Continuity, differentiability and limits practice questions — IB Maths AA

$g(0) = 1 > 0$, $g(\pi/2) = 0 - \pi/2 < 0$. Since $g$ is continuous, by the IVT there is $c \in (0, \pi/2)$ with $g(c) = 0$, i.e. $\cos c = c$.

Let $L$ be the trail length. Define

- $u(t)$: the monk's distance from the base on Monday at time $t \in [7, 17]$ - $d(t)$: her distance from the base on Tuesday at the same clock time

Both functions are continuous on $[7, 17]$ (a hiker doesn't teleport).

Now consider $f(t) = u(t) - d(t)$. By the problem statement:

- $u(7) = 0$ and $d(7) = L$, so $f(7) = -L < 0$. - $u(17) = L$ and $d(17) = 0$, so $f(17) = +L > 0$.

Since $f$ is continuous and changes sign on $[7, 17]$, the Intermediate Value Theorem guarantees some $t^* \in (7, 17)$ with $f(t^*) = 0$, i.e. $u(t^*) = d(t^*)$. At that clock time, the monk stands at the same point on the trail on both days. $\blacksquare$

The physical intuition ("imagine two monks: one ascending Monday, one descending Tuesday at the same time — they must meet") collapses into a one-line IVT argument once you let $f$ do the work.

$\displaystyle\lim_{x \to 0} \dfrac{\sin 2x}{x} = 2 \cdot \lim_{x \to 0} \dfrac{\sin 2x}{2x} = 2 \cdot 1 = 2$. So $a = 2$.

Left limit: $\displaystyle\lim_{x \to 2^{-}} (x + 1) = 3$.

Right limit: $\displaystyle\lim_{x \to 2^{+}} (x^{2} - 1) = 3$.

$f(2) = 2^{2} - 1 = 3$.

All three agree, so $f$ is continuous at $x = 2$.

$\lim_{x \to 0^-}f = -1$ and $\lim_{x \to 0^+}f = 1$ — finite but unequal limits, hence a jump discontinuity.

LHS: $2 \cdot 1 + 1 = 3$. RHS: $1 + a$. Setting equal: $a = 2$.

Continuity is preserved by sum, difference, product, and composition (provided $g$ is continuous at $x_0$ and $f$ is continuous at $g(x_0)$).

(a) A polynomial — continuous on all of $\mathbb{R}$ ✓.

(b) Discontinuous at $x = 0$ (not defined).

(c) Discontinuous at $x = \tfrac{\pi}{2} + k\pi$.

(d) Defined only for $x \ge 0$, so not continuous on negative reals.

$\dfrac{x^2 - 1}{x - 1} = x + 1 \to 2$ as $x \to 1$. For continuity, $a = 2$.

$f(0) = -1 < 0$ and $f(1) = 1 > 0$. Since $f$ is continuous, by the IVT there exists $c \in (0, 1)$ with $f(c) = 0$.