CAT 2023 Slot 2 QA Question & Solution
AlgebraMedium
Question
If $p^{2}+q^{2}-29=2pq-20=52-2pq$, then the difference between the maximum and minimum possible value of $(p^{3}-q^{3})$
Options
243
486
378
189
Solution
Given that 2pq - 20 = 52 - 2pq => 4pq = 72 => pq = 18 ----(1)
Now, $p^2+q^2-29=2pq-20$ => $p^2+q^2-2pq=9$ => $\left(p-q\right)^2=9$ => $p-q=\pm\ 3$
Also, $p^2+q^2-29=2pq-20$ => $p^2+q^2=2pq+9=2\left(18\right)+9=45$
Now, $p^3-q^3=\left(p-q\right)\left(p^2+pq+q^2\right)$ = $\left(p-q\right)\left(45+18\right)$ = $\left(p-q\right)\left(63\right)$
=> When p-q = -3 => The value is 63(-3) = -189 and when p-q = 3 => The value is 63(3) = 189.
=> The difference = 189 - (-189) = 378.
