- $0$
- $\frac{\pi}{4}$
- $\frac{\pi}{2}$
- $\pi$
Solution:
Let $\theta$ be the angle between $\vec{\text{a}}$ and $\vec{\text{b}}.$
We know
$\big|\vec{\text{a}}\times\vec{\text{b}}\big|=|\vec{\text{a}}|\big|\vec{\text{b}}\big|\sin\theta$
$\vec{\text{a}}.\vec{\text{b}}=|\vec{\text{a}}|\big|\vec{\text{b}}\big|\cos\theta$
$\Rightarrow\big|\vec{\text{a}}.\vec{\text{b}}\big|=\big||\vec{\text{a}}|\big|\vec{\text{b}}\big|\cos\theta\big|=|\vec{\text{a}}|\big|\vec{\text{b}}\big||\cos\theta|$
Given: $\big|\vec{\text{a}}.\vec{\text{b}}\big|=\big|\vec{\text{a}}\times\vec{\text{b}}\big|$
$\Rightarrow|\vec{\text{a}}|\big|\vec{\text{b}}\big||\cos\theta|=|\vec{\text{a}}|\big|\vec{\text{b}}\big|\sin\theta$
$\Rightarrow|\cos\theta|=\sin\theta$
$\Rightarrow\theta=\frac{\pi}{4}$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$(A)$ $f(x)$ is continuous at $x=-\frac{\pi}{2}$
$(B)$ $f(x)$ is not differentiable at $x=0$
$(C)$ $f(x)$ is differentiable at $x=1$
$(D)$ $ f(x)$ is differentiable at $x=-\frac{3}{2}$
$STATEMENT -1$ : For each real $\mathrm{t}$, there exists a point $\mathrm{c}$ in $[\mathrm{t}, \mathrm{t}+\pi]$ such that $\mathrm{f}^{\prime}(\mathrm{c})=0$. because
$STATEMENT -2$: $f(t)=f(t+2 \pi)$ for each real $t$.