Proof: past paper questions & worked solutions

Since $c=3a+1$ and $a+b+c=15$, substitute for $c$ to get $a+b+3a+1=15$.

This gives $b=14-4a$.

The integers $a,b,c$ are positive. Since $b=14-4a>0$, we need $a<3.5$. Therefore $a=1,2,3$.

If $a=1$, then $b=10$ and $c=4$, so $abc=40$. If $a=2$, then $b=6$ and $c=7$, so $abc=84$. If $a=3$, then $b=2$ and $c=10$, so $abc=60$.

The products $40$, $84$ and $60$ are all multiples of $4$. All possible cases have been checked, so $abc$ is always a multiple of $4$.

Since $c=b+2$ and $a+b+c=10$, substitute for $c$: $a+b+(b+2)=10$.

This gives $a+2b+2=10$, so $a=8-2b$.

The integers $a,b,c$ are all positive. Since $b>0$ and $a=8-2b>0$, the only possible values are $b=1,2,3$.

If $b=1$, then $a=6$ and $c=3$, so $abc=18$. If $b=2$, then $a=4$ and $c=4$, so $abc=32$. If $b=3$, then $a=2$ and $c=5$, so $abc=30$.

The list contains every possible case, and in every case the product is even. Therefore $abc$ is always even.

A positive integer that is not a multiple of $3$ must have the form $m=3q+1$ or $m=3q+2$.

If $m=3q+1$, then $m^2+3m+2=(3q+1)^2+3(3q+1)+2=9q^2+15q+6=3(3q^2+5q+2)$.

If $m=3q+2$, then $m^2+3m+2=(3q+2)^2+3(3q+2)+2=9q^2+21q+12=3(3q^2+7q+4)$.

Both possible cases give a factor of $3$, so $m^2+3m+2$ is always a multiple of $3$.

Every integer is exactly one of the forms $3k$, $3k+1$ or $3k+2$, where $k$ is an integer.

For $n=3k$, the expression becomes $3(6k^2+k)+1$, so it leaves remainder $1$ when divided by $3$.

If $n=3k+1$, then $2n^2+n+1=2(3k+1)^2+(3k+1)+1$.

$2(9k^2+6k+1)+3k+2=18k^2+15k+4=3(6k^2+5k+1)+1$, which is not divisible by $3$.

If $n=3k+2$, then $2n^2+n+1=2(3k+2)^2+(3k+2)+1$.

$2(9k^2+12k+4)+3k+3=18k^2+27k+11=3(6k^2+9k+3)+2$, which is not divisible by $3$.

All possible integer forms have been checked, and none gives a multiple of $3$. Therefore $2n^2+n+1$ is not divisible by $3$.

For positive $x$ and $y$, both $\sqrt{x}$ and $\sqrt{y}$ are real, so $(\sqrt{x}-\sqrt{y})^2\geq0$.

$(\sqrt{x}-\sqrt{y})^2=x-2\sqrt{xy}+y\geq0$.

Thus $x+y\geq2\sqrt{xy}$, and dividing by $2$ gives $\dfrac{x+y}{2}\geq\sqrt{xy}$.

Take $x=-1$ and $y=-1$. Then $\dfrac{x+y}{2}=-1$, while $\sqrt{xy}=\sqrt1=1$.

Since $-1\geq1$ is false, the inequality does not hold for all negative $x$ and $y$.

For $x\geq0$, $\sqrt{3x}$ is real and $(\sqrt{3x}-1)^2\geq0$.

$(\sqrt{3x}-1)^2=3x-2\sqrt{3x}+1\geq0$.

Therefore $3x+1\geq2\sqrt{3x}$, as required.

The consecutive prime numbers $5$, $7$, and $11$ have sum $5+7+11=23$.

$23$ is not a multiple of $5$, so the statement is not true.

Let $x=\sqrt2$ and $y=\dfrac1{\sqrt2}$. Both are irrational and $x\ne y$, but $xy=1$, which is rational.

Let $n=2q+1$, where $q$ is an integer.

$n^3+3n+2=(2q+1)^3+3(2q+1)+2$.

$n^3+3n+2=8q^3+12q^2+12q+6=2(4q^3+6q^2+6q+3)$.

The expression has a factor of $2$, so it is even.

The bracket $4q^3+6q^2+6q+3$ is odd, because the first three terms are even and $3$ is odd. Therefore the expression is $2$ times an odd integer, so it is not divisible by $4$.

$7$ and $11$ are consecutive prime numbers.

$7^2+11^2=49+121=170$, which is a multiple of $10$.

This single counterexample proves the student's statement is not true.

