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JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

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

Choose the correct answer:

Explanation

Detailed Solution

Step 1: Binomial Expansion ka upyog

Humein pata hai ki:

$(a+b)^n + (a-b)^n = 2[ ^nC_0 a^n + ^nC_2 a^{n-2}b^2 + ^nC_4 a^{n-4}b^4 + \dots ]$

Yahan $n=6$ hai, toh expansion hoga:

$2[a^6 + 15a^4b^2 + 15a^2b^4 + b^6]$

Step 2: Numerator ko simplify karein

Let $a = \sqrt{3x+1}$ aur $b = \sqrt{3x-1}$ .

Numerator ( $N$ ) ho jayega:

$N = 2[(\sqrt{3x+1})^6 + 15(\sqrt{3x+1})^4(\sqrt{3x-1})^2 + 15(\sqrt{3x+1})^2(\sqrt{3x-1})^4 + (\sqrt{3x-1})^6]$

$N = 2[(3x+1)^3 + 15(3x+1)^2(3x-1) + 15(3x+1)(3x-1)^2 + (3x-1)^3]$

Jab $x \to \infty$ hota hai, toh hum sirf highest power of $x$ (leading term) par dhyan dete hain:

$(3x+1)^3 \approx (3x)^3 = 27x^3$

$15(3x+1)^2(3x-1) \approx 15(3x)^2(3x) = 15(9x^2)(3x) = 405x^3$

$15(3x+1)(3x-1)^2 \approx 15(3x)(3x)^2 = 405x^3$

$(3x-1)^3 \approx (3x)^3 = 27x^3$

Numerator ki leading term: $2[27 + 405 + 405 + 27]x^3 = 2[864]x^3 = 1728x^3$ .

Step 3: Denominator ko simplify karein

Let $c = x$ aur $d = \sqrt{x^2-1}$ .

Denominator ( $D$ ) ho jayega:

$D = 2[x^6 + 15x^4(\sqrt{x^2-1})^2 + 15x^2(\sqrt{x^2-1})^4 + (\sqrt{x^2-1})^6]$

$D = 2[x^6 + 15x^4(x^2-1) + 15x^2(x^2-1)^2 + (x^2-1)^3]$

Leading terms (highest power $x^6$ ):

$x^6$

$15x^4(x^2) = 15x^6$

$15x^2(x^2)^2 = 15x^6$

$(x^2)^3 = x^6$

Denominator ki leading term: $2[1 + 15 + 15 + 1]x^6 = 2[32]x^6 = 64x^6$ .

Step 4: Limit find karein

Ab limit ki value nikalte hain:

$L = \lim_{x \to \infty} \frac{1728x^3}{64x^6} \cdot x^3$

$L = \lim_{x \to \infty} \frac{1728x^6}{64x^6}$

$L = \frac{1728}{64}$

Divide karne par:

$1728 / 64 = 27$

Final Answer:

The correct option is (2) is equal to 27.

Choose the correct answer:

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