MAT Mock Section A: Multiple Choice

Fastest approach

$1, -4, 6, -4, 1$ are the coefficients of $(x-1)^{4}$.

The constant here is $5$, not $1$, so: $p(x) = (x-1)^{4} + 4$

A fourth power is never negative, and is zero at $x = 1$.

Minimum value: $4$, attained at $x = 1$.

Fastest approach

$\sin^{2}x\cos^{2}x = (\sin x \cos x)^{2} = \left(\tfrac{1}{2}\sin 2x\right)^{2} = \tfrac{1}{4}\sin^{2}2x$

Over $[0, 2\pi]$ the function $\sin^{2}2x$ completes whole periods, and $\sin^{2}$ averages $\tfrac{1}{2}$ over a whole period.

$\int_{0}^{2\pi} \tfrac{1}{4}\sin^{2}2x \,dx = \tfrac{1}{4} \cdot \tfrac{1}{2} \cdot 2\pi = \tfrac{\pi}{4}$

Fastest approach

$n^{2} + n = n(n+1)$.

One of two consecutive integers is even, so $2 \mid n(n+1)$ always. The condition reduces to $3 \mid n(n+1)$.

That holds when $n \equiv 0$ or $n \equiv 2 \pmod 3$, and fails only when $n \equiv 1$.

Among $1$ to $100$: the values $\equiv 1 \pmod 3$ are $1, 4, \ldots, 100$, which is $34$ numbers.

$100 - 34 = 66$

Fastest approach

By inspection: $2^{2} = 4 = 2^{2}$ and $2^{4} = 16 = 4^{2}$. Two solutions at $x = 2$ and $x = 4$.

Now the negative side. At $x = 0$: $2^{0} = 1 > 0 = x^{2}$. At $x = -1$: $2^{-1} = \tfrac{1}{2} < 1 = x^{2}$.

The sign of $2^{x} - x^{2}$ changes between $-1$ and $0$, so there is a root there.

For $x < -1$ the parabola only grows and the exponential only shrinks, so there are no further crossings. Three in total.

Fastest approach

$\sin 5x - \sin x = 2\cos 3x \sin 2x = 0$

$\cos 3x = 0$: $3x = \tfrac{\pi}{2}, \tfrac{3\pi}{2}, \tfrac{5\pi}{2}$, giving $x = \tfrac{\pi}{6}, \tfrac{\pi}{2}, \tfrac{5\pi}{6}$.

$\sin 2x = 0$: $2x = 0, \pi, 2\pi$, giving $x = 0, \tfrac{\pi}{2}, \pi$.

$x = \tfrac{\pi}{2}$ appears in both lists. The union is $\left\{0, \tfrac{\pi}{6}, \tfrac{\pi}{2}, \tfrac{5\pi}{6}, \pi\right\}$: five solutions.

Fastest approach

$\frac{1}{a_{n+1}} = \frac{1 + a_{n}}{a_{n}} = \frac{1}{a_{n}} + 1$

So $\frac{1}{a_n}$ is arithmetic with common difference $1$.

$\frac{1}{a_{1}} = 1$, hence $\frac{1}{a_{n}} = n$ and $a_{n} = \frac{1}{n}$.

$a_{100} = \frac{1}{100}$

Fastest approach

Write the line as $mx - y + 10 = 0$. Distance from the origin:

$d = \frac{|10|}{\sqrt{m^{2}+1}}$

Meeting the circle requires $d \leqslant 5$:

$\frac{10}{\sqrt{m^{2}+1}} \leqslant 5 \implies \sqrt{m^{2}+1} \geqslant 2 \implies m^{2} \geqslant 3$

The smallest value is $m^{2} = 3$, at which the line is tangent.

Fastest approach

The denominator has discriminant $1 - 4 < 0$, so it is never zero and the function is defined everywhere.

Set $\frac{x^{2}+2}{x^{2}+x+1} = k$:

$(1-k)x^{2} - kx + (2-k) = 0$

For a real $x$ to exist, the discriminant must be non-negative:

$k^{2} - 4(1-k)(2-k) \geqslant 0$

$k^{2} - 4(2 - 3k + k^{2}) \geqslant 0 \implies -3k^{2} + 12k - 8 \geqslant 0$

$3k^{2} - 12k + 8 \leqslant 0 \implies k = \frac{12 \pm \sqrt{48}}{6} = 2 \pm \frac{2\sqrt{3}}{3}$

The range is between those roots, so the largest value is $2 + \frac{2\sqrt{3}}{3}$.

Fastest approach

Roots at $1, 3, 5, 7$, all simple, leading coefficient positive.

Signs from the right: $+$ on $(7, \infty)$, then alternating inwards: $(5,7)$ negative, $(3,5)$ positive, $(1,3)$ negative, $(-\infty,1)$ positive.

Negative on $(1,3) \cup (5,7)$.

Integers strictly inside: $n = 2$ and $n = 6$. The roots themselves give zero, not a negative value, so they do not count.

Two integers.

Fastest approach

Roots at $0, 1, 2$. On $(0,1)$ the product is positive; on $(1,2)$ it is negative.

Expand: $x(x-1)(x-2) = x^{3} - 3x^{2} + 2x$, with antiderivative $\frac{x^{4}}{4} - x^{3} + x^{2}$.

$\int_{0}^{1} = \tfrac{1}{4} - 1 + 1 = \tfrac{1}{4}$

$\int_{1}^{2} = (4 - 8 + 4) - \tfrac{1}{4} = -\tfrac{1}{4}$

Total area $= \tfrac{1}{4} + \left|-\tfrac{1}{4}\right| = \tfrac{1}{2}$

The cubic is antisymmetric about $x = 1$, which is why the two loops match exactly.