If $\Delta = \begin{vmatrix} 1 & \cos \theta & 1 \\ -\cos \theta & 1 & \cos \theta \\ -1 & -\cos \theta & 1 \end{vmatrix}$,then $\Delta$ lies in the interval

  • A
    $[2, 4]$
  • B
    $(2, 4)$
  • C
    $[1, 4]$
  • D
    $[-1, 1]$

Explore More

Similar Questions

If ${a^{ - 1}} + {b^{ - 1}} + {c^{ - 1}} = 0$ such that $\left| {\begin{array}{*{20}{c}}{1 + a}&1&1\\1&{1 + b}&1\\1&1&{1 + c}\end{array}} \right| = \lambda $,then the value of $\lambda $ is

Difficult
View Solution

If the system of equations
$(k+1)^3 x + (k+2)^3 y = (k+3)^3$
$(k+1) x + (k+2) y = k+3$
$x + y = 1$
is consistent,then the value of $k$ is

The area of $\triangle PQR$ with the vertices $P(k, 1)$,$Q(2, 4)$,and $R(1, 1)$ is $3$ sq. units. Then,$k = $ . . . . . . .

The roots of the determinant equation (in $x$) $\left| \begin{array}{ccc} a & a & x \\ m & m & m \\ b & x & b \end{array} \right| = 0$ are:

If $px^4 + qx^3 + rx^2 + sx + t \equiv \left| \begin{array}{ccc} x^2 + 3x & x - 1 & x + 3 \\ x + 1 & 2 - x & x - 3 \\ x - 3 & x + 4 & 3x \end{array} \right|$,then $t =$

Difficult
View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo