_ x → 2 ( ∑ _ n = 1 ^ 9 x n ( n + 1 ) x ^ 2 + 2 ( 2n + 1 ) x + 4 ) is equal to :

Correct answer: D. (9)/(44) _ x → 2 ∑ _ n = 1 ^ 9 ( x n ( n + 1 ) x ^ 2 + 2 ( 2n + 1 ) x + 4 ) _ x → 2 ∑ _ n = 1 ^ 9 ( x n(n + 1) x ^ 2 + 2nx + 2(n + 1)x + 4 ) _ x → 2 ∑ _ n = 1 ^ 9 (([( n + 1 ) x + 2] - [nx + 2])/([( n + 1 ) x + 2][nx + 2])) _ x → 2 ∑ _ n = 1 ^ 9 ((1)/(( nx + 2 )) - (1)/(( n + 1 ) x + 2)) = (1)/(x + 2) - (1)/(2x + 2) (1)/(2x + 2) - (1)/(3x + 2) array ll . . . . array (+ (1)/(9x + 2) - (1)/(10x + 2))/((1)/(x + 2) - (1)/(10x + 2)) ⇒ _ x → 2 ((1)/(x + 2) - (1)/(10x + 2)) = (9)/(44) ⇒ _ x → 2 (10x + 2 - x - 2)/((x + 2)(10x + 2)) _ x → 2 (9x)/((x + 2)(10x + 2)) = (18)/(4 × 22) = (9)/(44)

JEE Main 202126 AugustEvening Shift

Exam: JEE Main
Year: 2021
Date: 26 August
Shift: Evening Shift
Subject: Mathematics
Chapter: Limits

JEE Main 2021 · 26 August · Evening Shift · Mathematics · Limits

_ x → 2 ( ∑ _ n = 1 ^ 9 x n ( n + 1 ) x ^ 2 + 2 ( 2n + 1 ) x + 4 )...

\lim_{x \to 2}

\left(\mathop{∑}_{n = 1}^{9}\dfrac{x}{n ( n + 1 ) {x}^{2} + 2 ( 2n + 1 ) x + 4}\right)

is equal to :

Options

A $\dfrac{5}{24}$
B $\dfrac{7}{36}$
C $\dfrac{1}{5}$
D $\dfrac{9}{44}$

Correct answer

D. $\dfrac{9}{44}$

Step-by-step solution

$\lim_{x \to 2}\mathop{∑}_{n = 1}^{9}\left(\dfrac{x}{n ( n + 1 ) {x}^{2} + 2 ( 2n + 1 ) x + 4}\right)$

$\lim_{x \to 2}\mathop{∑}_{n = 1}^{9}\left(\dfrac{x}{n\left(n + 1\right){x}^{2} + 2nx + 2\left(n + 1\right)x + 4}\right)$

$\lim_{x \to 2}\mathop{∑}_{n = 1}^{9}\left(\dfrac{\left[( n + 1 ) x + 2\right] - \left[nx + 2\right]}{\left[( n + 1 ) x + 2\right]\left[nx + 2\right]}\right)$

$\lim_{x \to 2}\mathop{∑}_{n = 1}^{9}\left(\dfrac{1}{( nx + 2 )} - \dfrac{1}{( n + 1 ) x + 2}\right)$

$= \dfrac{1}{x + 2} - \dfrac{1}{2x + 2}$

$\dfrac{1}{2x + 2} - \dfrac{1}{3x + 2}$

$\begin{array}{ll}. \\ . \\ . \\ .\end{array}$ \dfrac{+ \dfrac{1}{9x + 2} - \dfrac{1}{10x + 2}}{\dfrac{1}{x + 2} - \dfrac{1}{10x + 2}} $$ ⇒ \lim_{x \to 2}\left(\dfrac{1}{x + 2} - \dfrac{1}{10x + 2}\right) = \dfrac{9}{44} $$ ⇒ \lim_{x \to 2}\dfrac{10x + 2 - x - 2}{\left(x + 2\right)\left(10x + 2\right)} $$ \lim_{x \to 2}\dfrac{9x}{\left(x + 2\right)\left(10x + 2\right)} = \dfrac{18}{4 \times 22} = \dfrac{9}{44}$

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