CLASS XI | FOM - III

Fundamentals of Mathematics – III

Contents

• Modulus — definition, graph, geometry, piecewise (1/2/3 modulus)
• Modulus equations — one/two/three modulus
• Modulus inequalities — one/two/three modulus
• Indices — definition, properties
• Indices equations — solving
• Indices inequalities — solving
• Logarithm — definition, domain, values, properties
• Logarithmic equations — methods, 1/2/3+ logs
• Logarithmic inequalities — methods, 1/2/3+ logs
• Basic determinants — 1×1, 2×2, 3×3
• Consolidated practice

Modulus

Introduction and definition

For any real number $x$, the modulus (absolute value) is $$|x|=\begin{cases} x, & x\ge 0\\ -x, & x<0\end{cases}$$ More generally, $|x-a|$ is the distance between $x$ and $a$ on the real line.

Graph and geometrical meaning

The graph of $y=|x-a|$ is a “V” with vertex at $(a,0)$, slope $-1$ left of $a$, and $+1$ right of $a$.

Vertex a: 0.0 Visualizes $y=|x-a|$ with axes and arrowheads.

Opening point (vertex)

For $y=|x-a|$, the opening point is $x=a$. Shifting $a$ moves the vertex horizontally; vertical shifts $y=|x-a|+k$ move it to $(a,k)$.

Redefining one modulus

$$|ax+b|=\begin{cases} ax+b,& ax+b\ge 0\\ -(ax+b),& ax+b<0\end{cases}$$

Example: $|3x-6|=\begin{cases}3x-6,& x\ge 2\\ -3x+6,& x<2\end{cases}$

Redefining two modulus

For $|x-p|+|x-q|$, split the real line at $p$ and $q$ and write linear forms on each interval.

Example: $|x-1|+|x+2|=\begin{cases}-(x-1)-(x+2),& x<-2\\ -(x-1)+(x+2),& -2\le x<1\\ (x-1)+(x+2),& x\ge 1\end{cases}$

Redefining three modulus

For $|x-p|+|x-q|+|x-r|$ with $p\le q\le r$, split at $p,q,r$. The sum is minimized at the median (here $x=q$).

Practice

1. Write $|2x+3|$ in piecewise form and mark its vertex on the number line.
2. Expand $|x-2|+|x-5|$ into linear pieces on $\;(-\infty,2),\,[2,5),\,[5,\infty)$.
3. Find $\min\limits_{x\in\mathbb{R}}\,\big(|x-1|+|x-3|+|x-7|\big)$ and the $x$ where it occurs.

Answers

1. $|2x+3|=\begin{cases}2x+3,& x\ge -\tfrac32\\ -2x-3,& x<-\tfrac32\end{cases}$; vertex at $x=-\tfrac32$.
2. $x<2:\;-(x-2)-(x-5)=-2x+7.$ $2\le x<5:\;(x-2)-(x-5)=3.$ $x\ge 5:\;(x-2)+(x-5)=2x-7.$
3. Median is $x=3$, minimum value $=|3-1|+|3-3|+|3-7|=2+0+4=6$.

Modulus equations

Definition

An equation with absolute value, e.g., $|ax+b|=c$ or $|u(x)|=|v(x)|$.

General method

1. Identify critical points where inside-modulus expressions change sign.
2. Split into intervals (cases) and remove $|\,\cdot\,|$ accordingly.
3. Solve each linear case and keep only solutions valid in that interval.

One modulus equation

$$|ax+b|=c\;(c\ge 0)\;\Longleftrightarrow\; ax+b=c\;\text{ or }\; ax+b=-c$$

Example: $|5x+2|=3\Rightarrow 5x+2=3$ or $5x+2=-3\Rightarrow x=\tfrac15,\;x=-1$.

Two modulus equation

For $|ax+b|=|cx+d|$, either square both sides or solve piecewise on sign intervals of $(ax+b)$ and $(cx+d)$.

