JEE Main 2023 — Mathematics Calculus

Given,

$$\alpha (m,n)={\int }_{0}^{2}{t}^{m}{(1+3t)}^{n}dt$$α(m,n)=∫02​tm(1+3t)ndt

Now using integration by parts we get,

$$\Rightarrow \alpha (m,n)={{(1+3t)}^{n}\cdot \frac{{t}^{m+1}}{m+1}|}_{0}^{2}-{\int }_{0}^{2}n{(1+3t)}^{n-1}\times 3\cdot \frac{{t}^{m+1}}{m+1}dt$$⇒α(m,n)=(1+3t)n⋅m+1tm+1​∣02​−∫02​n(1+3t)n−1×3⋅m+1tm+1​dt

$$\Rightarrow (m+1)\alpha (m,n)={{(1+3t)}^{n}\cdot ({t}^{m+1})|}_{0}^{2}-3n(m+1){\int }_{0}^{2}{(1+3t)}^{n-1}\times {t}^{m+1}dt$$⇒(m+1)α(m,n)=(1+3t)n⋅(tm+1)∣02​−3n(m+1)∫02​(1+3t)n−1×tm+1dt

$$\Rightarrow (m+1)\alpha (m,n)={(1+3\times 2)}^{n}\cdot ({2}^{m+1})-3n(m+1)\alpha (m+1,n-1)$$⇒(m+1)α(m,n)=(1+3×2)n⋅(2m+1)−3n(m+1)α(m+1,n−1)

$$\Rightarrow (m+1)\alpha (m,n)={7}^{n}\cdot {2}^{m+1}-3n\alpha (m+1,n-1)$$⇒(m+1)α(m,n)=7n⋅2m+1−3nα(m+1,n−1)

$$\Rightarrow (m+1)\alpha (m,n)+3n\alpha (m+1,n-1)={7}^{n}\cdot {2}^{m+1}$$⇒(m+1)α(m,n)+3nα(m+1,n−1)=7n⋅2m+1

Now put $m=10,n=6$m=10,n=6 in above equation we get,

$$11\alpha (10,6)+18\alpha (11,5)={7}^{6}\cdot {2}^{11}=32\times {(14)}^{6}$$11α(10,6)+18α(11,5)=76⋅211=32×(14)6

Hence, on comparing with $11\alpha (10,6)+18\alpha (11,5)=p\times {(14)}^{6}$11α(10,6)+18α(11,5)=p×(14)6 we get,

$$\Rightarrow p=32$$⇒p=32