IPMAT Indore 2024 · Quantitative Ability · Question 2
IPMAT Indore 2024 previous year question on Surds and Indices, with the answer and a step-by-step solution.
Answer & solution
Answer: 6
The governing idea. A power {B}^{E} equals 1 in exactly three ways. Check all three — missing one is how this question is lost.
Case 1 — the exponent is zero (with a non-zero base).
{x}^{2}-5x+6=0 \Rightarrow \left( x-2 \right)\left( x-3 \right)=0 \Rightarrow x=2, 3
Base at x=2: 4-30+55=29\ne 0. At x=3: 9-45+55=19\ne 0. Both valid. 2 roots.
Case 2 — the base is 1.
{x}^{2}-15x+55=1 \Rightarrow{x}^{2}-15x+54=0 \Rightarrow \left( x-6 \right)\left( x-9 \right)=0
x=6, 9. A base of 1 works for any exponent. 2 roots.
Case 3 — the base is -1 and the exponent is even.
{x}^{2}-15x+55=-1 \Rightarrow{x}^{2}-15x+56=0 \Rightarrow \left( x-7 \right)\left( x-8 \right)=0
x=7: exponent =49-35+6=20, even. ✔ x=8: exponent =64-40+6=30, even. ✔ 2 roots.
Total. x\in \{2,3,6,7,8,9\} — all distinct.
Answer: 6
Shortcut. Remember “0, 1, −1”. {B}^{E}=1 needs E=0 (with B\ne 0), or B=1, or B=-1 with E even. Each case here is a neat quadratic giving two roots, so the count is 2+2+2=6 — you can answer without even checking the parity, since the symmetric set-up all but guarantees it. Students who solve only Case 1 answer 2; those who skip Case 3 answer 4. Both wrong answers are the ones the setter planted.
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