Solution:
Commutativity:
Let $\text{a, b}\in\text{R}$
a * b = ab + 1
= ba + 1
= b * a
Therefore,
a * b = b * a, $\forall\text{ a, b}\in\text{R}$
Therefore, * is commutative on R.
Associativity:
Let $\text{ a, b, c}\in\text{R}$
a * (b * c) = a * (bc + 1)
= a(bc + 1) + 1
= abc + a + 1
(a * b) * c = (ab + 1) * c
= (ab + 1)c + 1
= abc + c + 1
$\therefore$ a * (b * c) $\neq$ (a * b) * c
For example: a = 1, b = 2 and c = 3 [which belong to R]
Now,
1 * (2 * 3) = 1 * (6 + 1)
= 1 * 7
= 7 + 1
= 8
(1 * 2) * 3 = (2 + 1) * 3
= 3 * 3
= 9 + 1
= 10
⇒ 1 * (2 * 3) $\neq$ (1 * 2) * 3
Therefore, $\exists$ a = 1, b = 2 and c = 3 which belong to R such that
a * (b * c) $\neq$ (a * b) * c
Hence, * is not associative on R.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$(i)$ $f (x)$ 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 $f ' (c) < 0.$