Definition of Polynomial

A polynomial in a variable x is an algebraic expression of the form:

$ f(x) = a_nx^n + a_{n-1}x^{n-1} + a_{n-2}x^{n-2} + \dots + a_1x + a_0 $

where $ a_0, a_1, a_2, \dots, a_n $ are real or complex constants called coefficients, $ n $ is a non-negative integer called the degree of the polynomial, and $ a_n \neq 0 $.

Examples:

$ f(x) = 3x^2 + 2x + 1 $ — Quadratic Polynomial
$ f(x) = 5x^3 - 4x^2 + 2 $ — Cubic Polynomial
$ f(x) = 7 $ — Constant Polynomial

Note: Expressions containing terms like $ x^{-1} $ or $ \sqrt{x} $ are not polynomials.

Degree of a Polynomial

The degree of a polynomial is the highest power of the variable $x$ that appears with a non-zero coefficient.

For a general polynomial $f(x) = a_nx^n + a_{n-1}x^{n-1} + \dots + a_1x + a_0$, the degree is $n$, provided $a_n \neq 0$.

Examples:

$f(x) = 3x^4 + 2x^3 - 5x + 1$ → Degree $= 4$
$g(x) = 7x^2 + 9$ → Degree $= 2$
$h(x) = 5$ → Degree $= 0$ (Constant polynomial)

Special Cases:

If all coefficients are $0$, it is called the zero polynomial, and its degree is not defined.
For non-zero constant polynomials, the degree is $0$.

Types of Polynomials

1. Based on Degree:

Zero Polynomial: All coefficients are $0$. Degree is not defined.
Constant Polynomial: Degree $= 0$, example: $f(x) = 7$.
Linear Polynomial: Degree $= 1$, example: $f(x) = 3x + 2$.
Quadratic Polynomial: Degree $= 2$, example: $f(x) = x^2 - 4x + 3$.
Cubic Polynomial: Degree $= 3$, example: $f(x) = 2x^3 + x^2 - 5$.
Quartic Polynomial: Degree $= 4$, example: $f(x) = x^4 + 2x^3 - x + 1$.

2. Based on Number of Terms:

Monomial: A polynomial with one term. Example: $5x^3$.
Binomial: A polynomial with two terms. Example: $3x^2 + 2x$.
Trinomial: A polynomial with three terms. Example: $x^2 - 3x + 2$.
Multinomial: A polynomial having more than three terms. Example: $x^4 + 2x^3 + x^2 + x + 1$.

Important Note:

A polynomial can be categorized both by its degree and by its number of terms. For example, $x^2 + 3x + 2$ is a quadratic trinomial.

Roots of a Polynomial

Definition:

A root (or zero) of a polynomial $f(x)$ is the value of $x$ for which $f(x) = 0$.

In other words, if $f(a) = 0$, then $x = a$ is called a root of the polynomial $f(x)$.

Example:

Consider the polynomial $f(x) = x^2 - 5x + 6$.

To find its roots, set $f(x) = 0$: $x^2 - 5x + 6 = 0$

Factorize: $(x - 2)(x - 3) = 0$

Therefore, the roots are $x = 2$ and $x = 3$.

Number of Roots:

A polynomial of degree $n$ can have at most $n$ roots.

Example: - A linear polynomial $(ax + b = 0)$ has 1 root. - A quadratic polynomial $(ax^2 + bx + c = 0)$ has 2 roots. - A cubic polynomial $(ax^3 + bx^2 + cx + d = 0)$ has 3 roots.

Important Notes:

Every polynomial of degree $n$ has exactly $n$ roots (real or complex) when counted with multiplicity.
If a root is repeated, it is called a multiple root.
If $(x - a)$ is a factor of $f(x)$, then $a$ is a root of $f(x)$.
The relation between roots and factors is given by the Factor Theorem.

Verification Example:

Let $f(x) = x^2 - 4x + 3$. Check if $x = 1$ is a root.

$f(1) = (1)^2 - 4(1) + 3 = 0$

Therefore, $x = 1$ is a root of $f(x)$.

Theorems of Polynomials

1. Factor Theorem:

If $f(x)$ is a polynomial and $f(a) = 0$, then $(x - a)$ is a factor of $f(x)$.

Conversely, if $(x - a)$ is a factor of $f(x)$, then $f(a) = 0$.

Example:

Let $f(x) = x^2 - 5x + 6$. Since $f(2) = 0$, $(x - 2)$ is a factor of $f(x)$.

Therefore, $f(x) = (x - 2)(x - 3)$.

2. Remainder Theorem:

If a polynomial $f(x)$ is divided by $(x - a)$, the remainder is $f(a)$.

Example:

Divide $f(x) = x^3 - 2x^2 + 4x - 8$ by $(x - 2)$.

