- A${{177} \over {11}}$
- B$ - {{177} \over {11}}$
- ✓$ - {{23} \over {11}}$
- D${{23} \over {11}}$
$f'(x) = - 20 + 22x$
Put $f'(x) = 0$ $i.e.,$ $ - 20 + 22x = 0$
==> $x = 10/11$ and $f''(x) = 22 > 0$
Hence at $x = 10/11,\;\;\;f(x)$ will have minimum value,
$\therefore f\,\left( {\frac{{10}}{{11}}} \right) = 7 - \frac{{200}}{{11}} + \frac{{100 \times 11}}{{121}}$$ = 7 - \frac{{200}}{{11}} + \frac{{100}}{{11}}$
$ = - \frac{{23}}{{11}}$.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
${{b}_{1}}=b-\frac{b.a}{|a{{|}^{2}}}a,\,{{b}_{2}}=b+\frac{b.a}{|a{{|}^{2}}}a$ , ${{b}_{1}}=b-\frac{b.a}{|a{{|}^{2}}}a,\,{{b}_{2}}=b+\frac{b.a}{|a{{|}^{2}}}a$
, ${{c}_{2}}=c-\frac{c.a}{|a{{|}^{2}}}a-\frac{c.{{b}_{1}}}{|{{b}_{1}}{{|}^{2}}}{{b}_{1}}$
,${{c}_{3}}=c-\frac{c.a}{|a{{|}^{2}}}a-\frac{c.{{b}_{2}}}{|{{b}_{2}}{{|}^{2}}}{{b}_{2}}$,
${{c}_{4}}=a-\frac{c.a}{|a{{|}^{2}}}a$.
Then which of the following is a set of mutually orthogonal vectors is
$2 x+y \leq 10, x+3 y \leq 15, x, y \geq 0$ are $(0,0),(5,0),(3,4)$ and $(0,5) .$ Let $Z =p x+q y,$ where $p, q\,>\,0 .$ Condition on $p$ and $q$ so that the maximum of $Z$ occurs at both $(3,4)$ and $(0,5)$ is $....$