MCQ
$\left| {\,\begin{array}{*{20}{c}}1&a&{{a^2} - bc}\\1&b&{{b^2} - ac}\\1&c&{{c^2} - ab}\end{array}\,} \right| = $
- ✓$0$
- B${a^3} + {b^3} + {c^3} - 3abc$
- C$3abc$
- D${(a + b + c)^3}$
by $\left\{ \begin{array}{l}{R_1} \to {R_1} - {R_2}\\{R_2} \to {R_2} - {R_3}\end{array} \right.$
= $(a - b)\,(b - c)\,\left| {\,\begin{array}{*{20}{c}}0&1&{a + b + c}\\0&1&{a + b + c}\\1&c&{{c^2} - ab}\end{array}\,} \right| = 0$,
$\{\because\,\,{R_1} \equiv {R_2}\} $
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$PROPERTY 1$ if $\lim _{ h \rightarrow 0} \frac{ f ( h )- f (0)}{\sqrt{| h |}}$ exists and is finite, and $PROPERTY 2$ if $\lim _{h \rightarrow 0} \frac{f(h)-f(0)}{h^2}$ exists and is finite.
Then which of the following options is/are correct ?
$(1)$ $f(x)=x|x|$ has $PROPERTY$ $2$ $(2)$ $f(x)=x^{2 / 3}$ has $PROPERTY$ $1$ $(3)$ $f(x)=\sin x$ has $PROPERTY$ $2$ $(4)$ $f(x)=|x|$ has $PROPERTY$ $1$