(a) L.C.M. of $\left(\frac{a}{b},\frac{p}{q},\frac{l}{m}\right)=\frac{\text{ L.C.M. of }(a,p,l)}{\text{ H.C.F. of }(b,q,m)}$

(b) H.C.F. of $\left(\frac{a}{b},\frac{p}{q},\frac{\ell }{m}\right)=\frac{\text{ H.C.F. of }(a,p,\ell )}{\text{ L.C.M. of }(b,q,m)}$

(c) L.C.M. of rational and irrational number is not defined.

If a polynomial $a_1{x}^n+a_2{x}^{\nu -1}+a_3{x}^{n-2}+\dots +a_n$ is divided by $x-p$, then the remainder is obtained by putting $x=p$ in the polynomial.

A polynomial $a_{1} x^{n}+a_{2} x^{n-1}+a_{3} x^{n-2}+\ldots+$ $a_{n}$ is divisible by $x-p$, if the remainder is zero i.e., if $a_{1} p^{n}+$ $a_{2} p^{n-1}+\ldots+a_{n}=0$ then $x-p$ will be a factor of polynomial.

If an expression remain same after replacing $a$ by $b, b$ by $c$ and $c$ by $a$, then it is called cyclic expression and its factors are called cyclic factors.

e.g. $a(b-c)+b(c-a)+c(a-b)$

(a) $(a+b{)}^2=a^2+2ab+b^2=(a-b{)}^2+4ab$

(b) $(a-b{)}^2=a^2-2ab+b^2=(a+b{)}^2-4ab$

(c) $a^2-b^2=(a+b)(a-b)$

(d) $(a+b{)}^3=a^3+b^3+3ab(a+b)$

(e) $(a-b{)}^3=a^3-b^3-3ab(a-b)$

(f) $a^3+b^3=(a+b{)}^3-3ab(a+b)=(a+b)\left(a^2+b^2-ab\right)$

(g) $a^3-b^3=(a-b{)}^3+3ab(a-b)=(a-b)\left(a^2+b^2+ab\right)$

(h) $\begin{array}{rl}(a+b+c{)}^2& =a^2+b^2+c^2+2ab+2bc+2ca\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\\ & =a^2+b^2+c^2+2abc\end{array}$

(i) $a^2+b^2+c^2-ab-bc-ca=\frac{1}{2}\left[(a-b{)}^2+(b-c{)}^2+(c-a{)}^2\right]$

(j) $a^3+b^3+c^3-3abc=(a+b+c)\left(a^2+b^2+c^2-ab-bc-ca\right)$ $=\frac{1}{2}(a+b+c)\left[(a-b{)}^2+(b-c{)}^2+(c-a{)}^2\right]$

If $a+b+c=0$, then $a^3+b^3+c^3=3abc$

(k) $a^4-b^4=(a+b)(a-b)\left(a^2+b^2\right)$

(I) $a^4+a^2+1={\left(a^2+1\right)}^2-a^2=\left(1+a+a^2\right)\left(1-a+a^2\right)$

(m) $(a+b+c{)}^3=a^3+b^3+c^3+3(a+b)(b+c)(c+a)$

(a) A number will be divisible by 2 iff the digit at the unit place is divisible by 2 .

(b) A number will be divisible by 3 iff the sum of its digits of the number is divisible by 3 .

(c) A number will be divisible by 4 iff last two digits of the number together are divisible by 4 .

(d) A number will be divisible by 5 iff digit at the unit place is either 0 or 5 .

(e) A number will be divisible by 6 iff the digit at the unit place of the number is divisible by 2 and sum of all digits of the number is divisible by 3 .

(f) A number will be divisible by 8 iff the last 3 digits, all together, is divisible by 8 .

(g) A number will be divisible by 9 iff sum of all its digits is divisible by 9 .

(h) A number will be divisible by 10 iff it's last digit is zero.

(i) A number will be divisible by 11 iff the difference between the sum of the digits at even places and sum of the digits at odd places is a multiple of 11.

(a) $(\sqrt[n]{a}{)}^n=\sqrt[n]{a^n}=a$

(b) $\sqrt[n]{a}\times \sqrt[n]{b}=\sqrt[n]{ab}$ [Here order should be same.]

