WBJEE 2025 Math Paper

To find the maximum and minimum values of $f(θ)$, we first need to calculate the determinant:$f(θ)=\begin{vmatrix} 1 & \cos\theta & -1 \\ -\sin\theta & 1 & -\cos\theta \\ -1 & \sin\theta & 1 \end{vmatrix}$ We can expand the determinant along the first row: Now, calculate each $2 \times 2$ determinant: Substitute these back into the expression for $f(θ)$ : $f(θ)=1+2 \sin\theta \cos\theta$ We know the trigonometric identity $2 \sin\theta \cos\theta = \sin(2\theta)$.So, $f(θ)=1+ \sin(2\theta)$.Now we need to find the maximum and minimum values of $f(θ)$.The range of the sine function $\sin(x)$ is $[-1,1]$. Therefore, the range of $\sin(2\theta)$ is als $[-1,1]$. The minimum value of $f(θ)$ occurs when $\sin(2\theta)$ is at its minimum, which is -1 .Minimum value $B=1+(-1)=0$.The maximum value of $f(θ)$ occurs when $\sin(2\theta)$ is at its maximum, which is 1 .Maximum value $A=1+(1)=2$. So, $A=2$ and $B=0$. The pair $(A, B)$ is $(2,0)$.The calculated pair $(2,0)$ matches Option B.The final answer is $(2,0)$