MCQ
$f(x) = {x^2} - 3x$, then the points at which $f(x) = f'(x)$ are
- A$1, 3$
- B$1, -3$
- C$-1, 3$
- ✓None of these
But $f(x) = f'(x) \Rightarrow {x^2} - 3x = 2x - 3$
==> ${x^2} - 5x + 3 = 0$
$\therefore x = \frac{{5 \pm \sqrt {25 - 12} }}{2} = \frac{{5 \pm \sqrt {13} }}{2}$.
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$A_1=\left\{(x, y): x^2+2 y^2 \leq 1\right\}$
$A_2=\left\{(x, y):\left|x^3\right|+2 \sqrt{2} \mid y^3 \leq 1\right\}$
$A_3=\{(x, y): \max (|x|, \sqrt{2}|y|) \leq 1\}$ Then,