(c) $\sqrt[n]{a}÷\sqrt[n]{b}=\sqrt[n]{\frac{a}{b}}$

(d) $\sqrt[n]{\sqrt[m]{a}}=\sqrt[nm]{a}=\sqrt[m]{\sqrt[n]{a}}$

e.g., $\sqrt{\sqrt{\sqrt{2}}}=\sqrt[8]{2}$

(e) $\sqrt[n]{a}=\sqrt[n\times p]{a^p}$ or $\sqrt[n]{a^m}=\sqrt[n\times p]{a^{m\times p}}$

(f) $a\times a\times a\times \dots \times a(m$ times $)=a^m$

[Important for changing order of surds]

(g) $a^m\times a^n=a^{m+n}$

(h) $a^m÷a^n=a^{m-n}$

(i) ${\left(a^m\right)}^n=a^{mn}$

(j) $a^{-m}=\frac{1}{a^m}$

(k) ${\left(\frac{x}{y}\right)}^m=\frac{x^m}{y^m}$

(I) $(xy{)}^m=x^m\cdot y^m$

(m) $\sqrt[n]{x}=x^{1/n};n\ge 2,n\in N$

(n) $a^0=1$

(o) $a^x=a^y\Rightarrow x=y$ or $a=1$ or $a=0$ if $x>0$ and $y>0$

(p) $a^x=b^x\Rightarrow a=b$ or $\mathrm{x}=0$

(q) $a^{p/q}={\left(a^p\right)}^{1/q}={\left(a^{1/q}\right)}^p$

(r) ${\left(x^a\right)}^b\ne x^{a^b}$ but $=x^{ab}$

(a) If $a, b$ are positive real numbers, then

- $x^{2} \leq a^{2} \Leftrightarrow|x| \leq a \Leftrightarrow-a \leq x \leq a$

- $x^{2} \geq a^{2} \Leftrightarrow|x| \geq a \Leftrightarrow x \leq-a$ or, $x \geq a$

- $x^{2} < a^{2} \Leftrightarrow|x| < a \Leftrightarrow-a < x < a$

- $x^{2}>a^{2} \Leftrightarrow|x|>a \Leftrightarrow x<-a$ or, $x>a$

- $a^{2} \leq x^{2} \leq b^{2} \Leftrightarrow a \leq|x| \leq b \Leftrightarrow x \in[-b,-a] \cup[a, b]$

- $a^{2} < x^{2} < b^{2} \Leftrightarrow a< |x| < b \Leftrightarrow \in(-b,-a) \cup(a, b)$

(b) For real numbers $x$ and $y$, we have

- $|x+y|=|x|+|y|$, if $(x \geq 0$ and $y \geq 0)$ or, $(x,<0$ and $y<0)$

- $|x-y|=|x|-|y|$, if $(x \geq 0$ and $|x| \geq|y|)$ or, $(x \leq 0, y \leq 0$ and $|x| \geq|y|)$

- $|x \pm y| \leq|x|+|y|$

- $|x \pm y|>|| x|-| y||$

(A) $$[I] = I$$

$$[ - I] = - [I]$$

(B) $$[x + n] = [x] + n$$

Ex : $$[x + 2] = [x] + 2$$

$$[x - 3] = [x] - 3$$

(C) If $$[x] = n$$

then $$n \le x < n + 1$$

Ex : 1) $$[x] = 4$$

$$ \Rightarrow 4 \le x < 4 + 1$$

$$ \Rightarrow 4 \le x < 5$$

2) $$[x] = - 1$$

then $$ - 1 \le x < - 1 + 1$$

$$ \Rightarrow - 1 \le x < 0$$

(D) $$[x] + [ - x] = \left\{ {\matrix{ 0 & {x = I} \cr { - 1} & {x \ne I} \cr } } \right.$$

Ex : (a) $$x = 2$$

$$[2] + [ - 2] = 2 + ( - 2) = 0$$

(b) $$x = 2.8$$

$$[2.8] + [ - 2.8] = 2 + ( - 3) = - 1$$

The following are some very useful points to remember.

(a) $a \leq b \Rightarrow$ either $a < b$ or $a=b$

(b) $a < b$ and $b < c \Rightarrow a < c$

(c) $a < b \Rightarrow-a > -b$, i.e., the inequality sign reverses if both sides are multiplied by a negative number.

(d) $a < b$ and $c < d \Rightarrow a+c < b+d$ and $a-d < b-c$

(e) $a < b \Rightarrow k a < k b$ if $k > 0$, and $k a > k b$ if $k < 0$

(f) $0 < a < b \Rightarrow a^{r} < b^{r}$ of $r > 0$, and $a^{r} > b^{r}$ if $r < 0$

(g) $a+\frac{1}{a} \geq 2$ for $a > 0$ and equality holds for $a=1$

(h) $a+\frac{1}{a} \leq-2$ for $a < 0$ and eqaulity holds for $a=-1$

(i) If $a > 2 \Rightarrow 0< \frac{1}{x} < \frac{1}{2}$

(j) If $x < -3 \Rightarrow-\frac{1}{3} < \frac{1}{x} < 0$

(k) If $x < 2$, then we must consider $-\infty < x < 0$ or $0 < x < 2$

(I) Squaring an inequality: If $a < b$, then $a^{2} < b^{2}$ does not follow always.

