Let f(theta) = left| begin{array}{ccc} 1 & co | vedclass

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

Question

(B) To find $f(	heta)$,we expand the determinant along the first row: $f(	heta) = 1(1 - (-\sin 	heta \cdot -\cos 	heta)) - \cos 	heta(-\sin 	het

Vedclass · Accepted Answer

$(2, 0)$

Vedclass · Answer

(B) To find $f(	heta)$,we expand the determinant along the first row: $f(	heta) = 1(1 - (-\sin 	heta \cdot -\cos 	heta)) - \cos 	heta(-\sin 	heta - 1) - 1(-\sin^2 	heta + 1)$ $f(	heta) = 1(1 - \sin 	heta \cos 	heta) + \sin 	heta \cos 	heta + \cos 	heta + \sin^2 	heta - 1$ $f(	heta) = 1 - \sin 	heta \cos 	heta + \sin 	heta \cos 	heta + \cos 	heta + \sin^2 	heta - 1$ $f(	heta) = \sin^2 	heta + \cos 	heta$ Wait,let us re-evaluate the expansion: $f(	heta) = 1(1 + \sin 	heta \cos 	heta) - \cos 	heta(-\sin 	heta - \cos 	heta) - 1(-\sin^2 	heta + 1)$ $f(	heta) = 1 + \sin 	heta \cos 	heta + \sin 	heta \cos 	heta + \cos^2 	heta + \sin^2 	heta - 1$ $f(	heta) = 2 \sin 	heta \cos 	heta + (\sin^2 	heta + \cos^2 	heta) = \sin 2	heta + 1$ Since $-1 \le \sin 2	heta \le 1$,the range of $f(	heta)$ is $[1-1, 1+1] = [0, 2]$. Thus,the maximum value $A = 2$ and the minimum value $B = 0$. Therefore,$(A, B) = (2, 0)$.

Let $f(\theta) = \left| \begin{array}{ccc} 1 & \cos \theta & -1 \\ -\sin \theta & 1 & -\cos \theta \\ -1 & \sin \theta & 1 \end{array} \right|$. Suppose $A$ and $B$ are respectively the maximum and minimum values of $f(\theta)$. Then $(A, B)$ is equal to

Similar Questions

Let $\omega = -\frac{1}{2} + i \frac{\sqrt{3}}{2}$,where $i = \sqrt{-1}$. Then the value of $\left| \begin{array}{ccc} 1 & 1 & 1 \\ 1 & -1-\omega^2 & \omega^2 \\ 1 & \omega^2 & \omega^4 \end{array} \right|$ is

The number of real values of $t$ such that the system of homogeneous equations
$\begin{aligned}
t x+(t+1) y+(t-1) z &=0 \\
(t+1) x+t y+(t+2) z &=0 \\
(t-1) x+(t+2) y+t z &=0
\end{aligned}$
has non-trivial solutions is

If $\alpha, \beta, \gamma$ $(\alpha < \beta < \gamma)$ are the values of $x$ such that $\begin{vmatrix} x-2 & 0 & 1 \\ 1 & x+3 & 2 \\ 2 & 0 & 2x-1 \end{vmatrix} = 0$ is a singular matrix,then $2\alpha + 3\beta + 4\gamma = $

If $A = \begin{bmatrix} 2-k & 2 \\ 1 & 3-k \end{bmatrix}$ is a singular matrix,then the value of $5k - k^2$ is equal to

If $(x_{1}, y_{1}), (x_{2}, y_{2})$ and $(x_{3}, y_{3})$ are the vertices of a triangle whose area is $k$ square units,then $\left|\begin{array}{ccc}x_{1} & y_{1} & 4 \\ x_{2} & y_{2} & 4 \\ x_{3} & y_{3} & 4\end{array}\right|^{2}$ is (in $k^{2}$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module