The least value of the product xyz for which | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The determinant is given by $\Delta = x(yz - 1) - 1(z - 1) + 1(1 - y) = xyz - x - z + 1 - 1 + 1 - y = xyz - (x + y + z) + 2$. Given $\Delta \ge 0$

Vedclass · Accepted Answer

$-8$

Vedclass · Answer

(D) The determinant is given by $\Delta = x(yz - 1) - 1(z - 1) + 1(1 - y) = xyz - x - z + 1 - 1 + 1 - y = xyz - (x + y + z) + 2$. Given $\Delta \ge 0$,we have $xyz - (x + y + z) + 2 \ge 0$,which implies $xyz + 2 \ge x + y + z$. By the Arithmetic Mean-Geometric Mean ($AM$-$GM$) inequality,for positive $x, y, z$,$x + y + z \ge 3(xyz)^{1/3}$. However,since $xyz$ can be negative,we consider $t = (xyz)^{1/3}$. The inequality becomes $t^3 + 2 \ge x + y + z$. For the minimum value,we consider the case where $x=y=z=t$. Then $t^3 - 3t + 2 \ge 0$. Factoring the expression,we get $(t - 1)^2(t + 2) \ge 0$. Since $(t - 1)^2 \ge 0$,we must have $t + 2 \ge 0$,so $t \ge -2$. Thus,$xyz = t^3 \ge (-2)^3 = -8$. The least value is $-8$.

The least value of the product $xyz$ for which the determinant $\left| \begin{array}{ccc} x & 1 & 1 \\ 1 & y & 1 \\ 1 & 1 & z \end{array} \right|$ is non-negative,is

Similar Questions

If the area of a triangle is $35$ $sq$ $units$ with vertices $(2,-6), (5,4)$ and $(k, 4)$,then $k$ is:

If $A = \begin{bmatrix} 1 & 0 & 1 \\ 2 & 1 & 0 \\ 3 & 2 & 1 \end{bmatrix}$,then $\det A$ =

If $A_{\lambda} = \begin{bmatrix} \lambda & \lambda - 1 \\ \lambda - 1 & \lambda \end{bmatrix}; \lambda \in N$,then $|A_1| + |A_2| + \dots + |A_{300}|$ is equal to

The area of the triangle whose vertices are $(3,5), (2,2)$ and $(k, 2)$ is $3$ sq. unit. Then,the value of $k$ is . . . . . . .

The roots of the equation $\left| \begin{array}{ccc} 1 & 4 & 20 \\ 1 & -2 & 5 \\ 1 & 2x & 5x^2 \end{array} \right| = 0$ are

Vedclass Test Series

Exam Paper Generator

Online Exam Module