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(g(x))=√(x²−4) requires x≤−2 or x≥2. The reverse composition g(f(x))=x−4 still requires x≥0 from its inner root.
Solutions 7π/6 and 11π/6. The principal arcsine value is −π/6, within [−π/2,π/2] . A principal inverse does not enumerate every angle having the same sine.
Cap threshold 3x+10=100 gives x=30. Use d=3x+10 for 0≤x≤30 and d=100 for 30<x≤50. A reading of 70 gives x=20. A reading of 100 gives every x in [30,50] , not one recoverable input.
Unloaded output 6 V. Loaded lower equivalent 1 kΩ gives output 4.5 V and source current 4.5 mA. Each lower branch has 2.25 mA, adding to 4.5 mA. KVL gives 9−4.5−4.5=0.
With outgoing currents positive, (V−10)/2+(V−4)/4+V/4=0. Multiplying by 4 gives 4V−24=0, hence V=6 V. Outgoing currents are −2, 0.5 and 1.5 mA. Thus 2 mA enters from the 10 V node, balancing the two outgoing branches.
Choose the current driven by the 15 V rise. 15−6−I(3+6)=0 yields 1 mA, with resistor drops 3 V and 6 V. Reversing the reference makes the solved current −1 mA; the physical current is unchanged.