CAT 2022 Slot 2 · Quant · Question 17

A power equals 1 in exactly three situations, and each must be checked separately.

The first is when the base is 1, whatever the exponent.

That needs {x}^{2}-10=1, so {x}^{2}=11 — which has no integer solution.

The second is when the base is -1 and the exponent is even.

That needs {x}^{2}-10=-1, so {x}^{2}=9 and x=\pm 3.

At x=3 the exponent is 9-9-10=-10, which is even — so this works.

At x=-3 the exponent is 9+9-10=8, also even — so this works too.

The third is when the exponent is 0 and the base is non-zero.

That needs {x}^{2}-3x-10=0, which factors as \left( x-5 \right)\left( x+2 \right)=0, giving x=5 or x=-2.

At x=5 the base is 25-10=15, non-zero — so this works.

At x=-2 the base is 4-10=-6, non-zero — so this works too.

The integer solutions are -3, -2, 3 and 5, giving 4 in all.

Shortcut. Run the three standard cases in order and the whole question is three small equations. The easy trap is stopping at the exponent-zero case and reporting 2, or accepting a base of -1 without checking that the exponent is even.