Example: $|2x-1|=|x+3|\Rightarrow$ cases split at $x=\tfrac12$ and $x=-3$.

Three modulus equation

Chain pairwise equalities (e.g., $|u|=|v|$ and $|v|=|w|$) and solve with sign cases, checking consistency.

Practice

1. Solve $|2x-4|=6$.
2. Solve $|x-2|=|3x+1|$.
3. Solve $|x|=|x-2|=|x+4|$ (explain if none/all/finite solutions).

Answers

1. $2x-4=\pm 6\Rightarrow x=5$ or $x=-1$.
2. Split at $x=2$ and $x=-\tfrac13$; solutions $x=\tfrac12$ and $x=-\tfrac13$.
3. $|x|=|x-2|$ gives $x=1$. But $|1|=1\neq |1+4|=5$, so no $x$ satisfies all three; hence no solution.

Modulus inequalities

Definition

Inequalities involving $|\,\cdot\,|$, such as $|ax+b|

General method

• $|x-a|
• $|x-a|\ge k\;\Longleftrightarrow\; x\le a-k$ or $x\ge a+k$.
• For sums like $|u|+|v|\le k$ (one variable), split into sign-intervals of $u,v$ and solve linears; (two variables) this is a rotated diamond region.

One modulus inequality

Example: $|2x-1|>3\;\Longleftrightarrow\; 2x-1<-3$ or $2x-1>3\Rightarrow x<-1$ or $x>2$.

Two modulus inequality

For $|x-1|+|x+2|\le 5$, split at $x=-2$ and $x=1$ and solve in each region, then union the results.

Three modulus inequality

Proceed similarly, noting even-count roots do not flip the sign in polynomial analogies (helpful intuition).

Practice

1. Solve $|x+2|\le 5$ and write the answer as an interval.
2. Solve $|x-3|+|x|\lt 6$.
3. Solve $|x|+|x-2|+|x+1|\le 7$.

Answers

1. $-7\le x\le 3$.
2. Split at $x=0,3$: $x<0:\,-x-(x-3)<6\Rightarrow -2x+3<6\Rightarrow x> -\tfrac{3}{2}$; $0\le x<3:\, x+(3-x)<6\Rightarrow 3<6$ (always true); $x\ge 3:\, x+(x-3)<6\Rightarrow 2x<9\Rightarrow x<\tfrac{9}{2}$. Union $(-\tfrac32, \tfrac{9}{2})$.
3. Split at $x=-1,0,2$; solve piecewise and union the valid intervals.

Indices

Definition

For $a\ne 0$ and integer $n$, $a^n$ denotes repeated multiplication; $a^0=1$; $a^{-n}=1/a^n$. For $a>0$, $a^{m/n}=\sqrt[n]{a^m}$.

Properties (laws of exponents)

$a^m\cdot a^n=a^{m+n},\quad \dfrac{a^m}{a^n}=a^{m-n},\quad (a^m)^n=a^{mn},\quad (ab)^n=a^n b^n,\quad \left(\dfrac{a}{b}\right)^n=\dfrac{a^n}{b^n}$

Example: $(2^3)(2^4)=2^7=128$; $(3^2)^3=3^6=729$; $9^{3/2}=(\sqrt{9})^3=27$.

Practice

1. Simplify $5^3\cdot 5^{-1}$ and $(4^{1/2})^4$.
2. Express $27^{2/3}$ as an integer.

Answers

1. $5^{2}=25$; $(\sqrt{4})^4=2^4=16$.
2. $27^{2/3}=(\sqrt[3]{27})^2=3^2=9$.

Indices equations

Definition and method

Equations where the unknown is in the exponent. Use same-base comparison or logarithms.

Example: $5^{2x-1}=125=5^3\Rightarrow 2x-1=3\Rightarrow x=2$.

Practice

1. Solve $2^{x+1}=32$.
2. Solve $3^{2x-1}=27$.

Answers

1. $32=2^5\Rightarrow x+1=5\Rightarrow x=4$.
2. $27=3^3\Rightarrow 2x-1=3\Rightarrow x=2$.

