Functions
Piecewise function — discontinuity and non-differentiability count
MJAT_TS6_P2
Grade 12
Question:
Let $f(x)=\begin{cases}\dfrac{x^3}{x^5-4x^3+2\cos(\pi x)}&x\geq 0\\\dfrac{-\sin x-(x^2+8x+12)}{x^2}&x<0\end{cases}$. If $p$ = discontinuities, $q$ = non-differentiable points, then $p^2+pq+q^2$ equals:
Step-by-Step Solution
Key Concept: For $x<0$: denominator $x^2$, numerator has zeros at $x=-2$ and $x=-6$ from $(x+2)(x+6)$. For $x\geq 0$: check denominator zeros. Discontinuity at $x=0$; non-differentiabilities at $\pm 2,\pm 6$.
$p^2+pq+q^2=\mathbf{31}$.
Correct Answer: 31