Examples :

(1). $2 < 3 \Rightarrow 4 < 9$, but $-4 < 3 \Rightarrow 16 > 9$

(2). $x > 2 \Rightarrow x^{2} > 4$, but $x < 2 \Rightarrow x^{2} \geq 0$

(3). $2 < x < 4 \Rightarrow 4 < x^{2} < 16$

(4). $-2 < x < 4 \Rightarrow 0 \leq x^{2} < 16$

Let $y=\frac{f(x)}{g(x)}$ be an expression in $x$ where $f(x)$ and $g(x)$ are polynomials in $x$. Now, if it is given that $y>0$ (or $<0$). Then to solve the inequality following steps are taken.

Step I : Factorize $f(x)$ and $g(x)$ and generate the form :

$$ y=\frac{\left(x-a_{1}\right)^{n_{1}}\left(x-a_{2}\right)^{n_{2}} \ldots\left(x-a_{k}\right)^{n_{k}}}{\left(x-b_{1}\right)^{m_{1}}\left(x-b_{2}\right)^{m_{2}} \ldots\left(x-b_{p}\right)^{m_{p}}} $$

where $n_{1} n_{2} \ldots n_{k}, m_{1}, m_{2} \ldots m_{p}$ are natural $\ldots a_{k}, b_{1}, b_{2} \ldots b_{p}$ are real numbers. Clearly, here $a_{1}, a_{2} \ldots a_{k}, a_{2}$ roots of $f(x)=0$ roots of $f(x)=0$ and $b_{1}, b_{2}, \ldots b_{p}$ are roots of $g(x)=0$.

Step II : Identify the points where $y$ vanishes (i.e., where $f(x) = 0$). These roots $a_{1}, a_{2}, \ldots a_{k}$ are marked with black dots on a number line. Also identify the points where $y$ is undefined (i.e., where $g(x) = 0$). These points $b_{1}, b_{2}, \ldots b_{p}$ are marked with white dots and are excluded from the solution set.

Step III : Check the value of $y$ for a real number greater than the rightmost marked number on the number line. If $y$ is positive in this region, then it will be positive for all real numbers greater than this point. If $y$ is negative, it will be negative for all real numbers greater than this point.

Step IV : The points corresponding to factors with odd exponents are called "simple points", and those with even exponents are called "double points".

From right to left, draw a wavy curve above the number line (if $y$ was positive in Step III) or below it (if $y$ was negative). The curve should intersect the number line at simple points and stay on the same side of the line at double points.

Step V : The intervals where the curve is above the number line correspond to positive values of $y$, and intervals where the curve is below the number line correspond to negative values of $y$. The appropriate intervals are chosen based on whether the original inequality was $y > 0$ or $y < 0$. The union of these intervals represents the solution to the inequality.

(a) Set of all natural numbers $=\{1,2,3,4,\dots \}=N$

(b) Set of all whole numbers $=\{0,1,2,3,\dots \}=\mathrm{W}$

(c) set of all integers $=\{\dots .-3,-2,-1,0,1,2,3,\dots \}=\mathrm{I}=\mathrm{Z}$

(d) Set of all +ve integers $=\{1,2,3,\dots \}={\mathrm{Z}}^{+}=\mathrm{N}={\mathrm{I}}^{+}$

(e) Set of all - ve integers $=(-1,-2,-3,\dots \}=Z^{-}=1^{-}$

(f) The set of all non-zero integers $=\{\pm 1,\pm 2,\pm 3,\dots \}=Z_0$

(g) The set of all rational numbers $=\left\{\frac{p}{q}:p,q\in 1,q\ne 0\right\}=Q$

(h) The set of all positive rational numbers $=\left\{\frac{p}{q}:p,q\in I^{+}\right\}=Q^{+}$

(i) The set of all negative rational numbers $=\left\{\frac{p}{q}:p\in I^{-},q\in I^{+}\right\}=Q^{-}$

(j) The set of all real numbers $=R$

(k) The set of all positive real numbers $={\mathrm{R}}^{+}$

$(\ell )$ The set of all negative real numbers $={\mathrm{R}}^{-}$

(m) The set of all irrational numbers $=R-Q$

e.g. $\sqrt{2},\sqrt{3},\sqrt{5},\dots .\pi ,\mathrm{e},{\log }_{10}2$ etc. are all irrational numbers.