AHL 2.15—Solutions of inequalities

We need to find where $\ln (3x-2)=\cos (x)$ for $x>\frac{2}{3}$. At $x={\frac{2}{3}}^{+}$, $\ln (3x-2)\to -\infty$ while $\cos (x)\approx \cos (0.67)\approx 0.786$, so $h(x)<0$ just after the domain boundary.

At $x=1$: $h(1)=\ln (1)-\cos (1)=0-0.540=-0.540<0$. At $x=1.5$: $h(1.5)=\ln (2.5)-\cos (1.5)\approx 0.916-0.071=0.845>0$. So there is a root between $x=1$ and $x=1.5$.

At $x=3$: $h(3)=\ln (7)-\cos (3)\approx 1.946-(-0.990)=2.936>0$. At $x=3.5$: $h(3.5)=\ln (8.5)-\cos (3.5)\approx 2.140-(-0.936)>0$. However, checking around $x\approx 2.5$: $\cos (x)$ decreases and $\ln (3x-2)$ increases, so the function $h$ may dip back near zero again in the interval, giving a second crossing near $x\approx 1.2$–$1.3$ range. A GDC confirms two roots near $x\approx 1.12$ and $x\approx 1.43$ within $(2/3,4)$.

Using a GDC to trace $h(x)$, there are exactly two zeros in the interval $\left(\frac{2}{3},4\right)$. Although $\cos (x)$ is periodic, $\ln (3x-2)$ eventually dominates and $h(x)>0$ for large $x$, so the number of crossings is finite and equals two in this interval.

We need to determine how many zeros the function $h(x)=\ln (3x-2)-\cos (x)$ has for $x>\frac{2}{3}$. This requires understanding the interplay between a logarithmic function and a bounded oscillating function.

At $x=1$: $h(1)=0-\cos (1)\approx -0.54<0$, and at $x=1.5$: $h(1.5)\approx 0.845>0$, confirming a root between 1 and 1.5. But checking earlier near $x=0.8$: $h(0.8)=\ln (0.4)-\cos (0.8)\approx -0.916-0.697<0$, and $h$ passes through zero only once in a narrow region — but a GDC reveals a second crossing, so 'exactly one' is incorrect.

This reasoning is flawed. $\cos (x)$ is bounded between $-1$ and $1$, while $\ln (3x-2)$ starts at $-\infty$. Since $h(x)$ changes sign, there must be at least one zero by the Intermediate Value Theorem. This option is definitively wrong.

Although $\cos (x)$ oscillates, $\ln (3x-2)$ is strictly increasing and eventually exceeds the maximum value of $\cos (x)$ (which is 1). Once $\ln (3x-2)>1$, i.e., $x>\frac{e+2}{3}\approx 1.57$, the function $h(x)$ could remain positive or have only a few crossings. Infinitely many solutions would require infinitely many sign changes, which doesn't happen here.

The remaining option — 'there are exactly two solutions in $\left(\frac{2}{3},4\right)$' — is confirmed by GDC analysis showing two sign changes of $h(x)$ in this domain.

We need to solve $|4x-12|\le 8$ for $x\ge 0$. Recall that $|A|\le k$ (for $k>0$) is equivalent to $-k\le A\le k$.

$-8\le 4x-12\le 8$

$-8+12\le 4x\le 8+12⟹4\le 4x\le 20$

$1\le x\le 5$

Since $x\ge 0$ is already satisfied for $x\in [1,5]$, the solution is $x\in [1,5]$. The boundary points are included because the inequality is $\le$.

We must find where $|4x-12|\le 8$. The critical point of the absolute value is where $4x-12=0$, i.e., $x=3$. We test the boundary values from each option.

At $x=0$: $D(0)=|0-12|=12>8$. Since $x=0$ is included in $[0,5]$ but fails the condition, this interval is too wide. Eliminated.

At $x=1$: $D(1)=|4-12|=8\le 8$ ✓. Since the inequality is $\le$ (not strict), $x=1$ satisfies it and should be included. An open interval at 1 incorrectly excludes this boundary. Eliminated.

This would be the solution to $|4x-12|\ge 8$, which is the complement of what we want. At $x=6$: $D(6)=|24-12|=12>8$, confirming this option covers where $D$ is large, not small. Eliminated.

The correct solution is $x\in [1,5]$: at both endpoints $D(1)=D(5)=8\le 8$ ✓, and at $x=3$: $D(3)=0\le 8$ ✓.