Inequalities & Modulus: CAT previous year questions

Solution

For $x \in [k, k + 1)$, $[x]^2 = k^2$, so we need $[x^2] = k^2$, i.e. $x^2 < k^2 + 1$.

$k = 3$: $x \in [3, \sqrt{10})$. $k = 4$: $x \in [4, \sqrt{17})$. $k = 5$: $x \in [5, \sqrt{26})$. Also $x = 6$.

Option A, $(3, \sqrt{10}) \cup [5, \sqrt{26}) \cup \{6\}$, lies entirely inside these intervals.

The others include points such as $\sqrt{10}$, $\sqrt{17}$ or values above $\sqrt{17}$, which fail.

Solution

We need $y \geq 3$, $x > y$ and $x + y \leq 13$.

$y = 3$: $x = 4$ to $10$ (7 pairs). $y = 4$: $x = 5$ to $9$ (5). $y = 5$: $x = 6$ to $8$ (3). $y = 6$: $x = 7$ (1). $y \geq 7$: none.

Total $= 7 + 5 + 3 + 1 = 16$.

Solution

Case $x \geq -9$: $x^2 - (x + 9) + x = x^2 - 9 > 0 \Rightarrow x < -3$ or $x > 3$, giving $[-9, -3) \cup (3, \infty)$.

Case $x < -9$: $x^2 + x + 9 + x = x^2 + 2x + 9 > 0$ always (discriminant $< 0$).

Combined: $(-\infty, -3) \cup (3, \infty)$.

Solution

Write $b = a + p$, $c = a + q$, $d = a + r$ (integers). Then $4a + p + q + r = 46$.

Since 46 is not a multiple of 4, $p + q + r \neq 0$; in fact $p + q + r \equiv 2 \pmod 4$, so $|p + q + r| \geq 2$.

We minimise $p^2 + q^2 + r^2$: with $p + q + r = 2$, take $p = q = 1$, $r = 0$ (e.g. $a = 11$: 11, 12, 12, 11).

Minimum $= 1 + 1 + 0 = 2$.

Solution

$p + q = 900 - r$ with $r \in \{169, 196, \dots, 484\}$.

$0.3q \leq p \leq 0.7q$ means $\dfrac{3}{13} \leq \dfrac{p}{p + q} \leq \dfrac{7}{17}$.

Maximum $p$: take $r = 169$, so $p + q = 731$ and $p \leq \dfrac{7}{17} \times 731 = 301$ ($q = 430$ ✓).

Minimum $p$: take $r = 484$, so $p + q = 416$ and $p \geq \dfrac{3}{13} \times 416 = 96$ ($q = 320$ ✓).

Sum $= 301 + 96 = 397$.

Solution

$\dfrac{1}{x + 5} - \dfrac{1}{2x - 3} \leq 0 \Rightarrow \dfrac{x - 8}{(x + 5)(2x - 3)} \leq 0$.

Critical points: $-5$, $\dfrac{3}{2}$, $8$. Sign check:

$x < -5$: negative ✓. $-5 < x < \dfrac{3}{2}$: positive ✗. $\dfrac{3}{2} < x < 8$: negative ✓. $x = 8$: zero ✓. $x > 8$: positive ✗.

Answer: $x < -5$ or $\dfrac{3}{2} < x \leq 8$.

Solution

If $x < 0$: the first equation gives $y = 15$, then the second gives $x = 20$ ✗.

So $x \geq 0$: $2x + y = 15$. If $y \geq 0$, the second gives $x = 20$ and $y = -25$ ✗, so $y < 0$: $x - 2y = 20$.

Solving: $x = 10$, $y = -5$.

$x - y = 15$.

Solution

Left side $= |x - 2|$. Right side: $4$ for $x \geq 2$; $2x$ for $-2 \leq x \leq 2$; $-4$ for $x \leq -2$.

$x \geq 2$: $x - 2 = 4 \Rightarrow x = 6$ ✓.

$-2 \leq x \leq 2$: $2 - x = 2x \Rightarrow x = \dfrac{2}{3}$ ✓.

$x \leq -2$: $|x - 2| = -4$ is impossible.

Number of solutions $= 2$.

Solution

$|x + y| + |x - y| = 2\max(|x|, |y|)$, so we need $\max(|x|, |y|) = 1$.

These are the integer points on the boundary of the square with corners $(\pm 1, \pm 1)$:

$(\pm 1, 0), (0, \pm 1), (\pm 1, \pm 1)$. Total $= 8$.

Solution

$x = 0$ is a solution.

For $x \neq 0$, divide by $|x|$: $2(x^2 + 1) = 5|x| \Rightarrow 2t^2 - 5t + 2 = 0$ with $t = |x|$.

$t = 2$ or $\dfrac{1}{2}$; integer solutions $x = \pm 2$.

Total $= 3$.

Solution

$\dfrac{x + 3}{y - 3} - \dfrac{x}{y} > 0 \Rightarrow \dfrac{3(x + y)}{y(y - 3)} > 0$.

Option B: $y < 0$ makes $y(y - 3) > 0$, and $-x < y$ means $x + y > 0$. The fraction is positive ✓.

(The other options do not guarantee the sign.)

Solution

$5^{n-1} < 3^{n+1}$ holds for $n = 1, 2, 3, 4, 5$ (at $n = 5$: $625 < 729$) and fails from $n = 6$ ($3125 > 2187$).

We need $3^{n+1} < 2^{n+m}$ for each such $n$:

$n = 1$: $9 < 2^{1+m} \Rightarrow m \geq 3$. $n = 3$: $81 < 2^{3+m} \Rightarrow m \geq 4$. $n = 5$: $729 < 2^{5+m} \Rightarrow 5 + m \geq 10 \Rightarrow m \geq 5$.

