- A$(ab - a'b')(bc - b'c')(ca - c'a')$
- B$(ab + a'b')(bc + b'c')(ca + c'a')$
- ✓$(ab' - a'b)(bc' - b'c)(ca' - c'a)$
- D$(ab' + a'b)(bc' + b'c)(ca' + c'a)$
$a' = 2,\,b' = 2,\,c' = 1$
Then the determinant is $\left| {\,\begin{array}{*{20}{c}}0&{ - 1}&2\\0&1&2\\{ - 1}&0&4\end{array}\,} \right| = 4$
Option $ (c) $ also gives the same value.
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$f(x)=x \cos \frac{1}{x}, \quad x \geq 1,$
$(A)$ for at least one $x$ in the interval $[1, \infty), f(x+2)-f(x)<2$
$(B)$ $\lim _{x \rightarrow \infty} f^{\prime}(x)=1$
$(C)$ for all $x$ in the interval $[1, \infty), f(x+2)-f(x)>2$
$(D)$ $f^{\prime}(x)$ is strictly decreasing in the interval $[1, \infty)$
$A_1=\left\{(x, y): x^2+2 y^2 \leq 1\right\}$
$A_2=\left\{(x, y):\left|x^3\right|+2 \sqrt{2} \mid y^3 \leq 1\right\}$
$A_3=\{(x, y): \max (|x|, \sqrt{2}|y|) \leq 1\}$ Then,