JEE 2023 — Mathematics PYQ

Solution:

• ${\int }_0^a\frac{t^{50}}{1-t}dt={\int }_0^a\frac{1-(1-t^{50})}{1-t}dt={\int }_0^a(\frac{1}{1-t}-(1+t+t^2+\cdots +t^{49}))dt$.

• Integration karne par: $[-{\log }_e(1-t){]}_0^a-[t+\frac{t^2}{2}+\cdots +\frac{t^{50}}{50}{]}_0^a$.

• Iska result $-{\log }_e(1-a)-P_{50}(a)$ hota hai.

• Kyunki $\beta ={\log }_e(1-a)$, toh ye $-\beta -P_{50}(a)$ yaani $-(\beta +P_{50}(a))$ ban jata hai.

Correct Option: (4) $-(\beta +P_{50}(a))$