Differential Equations
Differential Equations
star_batch_jee_advanced_2025
Grade 12
Question:
Tangent is drawn at the point $(x_i, y_i)$ on the curve $y = f(x)$, which intersects the x-axis at $(x_{i+1}, 0)$. Now, again a tangent is drawn at $(x_{i+1}, y_{i+1})$ on the curve which intersects the x-axis at $(x_{i+2}, 0)$ and the process is repeated $n$ times, i.e., $i = 1, 2, 3, ........., n$. If $x_1, x_2, x_3, ........., x_n$ form an arithmetic progression with common difference equal to $\log_2 e$ and curve passes through $(0, 2)$. Now if curve passes through the point $(-2, k)$, then the value of $k$ is ______.
Step-by-Step Solution
Key Concept: Separation of variables in differential equations combined with initial conditions to determine constants.
Separate the integral into two parts: $\int \frac{x}{\sqrt{x^2-1}}dx + \int \frac{1}{x\sqrt{x^2-1}}dx$. The first integral equals $\sqrt{x^2-1}$ and the second equals $-\sec^{-1}x$, giving $\sqrt{y^2-1} = -\sqrt{x^2-1} + \sec^{-1}x + c$. Using the initial condition that the curve passes through $(1,1)$: $0 = 0 + 0 + c$, so $c=0$. Therefore $\sqrt{k^2-1} = -1 + \frac{\pi}{4}$, which gives $k^2 - 1 = 1 + \frac{\pi^2}{16} - \frac{\pi}{2}$, so $|k| = \sqrt{1.046} = 1$.
Correct Answer: 8