Question

If \(y^2 + \log_e(\cos^2 x) = y\), \(x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\), then :

• A. \(|y''(0)| = 2\)
• B. \(|y'(0)| + |y''(0)| = 3\)
• C. \(y''(0) = 0\)
• D. \(|y'(0)| + |y''(0)| = 1\)

Hint:

At \(x = 0\): \(y^2 = y\), so \(y(0) = 0\) or 1. Differentiate implicitly twice and evaluate at 0 for each case.

Answer 1

The Correct Option is A

At \(x = 0\): \(y^2 + 0 = y\), so \(y(0) = 0\) or \(1\).
Differentiate: \(2yy' - 2\tan x = y'\), i.e. \(y'(2y - 1) = 2\tan x\). At \(x = 0\): \(y'(0)(2y - 1) = 0\), and \(2y - 1 = \pm 1 \ne 0\), so \(y'(0) = 0\).
Differentiate again: \(y''(2y - 1) + 2y'^2 = 2\sec^2 x\). At \(x = 0\): \(y''(0)(2y(0) - 1) = 2\).
If \(y(0) = 1\): \(y''(0) = 2\). If \(y(0) = 0\): \(y''(0) = -2\).
Either way \(|y'(0)| = 0\) and \(|y''(0)| = 2\).
So \(|y''(0)| = 2\) is true; \(|y'(0)| + |y''(0)| = 2\), so options (2) and (4) are false, and (3) is false.
Answer: (1) \(|y''(0)| = 2\)