JEE MAIN - Mathematics (2023 - 11th April Morning Shift - No. 19)
For $$m, n > 0$$, let $$\alpha(m, n)=\int_\limits{0}^{2} t^{m}(1+3 t)^{n} d t$$. If $$11 \alpha(10,6)+18 \alpha(11,5)=p(14)^{6}$$, then $$p$$ is equal to ___________.
Answer
32
Explanation
We have, $\alpha(m, n)=\int\limits_0^2 t^m(1+3 t)^n d t$
Also, $11 \alpha(10,6)+18 \alpha(11,5)=P(14)^6$
$\Rightarrow 11 \int\limits_0^2 t^{10}(1+3 t)^6 d t+18 \int\limits_0^2 t^{11}(1+3 t)^5 d t=P(14)^6$
Using integration by part for expression $t^{10}(1+3 t)^6$
$$ \begin{array}{r} \left.\Rightarrow 11\left[(1+3 t)^6 \times \frac{t^{11}}{11}\right]_0^2-\int\limits_0^2 6(1+3 t)^5 \times 3 \times \frac{t^{11}}{11} d t\right] \\\\ +18 \int\limits_0^2 t^{11}(1+3 t)^5 d t=P(14)^6 \end{array} $$
$$ \begin{aligned} & \Rightarrow 7^6 \times 2^{11}-18 \int\limits_0^2 t^{11}(1+3 t)^5 d t+18 \int\limits_0^2 t^{11}(1+3 t)^5 d t=P(14)^6 \\\\ & \Rightarrow 7^6 \times 2^{11}=P(14)^6 \Rightarrow(7 \times 2)^6 \times 2^5=P(14)^6 \\\\ & \Rightarrow P=2^5=32 \end{aligned} $$
Also, $11 \alpha(10,6)+18 \alpha(11,5)=P(14)^6$
$\Rightarrow 11 \int\limits_0^2 t^{10}(1+3 t)^6 d t+18 \int\limits_0^2 t^{11}(1+3 t)^5 d t=P(14)^6$
Using integration by part for expression $t^{10}(1+3 t)^6$
$$ \begin{array}{r} \left.\Rightarrow 11\left[(1+3 t)^6 \times \frac{t^{11}}{11}\right]_0^2-\int\limits_0^2 6(1+3 t)^5 \times 3 \times \frac{t^{11}}{11} d t\right] \\\\ +18 \int\limits_0^2 t^{11}(1+3 t)^5 d t=P(14)^6 \end{array} $$
$$ \begin{aligned} & \Rightarrow 7^6 \times 2^{11}-18 \int\limits_0^2 t^{11}(1+3 t)^5 d t+18 \int\limits_0^2 t^{11}(1+3 t)^5 d t=P(14)^6 \\\\ & \Rightarrow 7^6 \times 2^{11}=P(14)^6 \Rightarrow(7 \times 2)^6 \times 2^5=P(14)^6 \\\\ & \Rightarrow P=2^5=32 \end{aligned} $$
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