If \(x^a y^b=e^m, x^c y^d=e^n, \Delta_1=\left|\begin{array}{ll}m & b \\ n & d\end{array}\right|, \Delta_2=\left|\begin{array}{ll}a & m \\ c & n\end{array}\right|, \Delta_3=\left|\begin{array}{ll}a & b \\ c & d\end{array}\right|\), then the values of \(x\) and \(y\) are respectively ( \(e\) is the base of natural logarithm)

If the roots of the equation \(x^3-13 x^2+\mathrm{K} x-27=0\) are in geometric progression then \(\mathrm{K}=\)

If the increase in the side of square is \(6 \%\), then the approximate percentage increase in its area ......

The area (in square units) bounded by the curves \(y=2 x^2\) and \(y=\max \{x-[x], x+|x|\}\) in between the lines \(x=0\) and \(x=2\) is

\(\lim _{x \rightarrow-a} \frac{x^7+a^7}{x+a}=7 \Rightarrow a=\)

The mean and standard deviation of a binomial variate \(X\) are 4 and \(\sqrt{3}\) respectively. Then \(P(X \geq 1)\) is equal to

If the median of the data \(6,7, x-2, x, 18\) and 21 written in ascending order is 16 , then the variance of that data is

Observe the following reactions
I. \(\mathrm{N}_2(\mathrm{~g})+3 \mathrm{H}_2(\mathrm{~g}) \xrightarrow[\substack{773 \mathrm{~K} \\ 200 \mathrm{~atm}}]{\mathrm{X}} 2 \mathrm{NH}_3(\mathrm{~g})\)
II. \(\mathrm{CO}(\mathrm{g})+\mathrm{H}_2 \mathrm{O}(\mathrm{g}) \frac{\mathrm{Y}}{673 \mathrm{~K}} \rightarrow \mathrm{CO}_2(\mathrm{~g})+\mathrm{H}_2(\mathrm{~g})\)
III. \(\mathrm{CH}_4(\mathrm{~g})+\mathrm{H}_2 \mathrm{O}(\mathrm{g}) \xrightarrow[1270 \mathrm{~K}]{\mathrm{Z}} \mathrm{CO}(\mathrm{g})+3 \mathrm{H}_2(\mathrm{~g})\)
Catalysts X, Y, Z respectively are

If \(\mathbf{a}, \mathbf{b}, \mathbf{c}\) and \(\mathbf{d}\) are vectors in which \(|\mathbf{d}|=1\) and given \(\mathbf{a}+\mathbf{b}+\mathbf{c}=s \mathbf{d}, \mathbf{b}+\mathbf{c}+\mathbf{d}=\mathbf{a}\), \(\mathbf{a} \cdot \mathbf{d}=4\), then \(s\) is equal to

Match the following List I with List II.
\(\begin{array}{|c|c|c|}
\hline & \text { List I } & \text { List II } \\
\hline \text { (A) } & \text { Transverse wave (i) } & \begin{array}{l}
\text {Vibrations parallel to the } \\
\text {direction of propagation }
\end{array} \\
\hline \text { (B) } & \begin{array}{l}
\text {Longitudinal wave }
\end{array} & \begin{array}{l}
\text {Vibrations perpendicular to } \\
\text {the direction of propagation }
\end{array} \\
\hline \text { (C) } & \text { Beats } & \begin{array}{l}
\text {Superposition of waves } \\
\text {travelling in the opposite directions }
\end{array} \\
\hline \text { (D) } & \text { Stationary waves (iv) } & \begin{array}{l}
\text {Superposition of waves } \\
\text {travelling in same direction }
\end{array} \\
\hline
\end{array}\)
The correct answer is

The value of \(\sum_{n=0}^{\infty}\left(\frac{2 i}{3}\right)^n\) is

If \(\frac{d}{d x}\left(A \log \left(\frac{\sqrt{1-x^3}+B}{\sqrt{1-x^3}+1}\right)\right)=\frac{1}{x \sqrt{1-x^3}}\), then \(A B=\)

In Solvay process, \(\mathrm{NH}_3\) is recovered when the solution containing \(\mathrm{NH}_4 \mathrm{Cl}\) is treated with compound ' X '. What is ' X ' ?