JEE 2023 — Mathematics PYQ

Q: How do I solve JEE 2023 Mathematics questions?

Practice JEE 2023 Mathematics previous year questions with step-by-step solutions on Mathem Solvex. Each question includes a detailed explanation and video solution to help you master the concept quickly.

Q: What is the difficulty level of JEE Mathematics questions?

JEE Mathematics questions range from moderate to advanced difficulty. Consistent practice with official previous year papers and topic-wise drills on Mathem Solvex is the most effective preparation strategy.

Q: Are JEE previous year papers available for free?

Yes. Mathem Solvex provides free access to JEE previous year question papers and topic-wise PDFs. You can download them or attempt them as timed live mock tests directly on the platform.

Question

JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

Evaluate the value of: $lim_{n \to \infty} \frac{1 + 2 - 3 + 4 + 5 - 6 + \dots + ( 3 n - 2 ) + ( 3 n - 1 ) - 3 n}{2 n ^{4} + 4 n + 3 - n ^{4} + 5 n + 4}$

Choose the correct answer:

Explanation

Solution:

1. Numerator ko simplify karna:

Numerator ek pattern follow kar raha hai jahan har 3 terms ka ek group ban raha hai.

\text{Numerator} = [1 + 2 - 3] + [4 + 5 - 6] + \dots + [(3n-2) + (3n-1) - 3n]

Har group ka sum nikalte hain:

Pehla group: $1 + 2 - 3 = 0$
Dusra group: $4 + 5 - 6 = 3$
Teesra group: $7 + 8 - 9 = 6$
$n$ -th group: $(3n-2) + (3n-1) - 3n = 3n - 3$

Yeh ek Arithmetic Progression (AP) ban rahi hai: $0, 3, 6, \dots, 3(n-1)$

\text{Sum of Numerator} = \frac{n}{2} [2(0) + (n-1)3] = \frac{3n(n-1)}{2} = \frac{3n^2 - 3n}{2}

2. Denominator ko simplify karna:

Denominator ko rationalize (parimeyakaran) karne ke liye conjugate se multiply karte hain:

\text{Denominator} = \frac{(\sqrt{2n^4 + 4n + 3} - \sqrt{n^4 + 5n + 4})(\sqrt{2n^4 + 4n + 3} + \sqrt{n^4 + 5n + 4})}{\sqrt{2n^4 + 4n + 3} + \sqrt{n^4 + 5n + 4}}

\text{Denominator} = \frac{(2n^4 + 4n + 3) - (n^4 + 5n + 4)}{\sqrt{2n^4 + 4n + 3} + \sqrt{n^4 + 5n + 4}} = \frac{n^4 - n - 1}{\sqrt{2n^4 + 4n + 3} + \sqrt{n^4 + 5n + 4}}

3. Limit apply karna:

Ab numerator aur denominator ko ek saath rakhte hain:

\lim_{n \to \infty} \frac{(3n^2 - 3n) / 2}{(n^4 - n - 1) / (\sqrt{2n^4 + 4n + 3} + \sqrt{n^4 + 5n + 4})}

\lim_{n \to \infty} \frac{3n^2(1 - 1/n) \times (\sqrt{2n^4 + 4n + 3} + \sqrt{n^4 + 5n + 4})}{2(n^4 - n - 1)}

Highest power $n^2$ ko root ke bahar nikalne par:

\lim_{n \to \infty} \frac{3n^2 \times n^2 (\sqrt{2 + 4/n^3 + 3/n^4} + \sqrt{1 + 5/n^3 + 4/n^4})}{2n^4(1 - 1/n^3 - 1/n^4)}

\frac{3 (\sqrt{2} + \sqrt{1})}{2(1)} = \frac{3(\sqrt{2} + 1)}{2}

JEE 2023 — Mathematics PYQ

Choose the correct answer:

Explanation

Solution:

Correct Option: (1) $\frac{3}{2}(\sqrt{2} + 1)$

Explanation

Solution:

Correct Option: (1) $\frac{3}{2}(\sqrt{2} + 1)$

Explore More