MCQ
If $|a \times b|\, = 4$ and $|a\,.\,b|\, = 2$, then $|a{|^2}\,\,|b{|^2} = $
- A$2$
- B$6$
- C$8$
- ✓$20$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
If $f(x) = {\rm{ }}\left\{ {\begin{array}{*{20}{c}}
{\frac{{\left( {{e^x} - 1} \right)^2}}{{\sin {\mkern 1mu} \left( {\frac{x}{k}} \right){\mkern 1mu} \log {\mkern 1mu} \left( {1 + \frac{x}{4}} \right)}}{\mkern 1mu} ,{\mkern 1mu} x \ne 0}\\
{{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} 12{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} ,x{\mkern 1mu} {\mkern 1mu} = 0{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} }
\end{array}} \right.$
is a continuous function then the value of $k$ is
$\left(1+\cos ^{2} \theta\right) x+\sin ^{2} \theta y+4 \sin 3 \theta z=0$
$\cos ^{2} \theta x+\left(1+\sin ^{2} \theta\right) y+4 \sin 3 \theta z=0$
$\cos ^{2} \theta x+\sin ^{2} \theta y+(1+4 \sin 3 \theta) z=0$
has a non-trivial solution, then the value of $\theta$ is :
$f(x)=\sin x+3 x-\frac{2}{\pi}\left(x^2+x\right) \text {, where } x \in\left[0, \frac{\pi}{2}\right] \text {, }$
consider the following two statements :
($I$) $\mathrm{f}$ is increasing in $\left(0, \frac{\pi}{2}\right)$.
($II$) $\mathrm{f}^{\prime}$ is decreasing in $\left(0, \frac{\pi}{2}\right)$.
Between the above two statements,
$\overrightarrow{ a }=\hat{ i }+\hat{ j }+ n \hat{ k }, \quad \overrightarrow{ b }=2 \hat{ i }+4 \hat{ j }- n \hat{ k } \quad$ and $\overrightarrow{ c }=\hat{ i }+ n \hat{ j }+3 \hat{ k } \quad( n \geq 0),$ is $158 cu. Units$, then