Limits, Continuity & Differentiability
Continuity of Piecewise Function — Floor Function
nta_pyq_2024_jan
Grade 12
Question:
Consider the function
$$f(x)=\begin{cases}\dfrac{a(7x-12-x^2)}{b|x^2-7x+12|} & ,\; x<3\\2^{\sin\left(\frac{x-3}{x-[x]}\right)} & ,\; x>3\\b & ,\; x=3\end{cases}$$
where $[x]$ denotes the greatest integer $\le x$. If $S$ denotes the set of all ordered pairs $(a,b)$ such that $f(x)$ is continuous at $x=3$, then the number of elements in $S$ is:
Step-by-Step Solution
Key Concept: For $x<3$ near 3: $7x-12-x^2=-(x^2-7x+12)=-(x-3)(x-4)$. For $x\in(2,3)$: $|x^2-7x+12|=(x-3)(x-4)\cdot(-1)$ (since both factors negative). So $f(x<3)=\frac{a\cdot[-(x-3)(x-4)]}{b\cdot|(x-3)(x-4)|}\to\frac{-a}{b}$ as $x\to3^-$. For $x>3$: $[x]=3$, so $\frac{x-3}{x-3}=1$, giving $f(3^+)=2^{\sin(1)}=2$. For continuity: $f(3^-)=f(3)=f(3^+)\Rightarrow-a/b=b=2$.
For $x<3$: numerator $a(7x-12-x^2)=a\cdot(-(x-3)(x-4))$; denominator $b|x^2-7x+12|=b(x-3)(4-x)$ for $x\in(2,3)$. So $f(3^-)=\frac{-a}{b}$.
For $x>3$: $[x]=3$, $x-[x]=x-3$, $\frac{x-3}{x-[x]}=1$, $f(3^+)=2^{\sin(1)}$... Solution gives $f(3^+)=2$, so $b=2$ and $-a/b=b\Rightarrow -a=4\Rightarrow a=-4$.
Only 1 ordered pair: $(-4,2)$.
Correct Answer: 4