If the function f:[a, b] rightarrow left[-fra | vedclass

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(A) Given $f(x) = \left| \begin{array}{ccc} 1 & 1 & 1 \ 1 & 1+\sin x & 1 \ 1+\cos x & 1 & 1 \end{array} ight|$.Applying $C_2 ightarrow C_2 - C_1

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$a = -\frac{\pi}{4}, b = \frac{\pi}{6}$

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(A) Given $f(x) = \left| \begin{array}{ccc} 1 & 1 & 1 \ 1 & 1+\sin x & 1 \ 1+\cos x & 1 & 1 \end{array} ight|$.Applying $C_2 ightarrow C_2 - C_1$ and $C_3 ightarrow C_3 - C_1$:$f(x) = \left| \begin{array}{ccc} 1 & 0 & 0 \ 1 & \sin x & 0 \ 1+\cos x & -\cos x & -\cos x \end{array} ight|$.Expanding along the first row: $f(x) = 1 \cdot (\sin x \cdot (-\cos x) - 0) = -\sin x \cos x = -\frac{1}{2} \sin 2x$.For $f$ to be one-one and onto, the range of $f(x)$ must be $\left[ -\frac{\sqrt{3}}{4}, \frac{1}{2} ight]$.So, $-\frac{\sqrt{3}}{4} \leq -\frac{1}{2} \sin 2x \leq \frac{1}{2}$.Multiplying by $-2$: $-1 \leq \sin 2x \leq \frac{\sqrt{3}}{2}$.This implies $2x \in \left[ -\frac{\pi}{2}, \frac{\pi}{3} ight]$.Dividing by $2$: $x \in \left[ -\frac{\pi}{4}, \frac{\pi}{6} ight]$.Thus, $a = -\frac{\pi}{4}$ and $b = \frac{\pi}{6}$.

If the function $f:[a, b] \rightarrow \left[-\frac{\sqrt{3}}{4}, \frac{1}{2}\right]$ defined by $f(x) = \left| \begin{array}{ccc} 1 & 1 & 1 \\ 1 & 1+\sin x & 1 \\ 1+\cos x & 1 & 1 \end{array} \right|$ is one-one and onto, then:

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