MCQ
If $|a|\,\, = 3,\,\,\,|b|\,\, = 4$ and $|a + b|\,\, = 5,$ then $|a - b|\,\, = $
- A$6$
- ✓$5$
- C$4$
- D$3$
$\therefore \,\,\,25 + |a - b{|^2} = 2(9 + 16) \Rightarrow |a - b| = 5$.
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$(i)$ $f (x)$ is bounded on $a \le x \le b.$
$(ii)$ The equation $f (x) = 0$ has at least one solution in $a < x < b.$
$(iii)$ The maximum and minimum values of $f (x)$ on $a \le x \le b$ occur at points where $f ' (c) = 0$.
$(iv)$ There is at least one point $c$ with $a < c < b$ where $f ' (c) > 0$.
$(v)$ There is at least one point $d$ with $a < d < b$ where $f ' (c) < 0.$