MCQ
Let $\left| {\,\begin{array}{*{20}{c}}{6i}&{ - 3i}&1\\4&{3i}&{ - 1}\\{20}&3&i\end{array}\,} \right| = x + iy$, then
- A$x = 3,y = 1$
- ✓$x = 0,y = 0$
- C$x = 0,y = 3$
- D$x = 1,y = 3$
$ \Rightarrow $ $6i( - 3 + 3) + 3i(4i + 20) + 1(12 - 60i) = x + iy$
$ \Rightarrow $ $(0+60i-12+12-60i)$ $ \Rightarrow $ $x = 0,\,y = 0$.
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$f(x)\left\{ \begin{gathered} = 1\,,\,{\text{if}}\,\,\,x > 0 \hfill \\ = - 1\,,\,{\text{if}}\,\,\,x < 0 \hfill \\ = 0\,,\,{\text{if}}\,\,\,x = 0 \hfill \\ \end{gathered} \right.$ then ${\left. {\frac{{dy}}{{dx}}} \right|_{x = \frac{{5\pi }}{4}}}$ is