MCQ
If $a,b,c$ be positive and not all equal, then the value of the determinant $\left| {\,\begin{array}{*{20}{c}}a&b&c\\b&c&a\\c&a&b\end{array}\,} \right|$ is
- ✓$- ve$
- B$=+ ve$
- CDepends on $a,b,c$
- DNone of these
= $ - (a + b + c)\,({a^2} + {b^2} + {c^2} - ab - bc - ca)$
$ = - \frac{1}{2}(a + b + c)\,[{(a - b)^2} + {(b - c)^2} + {(c - a)^2}]$,
which is clearly negative because of the given conditions.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
| List-$I$ | List-$II$ |
| ($I$) Probability of $\left(X_2 \geq Y_2\right)$ is | ($P$) $\frac{3}{8}$ |
| ($II$) Probability of $\left(X_2>Y_2\right)$ is | ($Q$) $\frac{11}{16}$ |
| ($III$) Probability of $\left(X_3=Y_3\right)$ is | ($R$) $\frac{5}{16}$ |
| ($IV$) Probability of $\left(X_3>Y_3\right)$ is | ($S$) $\frac{355}{864}$ |
| ($T$) $\frac{77}{432}$ |
The correct option is:
The corner points of the feasible region determined by the following system of linear inequalities:
2x + y ≤ 10, x + 3y ≤ 15, x, y ≥ 0 are (0, 0), (5, 0), (3, 4) and (0, 5).
Let Z = px + qy, where p.q > 0.
Condition on p and q so that the maximum of Z occurs at both (3, 4) and (0, 5) is: