Question

If \(f(x) = cot^{-1}(\frac{3x-x^3}{1-3x^2})\) and \(g(x) = cos^{-1}(\frac{1-x^2}{1+x^2})\), then \(\underset{x\rightarrow a}{lim}\frac{f(x)-f(a)}{g(x)-g(a)}\), \((0 < a < \frac{1}{2})\) is

• A. \(\frac{3}{2}\)
• B. \(\frac{1}{2}\)
• C. \(-\frac{3}{2}\)
• D. \(-\frac{1}{2}\)

Hint:

Differentiate implicitly twice and substitute x = 0, y = 1.

Answer 1

The Correct Option is C

Step 1: Solve for y' in general:
From $y'(e^y + x) = -y$, $y' = -\frac{y}{e^y + x}$.

Step 2: Differentiate the quotient:
At $x = 0$, $y = 1$: $y' = -\frac{1}{e}$. Then $y'' = -\frac{y'(e^y + x) - y(e^y y' + 1)}{(e^y + x)^2}$.
In the numerator, $y'(e^y + x) = -\frac1e\cdot e = -1$ and $y(e^y y' + 1) = 1\cdot(e\cdot(-\frac1e) + 1) = 0$. So the numerator is $-1 - 0 = -1$, and $y'' = -\frac{-1}{e^2} = \frac{1}{e^2}$. Option (B).

Final Answer:
$\left(-\frac1e, \frac{1}{e^2}\right)$. \[ \boxed{\left(-\frac{1}{e},\frac{1}{e^2}\right)} \]