The solution of the equation \(\left(x-4 y^3\right) \frac{d y}{d x}-y=0,(y>0)\) is

\(\int_0^{\pi / 2} \log _e(\sin 2 x) d x\)

Choose the correct option about the matrices given below
\(\begin{aligned} & A=\left[\begin{array}{ccc}\cos \frac{\pi}{4} & \sin \frac{\pi}{4} & 0 \\ -\sin \frac{\pi}{4} & \cos \frac{\pi}{4} & 0 \\ 0 & 0 & 1\end{array}\right] \\ & B=\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & \cos \frac{\pi}{3} & \sin \frac{\pi}{3} \\ 0 & -\sin \frac{\pi}{3} & \cos \frac{\pi}{3}\end{array}\right] \\ & C=\left[\begin{array}{ccc}\cos \frac{\pi}{6} & 0 & \sin \frac{\pi}{6} \\ 0 & 1 & 0 \\ -\sin \frac{\pi}{6} & \cos \frac{\pi}{6} & 0\end{array}\right]\end{aligned}\)
\(D=\left[\begin{array}{ccc}
\cos \frac{\pi}{2} & \sin \frac{\pi}{2} & 0 \\
-\sin \frac{\pi}{2} & \cos \frac{\pi}{2} & 0 \\
0 & 0 & 1
\end{array}\right]\)

The combined equation of the direct common tangents of the circles \(x^2+y^2-2 x-2 y-2=0\) and \(x^2+y^2+4 x+6 y+12=0\), is

If \(\vec{a}\) and \(\vec{b}\) are two vectors such that \(|\vec{a}|=|\vec{b}|=\sqrt{14}\) and \(\vec{a} \cdot \vec{b}=-7\), then \(\frac{|\vec{a} \times \vec{b}|}{|\vec{a} \cdot \vec{b}|}=\)

Match the items of List - I with those of List - II
\(\begin{array}{ll}
\text{List - I (Complex number)} & \text{List - II (Polar form)} \\
\text{(i)} \sqrt{3}-i & \text{(a)} 2 \operatorname{cis} \frac{\pi}{6} \\
\text{(ii)} \sqrt{3}+i & \text{(b)} 2 \operatorname{cis} \frac{5 \pi}{6} \\
\text{(iii)} -\sqrt{3}+i & \text{(c)} 2 \operatorname{cis}\left(\frac{-5 \pi}{6}\right) \\
\text{(iv)} -\sqrt{3}-i & \text{(d)} 2 \operatorname{cis}\left(-\frac{\pi}{6}\right) \\
& \text{(e)} 2 \operatorname{cis} \frac{9 \pi}{6} \\
\end{array}\)
The correct matching is

When the coordinate axes are rotated through an angle \(\frac{\pi}{4}\) in the positive direction, an equation is transformed to \(x^2+y^2-6 x+8 y+21=0\). Then that original equation is

\(\int \frac{\cos 2 x \cdot \sin 4 x}{\cos ^4 x\left(1+\cos ^2 2 x\right)} d x=\)

If \(\theta\) is an acute angle, \(\cos h x=K\) and \(\sin h x=\tan \theta\), then \(\sin \theta=\)

If \(\mathbf{a}\) and \(\mathbf{b}\) are two unit vectors and \(\theta\) is the angle between them, then the unit vector along the angular bisector of \(\mathbf{a}\) and \(\mathbf{b}\) is given by

When \(\mathrm{CaCl}_2\) is added to \(\mathrm{AgCl}\) crystal, the defect introduced is

In Ellingham diagram, the plot is drawn between

If \(u+i v=\frac{3 i}{x+i y+2}\), then \(y=\)