The even integers greater than $0$ and less than $6$ are $2$ and $4$. Therefore the possible ordered pairs are $(2,2),(2,4),(4,2),(4,4)$.

If $(x,y)=(2,2)$, then $x^2-\dfrac{xy}{4}=4-1=3$. If $(x,y)=(2,4)$, then $x^2-\dfrac{xy}{4}=4-2=2$.

If $(x,y)=(4,2)$, then $x^2-\dfrac{xy}{4}=16-2=14$. If $(x,y)=(4,4)$, then $x^2-\dfrac{xy}{4}=16-4=12$.

The possible values are $3,2,14,12$, and each lies strictly between $1$ and $15$. Hence $1<x^2-\dfrac{xy}{4}<15$ in every possible case.

Take $p=7$, which is prime.

$2p+1=2(7)+1=15$, and $15$ is not prime. So the statement is false.

If $n$ is even, let $n=2k$. Then $5n^2+n+12=5(2k)^2+2k+12=20k^2+2k+12=2(10k^2+k+6)$, which is even.

If $n$ is odd, let $n=2k+1$. Then $5n^2+n+12=5(2k+1)^2+(2k+1)+12$.

$5(2k+1)^2+(2k+1)+12=20k^2+22k+18=2(10k^2+11k+9)$, which is even.

Every integer is either even or odd, and both cases give an even result. Therefore $5n^2+n+12$ is always even.

The single digit prime numbers are $2$, $3$, $5$, and $7$.

$2^3+2=10$, $3^3+3=30$, $5^3+5=130$, and $7^3+7=350$.

Each result is a multiple of $10$, so $p^3+p$ is a multiple of $10$ for all single digit primes.

$(n+1)^3-n^3=(n^3+3n^2+3n+1)-n^3=3n^2+3n+1$.

$3n^2+3n+1=3n(n+1)+1$.

$3n(n+1)$ is a multiple of $3$, so $3n(n+1)+1$ is one more than a multiple of $3$. Therefore it is not a multiple of $3$.

Take $n=5$, which is prime.

$3^5+2=243+2=245=5\times49$, so it is not prime. This disproves the statement.

If integer $m$ is not divisible by $3$, then $m$ has remainder $1$ or $2$ on division by $3$.

If $m=3q+1$, then $m^2-1=(3q+1)^2-1=9q^2+6q=3(3q^2+2q)$, which is divisible by $3$.

If $m=3q+2$, then $m^2-1=(3q+2)^2-1=9q^2+12q+3=3(3q^2+4q+1)$, which is divisible by $3$.

Both possible cases for an integer not divisible by $3$ have been checked, so $m^2-1$ is divisible by $3$.

$(x-y)^3>x^3-y^3$.

$x^3-3x^2y+3xy^2-y^3>x^3-y^3$.

Subtract $x^3-y^3$ from both sides to get $-3x^2y+3xy^2>0$.

$3xy(y-x)>0$.

Since $x$ and $y$ are positive, $3xy>0$. Therefore $y-x>0$, so $y>x$.

Take $x=1$ and $y=-1$. Then $(x-y)^3=2^3=8$ and $x^3-y^3=1-(-1)=2$, so the given inequality is true.

But $y>x$ would mean $-1>1$, which is false. Therefore the result is not true for all real numbers.

$x^2+y^2+8x-10y=29$ becomes $(x+4)^2-16+(y-5)^2-25=29$.

$(x+4)^2+(y-5)^2=70$.

The centre of $C_1$ is $(-4,5)$ and its radius is $\sqrt{70}$.

$C_2$ has equation $(x-5)^2+(y+8)^2=52$, so its centre is $(5,-8)$ and its radius is $\sqrt{52}=2\sqrt{13}$.

The distance between $(-4,5)$ and $(5,-8)$ is $\sqrt{9^2+(-13)^2}=\sqrt{250}=5\sqrt{10}$.

$\sqrt{70}+2\sqrt{13}=15.58\ldots$, while $5\sqrt{10}=15.81\ldots$.

The distance between centres is greater than the sum of the radii, so the circles are separate. Therefore they neither touch nor intersect.

Take $n=6$.

$n^2+3n+1=6^2+3(6)+1=36+18+1=55$.

$55$ is not prime, so the statement is false.

Every natural number is one of $4k$, $4k+1$, $4k+2$ or $4k+3$.

If $n=4k$, then $n^2-2=16k^2-2=4(4k^2-1)+2$, not a multiple of $4$.

If $n=4k+1$, then $n^2-2=16k^2+8k-1=4(4k^2+2k-1)+3$, not a multiple of $4$.

If $n=4k+2$, then $n^2-2=16k^2+16k+2=4(4k^2+4k)+2$, not a multiple of $4$.

If $n=4k+3$, then $n^2-2=16k^2+24k+7=4(4k^2+6k+1)+3$, not a multiple of $4$.

All possible cases leave remainder $2$ or $3$ when divided by $4$, so $n^2-2$ is not a multiple of $4$.

Take $n=5$.

$(5+1)^3-5^3=216-125=91$.

$91=7\times13$, so the statement is false.

For $A(1,0)$ and $C(7,-6)$, the gradient of $CA$ is $\dfrac{0-(-6)}{1-7}=-1$.

For $B(3,-10)$ and $C(7,-6)$, the gradient of $CB$ is $\dfrac{-10-(-6)}{3-7}=1$.

The product of the gradients is $-1$, so $CA$ is perpendicular to $CB$ and $\angle ACB=90^\circ$.

An angle of $90^\circ$ at the circumference subtends a diameter, so $AB$ is a diameter of the circle.

Let $q=2n-1$ and $p=2n+1$, where $n$ is a positive integer. Then $p>q>0$ and the numbers are consecutive odd numbers.

$p^2-q^2=(2n+1)^2-(2n-1)^2$.

$p^2-q^2=(4n^2+4n+1)-(4n^2-4n+1)=8n$.

Since $n$ is an integer, $8n$ is a multiple of $8$, so $p^2-q^2$ is a multiple of $8$.

For $y=x^3+12x^2+49x+2$, $\dfrac{dy}{dx}=3x^2+24x+49$.

$3x^2+24x+49=3(x+4)^2+1$. This is always positive for real $x$.

Since $\dfrac{dy}{dx}$ is never $0$, the curve has no stationary points.

$H=\dfrac{1}{20}t^{1/2}(20+6t-t^2)+17$.

$H=\dfrac{1}{20}(20t^{1/2}+6t^{3/2}-t^{5/2})+17$.

$\dfrac{dH}{dt}=\dfrac{1}{20}\left(10t^{-1/2}+9t^{1/2}-\dfrac52t^{3/2}\right)$.

$\dfrac{dH}{dt}=\dfrac{-5t^2+18t+20}{40\sqrt{t}}$.

At $t=\alpha$, $\dfrac{dH}{dt}=0$. Since $\alpha>0$, the denominator is not zero, so $-5\alpha^2+18\alpha+20=0$.

Multiplying by $-1$ gives $5\alpha^2-18\alpha-20=0$.

$\alpha=\dfrac{18\pm\sqrt{18^2+4(5)(20)}}{10}=\dfrac{9\pm\sqrt{181}}{5}$.

Only the positive value lies in $0\le t\le7$, so $\alpha=\dfrac{9+\sqrt{181}}{5}=4.491$ to 3 decimal places.

$\dfrac{d^2H}{dt^2}=\dfrac{-15t^2+18t-20}{80t^{3/2}}$.

At $t=\alpha$, $5\alpha^2=18\alpha+20$. Therefore the numerator of $\dfrac{d^2H}{dt^2}$ is $-15\alpha^2+18\alpha-20=-3(18\alpha+20)+18\alpha-20=-36\alpha-80$.

Since $\alpha>0$, $-36\alpha-80<0$, and the denominator $80\alpha^{3/2}$ is positive. Thus $\dfrac{d^2H}{dt^2}<0$, so $H$ is a maximum.

The logarithms require $4x+5>0$ and $x+3>0$, so $x>-\dfrac54$.

$2\log_3(4x+5)-\log_3(x+3)=\log_3\left(\dfrac{(4x+5)^2}{x+3}\right)$.

$2=\log_3 9$, so $\dfrac{(4x+5)^2}{x+3}=9$.

$(4x+5)^2=9(x+3)$, so $16x^2+40x+25=9x+27$. Hence $16x^2+31x-2=0$.

$16x^2+31x-2=(16x-1)(x+2)=0$, so $x=\dfrac1{16}$ or $x=-2$.

$x=-2$ is outside the domain, so $x=\dfrac1{16}$.

$\log_{10}a+\log_{10}b=\log_{10}(a+b)$ gives $\log_{10}(ab)=\log_{10}(a+b)$.

Since the logarithm function is one-to-one, $ab=a+b$.

$ab-a=b$, so $a(b-1)=b$ and therefore $a=\dfrac{b}{b-1}$.

Given $a>0$ and $b>0$, the fraction $\dfrac{b}{b-1}$ must be positive. Since the numerator is positive, the denominator must also be positive, so $b>1$.