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

Let $k$ and $m$ be positive real numbers such that the function $f (x) = {3 x^{2} + k x + 1, m x^{2} + k^{2}, am p; 0 am p; x \geq 1 l t; x l t; 1$ is differentiable for all $x > 0$ . Then $\frac{8 f ^{'} ( 8 )}{f ^{'} ( \frac{1}{8} )}$ is equal to ________.

Choose the correct answer:

Explanation

f(x) = \begin{cases} 3x^2 + k\sqrt{x+1}, & 0 < x < 1 \\ mx^2 + k^2, & x \geq 1 \end{cases}

\text{At } x=1 \text{ (Continuity): } f(1^-) = f(1^+) \Rightarrow 3 + k\sqrt{2} = m + k^2 \quad \dots(i)

\text{At } x=1 \text{ (Differentiability): } f'(1^-) = f'(1^+) \Rightarrow 6(1) + \frac{k}{2\sqrt{1+1}} = 2m(1)

\Rightarrow 6 + \frac{k}{2\sqrt{2}} = 2m \Rightarrow m = 3 + \frac{k}{4\sqrt{2}} \quad \dots(ii)

\text{Substitute (ii) in (i): } 3 + k\sqrt{2} = 3 + \frac{k}{4\sqrt{2}} + k^2

\Rightarrow k^2 + k \left[ \frac{1}{4\sqrt{2}} - \sqrt{2} \right] = 0

\Rightarrow k^2 + k \left[ \frac{1}{4\sqrt{2}} - \sqrt{2} \right] = 0

Choose the correct answer:

Explore More