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QJEEM23Ma6d34: JEE Main 2023 Mathematics - Class 12 JEEM23Ma6d34 Solution: For , let α ( m , n ) = ∫ 0 2 t m ( 1 + 3 t ) n d t . If , then …

For m, n > 0 , let α(m,n)=∫02tm(1+3t)ndt. If 11α(10,6)+18α(11,5)=p(14)611 \alpha(10,6)+18 \alpha(11,5)=p(14)^{6} , then pp is equal to ___________.

Step-by-Step Verified Solution

We have, α(m,n)=∫02tm(1+3t)ndt\alpha(m, n)=\int\limits_0^2 t^m(1+3 t)^n d t

Also, 11α(10,6)+18α(11,5)=P(14)611 \alpha(10,6)+18 \alpha(11,5)=P(14)^6

⇒11∫02t10(1+3t)6dt+18∫02t11(1+3t)5dt=P(14)6\Rightarrow 11 \int\limits_0^2 t^{10}(1+3 t)^6 d t+18 \int\limits_0^2 t^{11}(1+3 t)^5 d t=P(14)^6

Using integration by part for expression t10(1+3t)6t^{10}(1+3 t)^6

⇒11[(1+3t)6×t1111]02−∫026(1+3t)5×3×t1111dt]+18∫02t11(1+3t)5dt=P(14)6\begin{array}{r}\left.\Rightarrow 11\left[(1+3 t)^6 \times \frac{t^{11}}{11}\right]_0^2-\int\limits_0^2 6(1+3 t)^5 \times 3 \times \frac{t^{11}}{11} d t\right] \\\\+18 \int\limits_0^2 t^{11}(1+3 t)^5 d t=P(14)^6\end{array}

amp;⇒76×211−18∫02t11(1+3t)5dt+18∫02t11(1+3t)5dt=P(14)6amp;⇒76×211=P(14)6⇒(7×2)6×25=P(14)6amp;⇒P=25=32\begin{aligned}& \Rightarrow 7^6 \times 2^{11}-18 \int\limits_0^2 t^{11}(1+3 t)^5 d t+18 \int\limits_0^2 t^{11}(1+3 t)^5 d t=P(14)^6 \\\\& \Rightarrow 7^6 \times 2^{11}=P(14)^6 \Rightarrow(7 \times 2)^6 \times 2^5=P(14)^6 \\\\& \Rightarrow P=2^5=32\end{aligned}

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