MCQ
Let $\left| {\vec a} \right| = \left| {\vec b} \right| = \left| {\vec a - \vec b} \right| = \,1$ then angle between $\vec a$ and $\vec b$ is
- A$\frac{\pi }{6}$
- ✓$\frac{\pi }{3}$
- C$\frac{\pi }{4}$
- D$\frac{\pi }{2}$
$ \Rightarrow |\vec a{|^2} + |\vec b{|^2} - 2\left| {\vec a} \right|\left| {\vec b} \right|\cos \theta = 1$
$ \Rightarrow \cos \theta=1 / 2 \Rightarrow \cos \theta=1 / 2 $
$\Rightarrow \theta=\pi / 3$
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$I$. $f$ is continuous on the closed interval $[a, b]$
$II.$ $f$ is bounded on the open interval $(a, b)$
$III.$ If $a$ $< a_1< b_1< b$, and $f (a_1)<0< f (b_1)$, then there is $a$ number $c$ such that $a_1 < c < b_1$ and $f (c)=0$