By Remainder Theorem, remainder $= f(2) = 8 - 8 + 8 - 8 = 0$.

Hence, $(x - 2)$ is a factor of $f(x)$.

3. Fundamental Theorem of Algebra:

Every non-zero polynomial of degree $n$ has exactly $n$ roots in the complex number system (counting multiplicities).

Example:

The cubic equation $x^3 - 1 = 0$ has 3 roots: $x = 1$, $x = \omega$, and $x = \omega^2$, where $\omega = e^{i2\pi/3}$.

4. Relation Between Coefficients and Roots:

For a quadratic equation $ax^2 + bx + c = 0$ with roots $\alpha$ and $\beta$:

$\alpha + \beta = -\dfrac{b}{a}$
$\alpha \beta = \dfrac{c}{a}$

For a cubic equation $ax^3 + bx^2 + cx + d = 0$ with roots $\alpha, \beta, \gamma$:

$\alpha + \beta + \gamma = -\dfrac{b}{a}$
$\alpha\beta + \beta\gamma + \gamma\alpha = \dfrac{c}{a}$
$\alpha\beta\gamma = -\dfrac{d}{a}$

5. Polynomial Identity Theorem:

If two polynomials are equal for infinitely many values of $x$, then their corresponding coefficients are equal.

Example:

If $2x^2 + 3x + k = 2x^2 + 3x + 5$ for all $x$, then $k = 5$.

Important Notes:

Use the Factor Theorem to find roots quickly by substitution.
The Remainder Theorem helps in checking divisibility.
The Fundamental Theorem guarantees the total number of roots (real + complex).
Relations between coefficients and roots are frequently used in JEE problems.

Relations Between Coefficients and Roots of a Polynomial

If a polynomial equation of degree $n$ is given by:

$a_nx^n + a_{n-1}x^{n-1} + a_{n-2}x^{n-2} + \dots + a_1x + a_0 = 0$

and its roots are $\alpha_1, \alpha_2, \alpha_3, \dots, \alpha_n$, then the sum and products of the roots are related to the coefficients by fixed algebraic relations.

1. Quadratic Equation

For $ax^2 + bx + c = 0$ with roots $\alpha$ and $\beta$:

$\alpha + \beta = -\dfrac{b}{a}$
$\alpha \beta = \dfrac{c}{a}$

These are called the relations between roots and coefficients.

2. Cubic Equation

For $ax^3 + bx^2 + cx + d = 0$ with roots $\alpha, \beta, \gamma$:

$\alpha + \beta + \gamma = -\dfrac{b}{a}$
$\alpha\beta + \beta\gamma + \gamma\alpha = \dfrac{c}{a}$
$\alpha\beta\gamma = -\dfrac{d}{a}$

3. Quartic Equation

For $ax^4 + bx^3 + cx^2 + dx + e = 0$ with roots $\alpha, \beta, \gamma, \delta$:

$\alpha + \beta + \gamma + \delta = -\dfrac{b}{a}$
$\alpha\beta + \alpha\gamma + \alpha\delta + \beta\gamma + \beta\delta + \gamma\delta = \dfrac{c}{a}$
$\alpha\beta\gamma + \alpha\beta\delta + \beta\gamma\delta + \gamma\alpha\delta = -\dfrac{d}{a}$
$\alpha\beta\gamma\delta = \dfrac{e}{a}$

4. General Form (for any degree $n$)

For polynomial $a_nx^n + a_{n-1}x^{n-1} + a_{n-2}x^{n-2} + \dots + a_1x + a_0 = 0$ having roots $\alpha_1, \alpha_2, \dots, \alpha_n$, we have:

$\displaystyle \sum \alpha_i = -\dfrac{a_{n-1}}{a_n}$
$\displaystyle \sum \alpha_i\alpha_j = \dfrac{a_{n-2}}{a_n}$
$\displaystyle \sum \alpha_i\alpha_j\alpha_k = -\dfrac{a_{n-3}}{a_n}$
$\vdots$
$\alpha_1\alpha_2\cdots\alpha_n = (-1)^n \dfrac{a_0}{a_n}$

5. Example

Find the sum and product of roots of $2x^2 - 5x + 3 = 0$.

Comparing with $ax^2 + bx + c = 0$, we get $a = 2$, $b = -5$, $c = 3$.

$\alpha + \beta = -\dfrac{b}{a} = \dfrac{5}{2}$
$\alpha\beta = \dfrac{c}{a} = \dfrac{3}{2}$

Important Notes:

These relations are frequently used to form equations when roots are known.
They also help in solving symmetric functions of roots without finding actual roots.
Always divide the given polynomial by its leading coefficient before applying formulas.

If $\alpha$ and $\beta$ are the roots of $x^2 - 5x + 6 = 0$, find the value of $\alpha^2 + \beta^2$.

