If \(\left(1+2 x+3 x^2\right)^{10}\) \(=a_0+a_1 x+a_2 x^2+\ldots+a_{20} x^{20}\)
then \(\frac{a_2}{a_1}\) is equal to

If $x=\sin \left(2{\tan }^{-1}2\right), y=\cos \left(2{\tan }^{-1}3\right), z=\sec \left(3{\tan }^{-1}4\right)$ then

Let \(a\gt1\) and \(0 \lt b \lt 1\). If \(f: \mathbf{R} \rightarrow[0,1]\) is defined by
\(f(x)=\left\{\begin{array}{l}a^x,-\infty \lt x \lt 0 \\ b^x, 0 \leq x \lt \infty\end{array}\right.\) then \(f(x)\) is

The area (in square units) of the triangle formed by the lines \(6 x^2+13 x y+6 y^2=0\) and \(x+2 y+3=0\) is

If \(X\) is a binomial variate with the range \(\{0,1,2,3,4,5,6\}\) and \(P(X=2)=4 P(X=4)\), then the parameter \(p\) of \(X\) is

If \(z=x+i y\) and \(z^2=(i \bar{z})^2\), then

The circumference of a circle is measured as \(56 \mathrm{~cm}\) with an error \(0.02 \mathrm{~cm}\). The percentage error in its area is

Which of the following pair of ions have same paramagnetic moment?

As shown in the figure, two infinitely long straight parallel wires \(P\) and \(Q\) carrying equal currents in opposite directions are arranged parallel to \(\mathrm{Y}\)-axis. If the magnetic field due to wire \(\mathrm{P}\) at the origin ' \(\mathrm{O}\) ' of the co-ordinate system is B, then match the resultant magnetic fields at various points given in column \(A\) with the points given in column \(B\).
\(\begin{array}{ll}
\text{Column A} & \text{Column B} \\
\text{A)} \frac{B}{4} & \text{i)} (0,0) \\
\text{B)} \frac{B}{2} & \text{ii)} (a, 0) \\
\text{C)} \frac{2 \mathrm{~B}}{3} & \text{iii)} (2a, 0) \\
\text{D)} 2 \mathrm{~B} & \text{iv)} (3 \mathrm{a}, 0) \\
\end{array}\)


The correct answer is

If \(\tan \left(\frac{\pi}{4}+\alpha\right)=\tan ^3\left(\frac{\pi}{4}+\beta\right)\), then \(\tan (\alpha+\beta) \cot (\alpha-\beta)=\)

Kepler's second law (law of areas) is nothing but a statement of

If two unbiased six-faced dice are thrown simultaneously until a sum of either $7$ or $11$ occurs, then the probability that $7$ comes before $11$ is

The radius of orbit of an electron and the speed of electron in the ground state of hydrogen atom are \(5.5 \times 10^{-11} \mathrm{~m}\) and \(4 \times 10^6 \mathrm{~ms}^{-1}\) respectively. Then, the orbital period of this electron in the first excited state will be
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