwards through the original rule: $\frac{131+24}{5}=31$, which is $u_2$, and a further $2$ forward steps return $u_4=7$ as stated.

The key is $131$.

Why the Other Options Are Wrong (❌):

• B. 6 · Fixed point mistaken for the term
  The fixed point was quoted as the answer. Solving $L=\frac{L+24}{5}$ gives $5L=L+24$, so $(5-1)L=24$ and $L=6$. That is the value the sequence approaches as it runs forwards, not its first term. If $u_1$ were $6$ then every term would equal $6$, and the fourth term would be $6$ rather than the stated $7$.
• C. 11 · Backward steps undercounted
  The inverse rule $u_n=5u_{n+1}-24$ was applied two times too few, stopping at $u_3=11$ instead of $u_1$. The gap from index $4$ to index $1$ is $3$, so two further applications are still owed.
• D. 125 · Fixed point not restored
  The multiply-by-$5$ shortcut was used but the fixed point was never added back on. $5^3\times (1)=125$ is the difference between $u_1$ and the fixed point $6$, not $u_1$ itself, so $L$ still has to be added to it.
• E. 31 · Index gap off by one
  Only $2$ applications of the inverse rule were made, so the walk stopped one index short at $u_2=31$. Going from index $4$ to index $1$ needs $4-1=3$ applications, one more than were made, and it is the subscripts that fix that count rather than the terms written down.

Common Mistake (⚠️):
Counting the steps by the number of terms rather than the gap between the indices. $4$ terms are involved, so $2$ or $4$ applications of the inverse rule feel natural, but moving from $u_4$ to $u_1$ is exactly $4-1=3$ of them.

Takeaway (📌):
For $u_{n+1}=\frac{u_n+c}{k}$, solve $L=\frac{L+c}{k}$ for the fixed point first. The difference from $L$ is divided by $k$ at every forward step and multiplied by $k$ at every backward one, so $u_1=L+k^m(u_{1+m}-L)$ replaces the whole chain of substitutions. Here that reads $u_1=6+5^3(7-(6))=131$.