If the interval in which the real-valued function \[ f(x) = \log\left(\frac{1+x}{1-x}\right) - 2x - \frac{x^{3}}{1-x^{2}} \] is decreasing in \( (a,b) \), where \( |b-a| \) is maximum, then {a}⁄{b} =
Find \( \frac{dy}{dx} \) for the given function:
\[ y = \tan^{-1} \left( \frac{\sin^3(2x) - 3x^2 \sin(2x)}{3x \sin(2x) - x^3} \right). \]
If \( m \) and \( M \) are respectively the absolute minimum and absolute maximum values of a function \( f(x) = 2x^3 + 9x^2 + 12x + 1 \) defined on \([-3,0]\), then \( m + M \) is:
\[ y = \sqrt{\sin(\log(2x)) + \sqrt{\sin(\log(2x)) + \sqrt{\sin(\log(2x))} + \dots \infty}} \]
Evaluate the integral: \[ \int \frac{\sec x}{3(\sec x + \tan x) + 2} \,dx \]
Evaluate the integral: \[ I = \int_{-\frac{\pi}{15}}^{\frac{\pi}{15}} \frac{\cos 5x}{1 + e^{5x}} \, dx \]
The area of the region (in square units) enclosed by the curves \( y = 8x^3 - 1 \), \( y = 0 \), \( x = -1 \), and \( x = 1 \) is:
The general solution of the differential equation: \[ (6x^2 - 2xy - 18x + 3y) dx - (x^2 - 3x) dy = 0 \]
If the function
$ f(x) = \begin{cases} \frac{\cos ax - \cos 9x}{x^2}, & \text{if } x \neq 0 \\ 16, & \text{if } x = 0 \end{cases} $
is continuous at $ x = 0 $, then $ a = ? $
If
\[ A = \{ P(\alpha, \beta) \mid \text{the tangent drawn at P to the curve } y^3 - 3xy + 2 = 0 \text{ is a horizontal line} \} \]
and
\[ B = \{ Q(a, b) \mid \text{the tangent drawn at Q to the curve } y^3 - 3xy + 2 = 0 \text{ is a vertical line} \} \]
then \( n(A) + n(B) = \)
$ \lim_{x \to -\frac{3}{2}} \frac{(4x^2 - 6x)(4x^2 + 6x + 9)}{\sqrt{2x - \sqrt{3}}} $
\[ f(x) = \begin{cases} \frac{(4^x - 1)^4 \cot(x \log 4)}{\sin(x \log 4) \log(1 + x^2 \log 4)}, & \text{if } x \neq 0 \\ k, & \text{if } x = 0 \end{cases} \]
Find \( e^k \) if \( f(x) \) is continuous at \( x = 0 \).