JEE Main 2023 Mathematics - Class 12 JEEM23Mad614 Solution: Let f ( x ) = ∑ k = 1 10 k x k , x ∈ R . If then is equal to…

Given,

f(x)=\sum_\limits{k=1}^{10} k x^{k}

\begin{aligned}&amp; f(x)=x+2 x^2+\ldots \ldots \ldots+10 x^{10} \\\\&amp; f(x) . x=x^2+2 x^3+\ldots \ldots \ldots+9 x^{10}+10 x^{11} \\\\&amp; f(x)(1-x)=x+x^2+x^3+\ldots \ldots \ldots+x^{10}-10 x^{11} \\\\&amp; \therefore f(x)=\frac{x\left(1-x^{10}\right)}{(1-x)^2}-\frac{10 x^{11}}{(1-x)}\end{aligned}

\Rightarrow

f(x)=\frac{x-x^{11}-10 x^{11}+10 x^{12}}{(1-x)^2} = \frac{10 x^{12}-11 x^{11}+x}{(1-x)^2}

So,

\begin{aligned}&amp; f(2)=2\left(1-2^{10}\right)+10 \cdot 2^{11} \\\\&amp; =2+18 \cdot 2^{10}\end{aligned}

f'\left( x \right) = {{{{\left( {1 - x} \right)}^2}\left( {120{x^{11}} - 121{x^{10}} + 1} \right) + 2\left( {1 - x} \right)\left( {10{x^{12}} - {{11.x}^{11}} + 2} \right)} \over {{{\left( {1 - x} \right)}^4}}}

So,

f'\left( x \right) = {{1\left( {{{120.2}^{11}} - {{121.2}^{10}} + 1} \right) + 2\left( { - 1} \right)\left( {{{10.2}^{12}} - {{11.2}^{11}} + 2} \right)} \over {{{\left( { - 1} \right)}^4}}}

{2^{10}}\left( {83} \right) - 3

Hence

2 f(2)+f^{\prime}(2)=119.2^{10}+1

\Rightarrow \text { So, } \mathrm{n}=10

QJEEM23Mad614: JEE Main 2023 Mathematics - Class 12 JEEM23Mad614 Solution: Let f ( x ) = ∑ k = 1 10 k x k , x ∈ R . If then is equal to…