Indices inequalities

Definition and method

• If $a>1$, $a^x$ is increasing: $a^x\le a^y\iff x\le y$.
• If $0

Example: $2^x\ge 8\Rightarrow x\ge 3$; $(\tfrac12)^x<4=2^2\Rightarrow 2^{-x}<2^2\Rightarrow -x<2\Rightarrow x>-2$.

Practice

1. Solve $5^{x-1}\le 1$.
2. Solve $(\tfrac13)^{2x}> \tfrac{1}{9}$.

Answers

1. $1=5^0\Rightarrow x-1\le 0\Rightarrow x\le 1$.
2. $\tfrac{1}{9}=(\tfrac13)^2$. Since base $<1$, inequality reverses: $2x<2\Rightarrow x<1$.

Logarithm

Introduction and definition

$\log_a b=x\iff a^x=b$, with conditions $a>0$, $a\ne 1$, $b>0$.

Value calculation and change of base

$\log_a b=\dfrac{\log_c b}{\log_c a},\quad \log_a a^x=x,\quad a^{\log_a x}=x$

Example: $\log_3 9=2$; $\log_5 \sqrt{25}=\log_5 5^{1/2}=\tfrac12$.

Properties

$\log_a(xy)=\log_a x+\log_a y,\quad \log_a\!\left(\dfrac{x}{y}\right)=\log_a x-\log_a y,\quad \log_a(x^k)=k\log_a x,\quad \log_a b=\dfrac{1}{\log_b a}$

Practice

1. Simplify $\log_2(8\cdot 4)-\log_2 2$.
2. Evaluate $\dfrac{\log 100}{\log 10}$ (common logs).

Answers

1. $\log_2 32-\log_2 2=5-1=4$.
2. $\dfrac{2}{1}=2$.

Logarithmic equations

Introduction & method

1. Domain: each log argument $>0$; base valid ($>0$, $\ne 1$).
2. Use log rules to combine/compare, or convert to exponential form.
3. Solve algebraically; finally, recheck domain.

One logarithm

Example: $\log_2(x+1)=3\Rightarrow x+1=8\Rightarrow x=7$ (domain $x>-1$ OK).

Two logarithms

If $\log_a f(x)=\log_a g(x)$ with same valid base $a$, then $f(x)=g(x)$ subject to $f(x),g(x)>0$.

Example: $\log_2(x+4)=\log_2(3x-2)\Rightarrow x+4=3x-2\Rightarrow x=3$ and $3x-2>0$ $\Rightarrow$ OK.

Three or more logarithms

Equate pairwise to a common expression and intersect the resulting solution sets with the domain.

Practice

1. Solve $\log_3(x-2)=2$.
2. Solve $\log_5(x+1)=\log_5(4x-3)$.
3. Discuss $\log_2 x=\log_2(3x-4)=\log_2(5x+1)$.

Answers

1. $x-2=9\Rightarrow x=11$ (domain $x>2$).
2. $x+1=4x-3\Rightarrow x=\tfrac{4}{3}$ (domain $x>\tfrac34$ OK).
3. From first pair $x=2$, but then $5x+1=11$ $\ne 2$; no common $x$ satisfies all three.

Logarithmic inequalities

Introduction & method

• Ensure argument $>0$ and base valid. If $a>1$, $\log_a$ is increasing; if $0
• Convert to exponential or compare arguments directly (same base).

One logarithm

Example: $\log_2(x-1)>3\Rightarrow x-1>8\Rightarrow x>9$ (domain $x>1$).

Two logarithms

With same base $a>1$, $\log_a f(x)\le \log_a g(x)\iff f(x)\le g(x)$ given $f,g>0$.

Three or more

Chain the inequalities and intersect with the domain of all logs.

