_ x → ∞ ( 3x + 1 + 3x - 1 ) ^ 6 + ( 3x + 1 - 3x - 1 ) ^ 6 (x + x ^ 2 - 1 ) ^ 6 + (x - x ^ 2 - 1 ) ^ 6 x ^ 3

Correct answer: D. is equal to 27 _ x → ∞ ( 3x + 1 + 3x - 1 ) ^ 6 + ( 3x + 1 - 3x - 1 ) ^ 6 (x + x ^ 2 - 1 ) ^ 6 + (x - x ^ 2 - 1 ) ^ 6 We know the binomial expansion of (a + b) ^ n and (a - b) ^ n is given by, (a + b) ^ n = C0 n a ^ n b ^ 0 + C1 n a ^ n - 1 b ^ 1 + C2 n a ^ n - 2 b ^ 2 + . . . . . . + Cn n a ^ 0 b ^ n (a - b) ^ n = C0 n a ^ n b ^ 0 - C1 n a ^ n - 1 b ^ 1 + C2 n a ^ n - 2 b ^ 2 - . . . . . . + Cn n a ^ 0 b ^ n So, (a + b) ^ 6 + (a - b) ^ 6 = 2( T _ 1 + T _ 3 + T _ 5 + T _ 7 ) = 2(C0 6 a ^ 6 + C2 6 a ^ 4 b ^ 2 + c4 6 a ^ 2 b ^ 4 + C6 6 b ^ 6 ) From equation (1) we get, _ x → ∞ (2 (3x + 1) ^ 3 + 30 (3x + 1) ^ 2 (3x - 1) + 30(3x + 1) (3x - 1) ^ 2 + 2 (3x - 1) ^ 3 ) 2 x ^ 6 + 30 x ^ 4 ( x ^ 2 - 1) + 30 x ^ 2 ( x ^ 2 - 1) ^ 2 + ( x ^ 2 - 1) ^ 3 = _ x → ∞ x ^ 6 [2 (3 + (1)/(x)) ^ 3 + 30 (3 + (1)/(x)) ^ 2 (3 - (1)/(x)) + 30(3 + (1)/(x)) (3 - (1)/(x)) ^ 2 + 2 (3 - (1)/(x)) ^ 3 \ x ^ 6 [2 + 30(1 - 1 x ^ 2 + 30 (1 - 1 x ^ 2 ) ^ 2 + 2 (1 - 1 x ^ 2 ) ^ 3 ]\ = 2 (3) ^ 3 + 30 (3) ^ 3 + 30 (3) ^ 3 + 2 (3) ^ 3 2 + 30 + 30 + 2 = (3) ^ 3 = 27

JEE Main 202331 JanuaryEvening Shift

Exam: JEE Main
Year: 2023
Date: 31 January
Shift: Evening Shift
Subject: Mathematics
Chapter: Limits

JEE Main 2023 · 31 January · Evening Shift · Mathematics · Limits

_ x → ∞ ( 3x + 1 + 3x - 1 ) ^ 6 + ( 3x + 1 - 3x - 1 ) ^ 6 (x + x ^...

\lim_{x \to \infty}\dfrac{{\left(\sqrt{3x + 1} + \sqrt{3x - 1}\right)}^{6} + {\left(\sqrt{3x + 1} - \sqrt{3x - 1}\right)}^{6}}{{\left(x + \sqrt{{x}^{2} - 1}\right)}^{6} + {\left(x - \sqrt{{x}^{2} - 1}\right)}^{6}}{x}^{3}

Options

Ais equal to $\dfrac{27}{2}$
Bis equal to $9$
Cdoes not exist
Dis equal to $27$

Correct answer

D. is equal to $27$

Step-by-step solution

\lim_{x \to \infty}\dfrac{{\left(\sqrt{3x + 1} + \sqrt{3x - 1}\right)}^{6} + {\left(\sqrt{3x + 1} - \sqrt{3x - 1}\right)}^{6}}{{\left(x + \sqrt{{x}^{2} - 1}\right)}^{6} + {\left(x - \sqrt{{x}^{2} - 1}\right)}^{6}} $</p><p>We know the binomial expansion of$ {\left(a + b\right)}^{n} $and$ { $\left(a - b\right)}^{n}$ is given by,

${\left(a + b\right)}^{n} = C0 n {a}^{n}{b}^{0} + C1 n {a}^{n - 1}{b}^{1} + C2 n {a}^{n - 2}{b}^{2} + . . . . . . + Cn n {a}^{0}{b}^{n}$

${\left(a - b\right)}^{n} = C0 n {a}^{n}{b}^{0} - C1 n {a}^{n - 1}{b}^{1} + C2 n {a}^{n - 2}{b}^{2} - . . . . . . + Cn n {a}^{0}{b}^{n}$

So, ${\left(a + b\right)}^{6} + {\left(a - b\right)}^{6} = 2\left({T}_{1} + {T}_{3} + {T}_{5} + {T}_{7}\right)$

$= 2\left(C0 6 {a}^{6} + C2 6 {a}^{4}{b}^{2} + c4 6 {a}^{2}{b}^{4} + C6 6 {b}^{6}\right)$

From equation (1) we get,

$\lim_{x \to \infty}\dfrac{\left(2{\left(3x + 1\right)}^{3} + 30{\left(3x + 1\right)}^{2}\left(3x - 1\right) + 30\left(3x + 1\right){\left(3x - 1\right)}^{2} + 2{\left(3x - 1\right)}^{3}\right)}{2{x}^{6} + 30{x}^{4}\left({x}^{2} - 1\right) + 30{x}^{2}{\left({x}^{2} - 1\right)}^{2} + {\left({x}^{2} - 1\right)}^{3}}$

$= \lim_{x \to \infty}\dfrac{{x}^{6}\left[2{\left(3 + \dfrac{1}{x}\right)}^{3} + 30{\left(3 + \dfrac{1}{x}\right)}^{2}\left(3 - \dfrac{1}{x}\right) + 30\left(3 + \dfrac{1}{x}\right){\left(3 - \dfrac{1}{x}\right)}^{2} + 2{\left(3 - \dfrac{1}{x}\right)}^{3}\right\}{{x}^{6}\left[2 + 30\left(1 - \dfrac{1}{{x}^{2}} + 30{\left(1 - \dfrac{1}{{x}^{2}}\right)}^{2} + 2{\left(1 - \dfrac{1}{{x}^{2}}\right)}^{3}\right]\right\}$

$= \dfrac{2{\left(3\right)}^{3} + 30{\left(3\right)}^{3} + 30{\left(3\right)}^{3} + 2{\left(3\right)}^{3}}{2 + 30 + 30 + 2} = {\left(3\right)}^{3} = 27$

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