The value of \(\lim_{x \to 0} \frac{6\sin(x) - 2\sin(3x)}{3x^3}\) is equal to
\(\int \frac{1}{(1 - \cos x)}(1 + \sec x) dx = \)
\(\int \frac{10e^x}{(2e^x + 5)^3} dx\) is equal to
Let \(f(x) = \cos(5x)\cos(3x) - \sin(5x)\sin(3x), 0 \leq x \leq \frac{\pi}{4}\). Then \(f\) attains its minimum at \(x =\)
The point \(P\) lies on the curve with equation \[ y = x\sqrt{2\log_e(x)}, \qquad x>0. \] If \[ \frac{dy}{dx}=2 \] at \(x=k\), then the value of \(k\) is equal to:
If \[ f(x)= \begin{cases} x^2, & \text{if } x\leq 2,\\[4pt] 4x-\alpha, & \text{if } x>2, \end{cases} \] is continuous at \(x=2\), then the value of \(\alpha\) is equal to: