JEE 2023 — Mathematics PYQ

Solution:

• Integration by Parts: $\alpha (11,5)={\int }_0^2{t}^{11}(1+3t{)}^{5}dt$ ko hal karein.

• Logic: Is prakar ke reduction formula mein $11\alpha (10,6)+18\alpha (11,5)$ ko simplify karne par hume integration ki boundary values milti hain.

• Result: Solving leads to $[t^{11}(1+3t{)}^6{]}_0^2=2^{11}(7{)}^6=(2\cdot 7{)}^6\cdot 2^5=32\cdot (14{)}^6$.

• Answer: 32