Solution:
If there is no point in the feasible set, there is no feasible solution of the linear programming model.
In linear programming, lack of points for a solution set is said to have no feasible solution.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$STATEMENT -1$ : 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 $t$.
| $X$ | $1$ | $2$ | $3$ | $4$ | $5$ |
| $P(X)$ | $K^2$ | $2K$ | $K$ | $2K$ | $5K^2$ |
Then $\mathrm{P}(\mathrm{X}> 2)$ is equal to
| column-$I$ | column-$II$ |
| $(A)$ A line from the origin meets the lines $\frac{x-2}{1}=\frac{y-1}{-2}=\frac{z+1}{1}$ and $\frac{x-\frac{8}{3}}{2}=\frac{y+3}{-1}=\frac{z-1}{1}$ at $P$ and $Q$ respectively. If length $P Q=d$, then $d^2$ is | $(p)$ $-4$ |
| $(B)$ The values of $x$ satisfying $\tan ^{-1}(x+3)-\tan ^{-1}(x-3)=\sin ^{-1}\left(\frac{3}{5}\right)$ are | $(q)$ $0$ |
| $(C)$ Non-zero vectors $\vec{a}, \vec{b}$ and $\overrightarrow{\mathrm{c}}$ satisfy $\vec{a} \cdot \vec{b}=0$, $(\vec{b}-\vec{a}) \cdot(\vec{b}+\vec{c})=0$ and $2|\vec{b}+\vec{c}|=|\vec{b}-\vec{a}|$. If $\vec{a}=\mu \vec{b}+4 \vec{c}$, then the possible values of $\mu$ are | $(r)$ $4$ |
| $(D)$ Let $f$ be the function on $[-\pi, \pi]$ given by $f(0)=9$ and $f(x)=$ $\sin \left(\frac{9 x}{2}\right) / \sin \left(\frac{x}{2}\right)$ for $x \neq 0$. The value of $\frac{2}{\pi} \int_{-\pi}^\pi f(x) d x$ is | $(s)$ $5$ |
| $(t)$ $6$ |