Practice

1. Solve $\log_3(2x+1)\le 2$.
2. Solve $\log_{1/2}(x-3)< -1$.
3. Solve $\log_5(x)\lt \log_5(2x-1)$.

Answers

1. $2x+1\le 9$ and $2x+1>0\Rightarrow x\le 4$ and $x>-\tfrac12$ $\Rightarrow (-\tfrac12,4]$.
2. Base $<1$ so reverse: $x-3> (1/2)^{-1}=2\Rightarrow x>5$.
3. $x>0$, $2x-1>0\Rightarrow x>\tfrac12$ and $x<2x-1\Rightarrow x>1$; final $(1,\infty)$.

Basic determinants

Introduction

A determinant is a scalar computed from a square matrix; it tests invertibility ($\det\ne 0$) and scales area/volume under linear maps.

1×1 determinant

If $A=[a]$, then $\det(A)=a$.

2×2 determinant

If $A=\begin{bmatrix} a & b\\ c & d\end{bmatrix}$, then $\det(A)=ad-bc$.

3×3 determinant

For $A=\begin{bmatrix} a & b & c\\ d & e & f\\ g & h & i\end{bmatrix}$, $$\det(A)=a(ei-fh)-b(di-fg)+c(dh-eg)$$

Practice

1. Compute $\det\begin{bmatrix}3&1\\4&2\end{bmatrix}$.
2. Evaluate $\det\begin{bmatrix}1&2&3\\4&5&6\\7&8&9\end{bmatrix}$.
$(1,\infty)$.

Basic determinants

Introduction

A determinant is a scalar computed from a square matrix; it tests invertibility ($\det\ne 0$) and scales area/volume under linear maps.

1×1 determinant

If $A=[a]$, then $\det(A)=a$.

2×2 determinant

If $A=\begin{bmatrix} a & b\\ c & d\end{bmatrix}$, then $\det(A)=ad-bc$.

3×3 determinant

For $A=\begin{bmatrix} a & b & c\\ d & e & f\\ g & h & i\end{bmatrix}$, $$\det(A)=a(ei-fh)-b(di-fg)+c(dh-eg)$$

Practice

1. Compute $\det\begin{bmatrix}3&1\\4&2\end{bmatrix}$.
2. Evaluate $\det\begin{bmatrix}1&2&3\\4&5&6\\7&8&9\end{bmatrix}$.
3. Find $k$ such that $\det\begin{bmatrix}1&1&1\\1&2&3\\1&3&k\end{bmatrix}=0$.

Answers

1. $3\cdot 2-1\cdot 4=2$.
2. $0$ (rows are in arithmetic progression → rank deficiency).
3. Compute: $1(2k-9)-1(k-3)+1(3-6)=2k-9-k+3-3= k-9=0\Rightarrow k=9$.

Consolidated practice

1. Solve $|2x-3|+|x+1|=7$ (piecewise).
2. Solve $4^{x-1}=8^{2-x}$.
3. Solve $\log2(x+3)-\log2(x-1)=2$.
4. Solve $\log_3(x-2)\ge 1$ and write the solution as an interval.
5. Compute $\det\begin{bmatrix}2&-1&0\\0&3&4\\1&2&1\end{bmatrix}$.

Answer sketch

1. Split at $x=-1,\,\tfrac32$; solve in each interval and union valid roots.
2. $4^{x-1}=(2^2)^{x-1}=2^{2x-2}$ and $8^{2-x}=(2^3)^{2-x}=2^{6-3x}$ $\Rightarrow 2x-2=6-3x\Rightarrow x=\tfrac{8}{5}$.
3. $\log_2\frac{x+3}{x-1}=2\Rightarrow \frac{x+3}{x-1}=4\Rightarrow x+3=4x-4\Rightarrow x=\tfrac{7}{3}$ (domain $x>1$ OK).
4. $x-2\ge 3\Rightarrow x\ge 5$ (domain $x>2$); answer $[5,\infty)$.
5. Expand by any row/column; value $=2(3\cdot 1-4\cdot 2)-(-1)(0\cdot 1-4\cdot 1)+0(\cdots)=2(3-8)+1( -4)= -10-4=-14$.