Applications of Derivatives
Differentiability of Piecewise Function — Finding Ratio of Derivatives
nta_pyq_2023_apr
Grade 12
Question:
Let $k$ and $m$ be positive real numbers such that the function $f(x)=\begin{cases}3x^2+k\sqrt{x+1}, & 0<x<1\\mx^2+k^2, & x\geq1\end{cases}$ is differentiable for all $x>0$. Then $\dfrac{8f'(8)}{f'\!\left(\tfrac{1}{8}\right)}$ is equal to
Step-by-Step Solution
Key Concept: For differentiability at $x=1$: continuity gives $3+k\sqrt{2}=m+k^2$ and equal derivatives give $6+\frac{k}{2\sqrt{2}}=2m$.
$k=\frac{7\sqrt{2}}{8},\ m=\frac{103}{96}$. $\frac{8f'(8)}{f'(1/8)}=309$.
Correct Answer: 309