Use the rubric on the exam page. Award method credit for a correct model carried through with a minor arithmetic error, but check domain exclusions, sign conventions and units explicitly.
f⁻¹(y)=√(y−2) for y≥2. Output 11 corresponds to x=3. The original nonnegative domain selects the positive root.
First require x>1. Combining logs gives x²−1=15, so candidates are ±4 and only x=4 is admissible. The second expression requires −4<x<4.
Threshold time t=5 ln 5≈8.047 s. Time constant 5 s corresponds to a factor 1/e. Half-value time 5 ln 2≈3.466 s corresponds to a factor 1/2. Different output ratios select different times.
220 μA=0.000220 A, power 1.10 mW; resistance 4700 Ω. M is mega and m is milli, differing by a factor of a billion.
Unloaded voltage 6 V. Lower equivalent 10×100/110≈9.0909 kΩ. Reading 12×9.0909/(10+9.0909)≈5.71429 V. Error (5.71429−6)/6×100%≈−4.7619%.
Current 6/1020≈5.88235 mA, compared with 6 mA unloaded, a −1.9608% change. Display error bound 0.01×3.20+2×0.01=0.052 V. Loading and the accuracy specification describe different effects.