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:
- A$a = -\frac{\pi}{4}, b = \frac{\pi}{6}$
- B$a = -\frac{\pi}{2}, b = \frac{\pi}{2}$
- C$a = -\frac{\pi}{6}, b = \frac{\pi}{4}$
- D$a = -\pi, b = \pi$
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Let $A$ be a $2 \times 2$ matrix with real entries. Let $I$ be the $2 \times 2$ identity matrix. $\operatorname{Tr}(A)$ denotes the sum of diagonal entries of $A$. Assume that $A^2=I$.
Statement $I$: If $A \neq I$ and $A \neq -I$,then $\operatorname{det}(A) = -1$.
Statement $II$: If $A \neq I$ and $A \neq -I$,then $\operatorname{Tr}(A) \neq 0$.
Statement $I$: If $A \neq I$ and $A \neq -I$,then $\operatorname{det}(A) = -1$.
Statement $II$: If $A \neq I$ and $A \neq -I$,then $\operatorname{Tr}(A) \neq 0$.
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If $A$ and $B$ are two square matrices such that $B = -A^{-1}BA$,then $(A + B)^2 = $
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