MCQ
If $|\vec{a}|=2,|\vec{b}|=5$ and $|\vec{a} \times \vec{b}|=8$, then $|\vec{a} \cdot \vec{b}|$ is equal to :
- A$5$
- B$4$
- ✓$6$
- D$3$
$|\vec{a} \times \vec{b}|=|\vec{a}||\vec{b}| \sin \theta=\pm 8$
$\sin \theta=\pm \frac{4}{5}$
$\therefore \vec{a} \cdot \vec{b}=|\vec{a}| \vec{b} \mid \cos \theta$
$=10 \cdot\left(\pm \frac{3}{5}\right)=\pm 6$
$|\vec{a} \cdot \vec{b}|=6$
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$D^*f(x) =\mathop {Limit}\limits_{h \to 0} \frac{{{f^2}(x + h) - {f^2}(x)}}{h}$ where $f^2(x)$ means $[f(x)]^2.$ If $f(x) = x lnx$ then
${\left. {D^*f(x)} \right|_{x = e}}$ has the value
$f(x)=\frac{\cos ^{-1}\left(\frac{x^{2}-5 x+6}{x^{2}-9}\right)}{\log _{e}\left(x^{2}-3 x+2\right)} \text { is }$