Least $m = 5$.

Solution

$f$ is a sum of absolute values, so on $[a, 100]$ its maximum is at an endpoint.

$f(a) = 0 + (100 - a) + 50 = 150 - a$. $f(100) = (100 - a) + 0 + |50 - a|$.

Check $a = 50$: $f(a) = 100$ and $f(100) = 50$, so the maximum is 100 ✓.

(For $a = 0$ or $25$ the maximum exceeds 100; for $a = 100$ it is 50.)

Solution

$|x + a| + |x - 1| \geq |1 + a|$, with equality for every $x$ between $-a$ and $1$.

Infinitely many solutions need this minimum to equal 2: $|a + 1| = 2 \Rightarrow a = 1$ or $a = -3$.

Largest value: $a = 1$.

Solution

$a = 6 - 2b \geq 0 \Rightarrow 0 \leq b \leq 3$.

$a + b = 6 - b$, which ranges from 3 (at $b = 3$) to 6 (at $b = 0$).

Average $= \dfrac{3 + 6}{2} = 4.5$.

Solution

Let $t = \dfrac{x}{y}$ (any non-zero real). $c = 16t + \dfrac{49}{t}$.

$t > 0$: by AM–GM, $c \geq 2\sqrt{16 \times 49} = 56$. $t < 0$: $c \leq -56$.

So $c$ cannot lie in $(-56, 56)$; of the options, $-50$ is impossible.

Solution

Let $t = 3 - x > 0$. Then $x^2 - 6x + 10 = (x - 3)^2 + 1 = t^2 + 1$.

Expression $= \dfrac{t^2 + 1}{t} = t + \dfrac{1}{t} \geq 2$, with equality at $t = 1$ ($x = 2$).

Minimum $= 2$.

Solution

$|n - 60| < |n - 100|$: $n$ is closer to 60 than to 100, so $n < 80$.

$|n - 100| < |n - 20|$: $n$ is closer to 100 than to 20, so $n > 60$.

$n = 61, \dots, 79$: $19$ integers.

Solution

$f(x) = \dfrac{(x + 5)(x - 3)}{(x - 9)(x + 2)}$. Critical points: $-5, -2, 3, 9$.

Signs from right to left: $x > 9$: $+$; $3 < x < 9$: $-$; $-2 < x < 3$: $+$; $-5 < x < -2$: $-$; $x < -5$: $+$.

Negative for $-5 < x < -2$ or $3 < x < 9$.

Solution

$0.25 \leq 2^{n+2} \leq 200 \Rightarrow -2 \leq n + 2 \leq 7 \Rightarrow -4 \leq n \leq 5$.

$3 + 3^{n+1}$ is an integer only when $n + 1 \geq 0$, i.e. $n = -1, 0, \dots, 5$.

Number of integer values $= 7$.

Solution

$|3x - 20| + |3x - 40| \geq |(3x - 20) - (3x - 40)| = 20$, with equality when $20 \leq 3x \leq 40$.

So the equation holds exactly for $\dfrac{20}{3} \leq x \leq \dfrac{40}{3}$ (about 6.67 to 13.33).

The interval in the options that lies fully inside is $7 < x < 12$.

Solution

$|1 + mn| < |m + n| \iff (1 + mn)^2 < (m + n)^2 \iff (m^2 - 1)(n^2 - 1) < 0$.

So one of $|m|, |n|$ is 0 and the other is at least 2.

With $|m + n| < 5$: $m = 0$, $n \in \{\pm 2, \pm 3, \pm 4\}$ (6 pairs), and the same with $m, n$ swapped (6 pairs).

Total $= 12$.

Solution

If $x < 0$: the second equation gives $y = \dfrac{1}{3}$, then the first gives $x = 2$ ✗. So $x \geq 0$: $2x + 3y = 1$.

If $y \geq 0$: $3x + 3y = 7$, giving $x = 6$, $y < 0$ ✗. So $y < 0$: $3x - y = 7$.

Solving: $x = 2$, $y = -1$.

$x + 2y = 0$.

Solution

$x = z - 9$ and $y = z - 1$. Then $x + y < z + 5 \Rightarrow 2z - 10 < z + 5 \Rightarrow z < 15$.

$2x + y = 3z - 19$, largest at $z = 14$: $2x + y = 23$ ($x = 5$, $y = 13$).

Solution

For $x > 0$: $\dfrac{x}{\sqrt{1 + x^4}} = \dfrac{1}{\sqrt{x^2 + \dfrac{1}{x^2}}}$.

$x^2 + \dfrac{1}{x^2} \geq 2$, so the expression $\leq \dfrac{1}{\sqrt{2}}$, at $x = 1$.

Maximum $= \dfrac{1}{\sqrt{2}}$.

Solution

$\left(1 + \dfrac{1}{x}\right)\left(1 + \dfrac{1}{y}\right) = 1 + \dfrac{x + y + 1}{xy} = 1 + \dfrac{103}{xy}$.

This is smallest when $xy$ is largest, i.e. $x = y = 51$, $xy = 2601$.

Minimum $= 2601\left(1 + \dfrac{103}{2601}\right) = 2601 + 103 = 2704$.

Solution

$0.2 < \dfrac{m}{20} < 0.5 \Rightarrow 4 < m < 10$.

$0.2 < \dfrac{n}{11} < 0.5 \Rightarrow 2.2 < n < 5.5$, and $n$ even, so $n = 4$.

$0.2 < \dfrac{4}{m} < 0.5 \Rightarrow 8 < m < 20$. With $4 < m < 10$: $m = 9$.

$m - 2n = 9 - 8 = 1$.