Continuity, differentiability and limits practice questions — IB Maths AA

$g(0)=1>0$, $g(\pi \mathrm{/}2)=0-\pi \mathrm{/}2<0$. Since $g$ is continuous, by the IVT there is $c\in (0,\pi \mathrm{/}2)$ 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 $t^{\ast }\in (7,17)$ with $f(t^{\ast })=0$, i.e. $u(t^{\ast })=d(t^{\ast })$. At that clock time, the monk stands at the same point on the trail on both days. $\mathrm{■}$

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.

${\displaystyle \underset{x\to 0}{\lim }{\displaystyle \frac{\sin 2x}{x}}=2\cdot \underset{x\to 0}{\lim }{\displaystyle \frac{\sin 2x}{2x}}=2\cdot 1=2}$. So $a=2$.

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

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

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

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 $x=\frac{\pi }{2}+k\pi$.

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

${\displaystyle \frac{x^2-1}{x-1}}=x+1\to 2$ 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 $f(c)=0$.