Solution:

We know from the relation between coefficients and roots:

$\alpha + \beta = 5$, $\alpha\beta = 6$

We need to find $\alpha^2 + \beta^2$.

$\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$

Substitute the values:

$\alpha^2 + \beta^2 = 5^2 - 2(6) = 25 - 12 = 13$

⇒ $\alpha^2 + \beta^2 = 13$

Concept used: symmetric functions of roots and standard relations.

Form a quadratic polynomial whose roots are $2$ and $3$.

Solution:

Let the required polynomial be:

$x^2 - (\alpha + \beta)x + \alpha\beta = 0$

Here, $\alpha = 2$, $\beta = 3$.

$\alpha + \beta = 5$, $\alpha\beta = 6$

Substitute the values:

$x^2 - 5x + 6 = 0$

⇒ Required polynomial is $x^2 - 5x + 6 = 0$

Concept used: General form of polynomial with given roots.

Find the quadratic polynomial whose sum of roots is $p$ and product of roots is $q$.

Solution:

Let the required polynomial be:

$x^2 - (\alpha + \beta)x + \alpha\beta = 0$

Given:

$\alpha + \beta = p$, $\alpha\beta = q$

Substitute these values:

$x^2 - px + q = 0$

⇒ Required polynomial is $x^2 - px + q = 0$

Concept used: Standard form of a quadratic equation using sum and product of roots.

Let the cubic polynomial be $ax^3 + bx^2 + cx + d = 0$ whose roots are $\alpha, \beta, \gamma$.

We need to express:

Sum of roots: $\alpha + \beta + \gamma$
Sum of product of roots taken two at a time: $\alpha\beta + \beta\gamma + \gamma\alpha$
Product of roots: $\alpha\beta\gamma$

Solution:

Compare $ax^3 + bx^2 + cx + d = 0$ with $a(x - \alpha)(x - \beta)(x - \gamma) = 0$.

Expanding the RHS:

$a[x^3 - (\alpha + \beta + \gamma)x^2 + (\alpha\beta + \beta\gamma + \gamma\alpha)x - \alpha\beta\gamma] = 0$

Comparing coefficients with $ax^3 + bx^2 + cx + d = 0$, we get:

$\alpha + \beta + \gamma = -\dfrac{b}{a}$
$\alpha\beta + \beta\gamma + \gamma\alpha = \dfrac{c}{a}$
$\alpha\beta\gamma = -\dfrac{d}{a}$

⇒ Relations between roots and coefficients: $$ \begin{aligned} \alpha + \beta + \gamma &= -\frac{b}{a} \\ \alpha\beta + \beta\gamma + \gamma\alpha &= \frac{c}{a} \\ \alpha\beta\gamma &= -\frac{d}{a} \end{aligned} $$

Concept used: Comparison of coefficients from expanded and standard forms.

Determine the value(s) of $k$ for which the cubic $$x^{3} - 6x^{2} + kx - 6 = 0$$ has its three roots in arithmetic progression.

Solution (step-by-step):

Step 1 — Let the roots be in A.P.

Assume the three roots are $(a-d),\; a,\; (a+d)$.

Step 2 — Use sum of roots (coefficient comparison)

$\displaystyle (a-d)+a+(a+d)=3a \;=\; 6 \quad\Rightarrow\quad a=2$

Step 3 — Use product of roots (constant term)

For cubic $x^3 + p x^2 + q x + r=0$, product of roots = $-r$. Here $r=-6$, so product = $6$. Thus

$\displaystyle (a-d)\,a\,(a+d)=a(a^2-d^2)=6$

Substitute $a=2$:

$\displaystyle 2(4-d^2)=6 \;\Rightarrow\; 8-2d^2=6 \;\Rightarrow\; 2d^2=2 \;\Rightarrow\; d^2=1$

Step 4 — Use sum of pairwise products to find $k$

Coefficient of $x$ in $x^3-6x^2+kx-6$ equals the sum of products of roots taken two at a time (with sign +). So:

$\displaystyle \alpha\beta+\beta\gamma+\gamma\alpha \;=\; k$

For roots $(a-d),a,(a+d)$ the sum is:

$\displaystyle (a-d)a + a(a+d) + (a-d)(a+d) = 3a^2 - d^2$

Substitute $a=2$ and $d^2=1$:

$\displaystyle k = 3(2)^2 - 1 = 12 - 1 = 11$

⇒ $\boxed{k = 11}$

(You can verify: with $k=11$ the cubic becomes $x^3-6x^2+11x-6=(x-1)(x-2)(x-3)$ whose roots $1,2,3$ are in A.P.)

Determine all real values of $k$ for which the cubic $$x^{3} - kx^{2} + kx - 1 = 0$$ has three real roots in geometric progression (G.P.).

Solution (step-by-step):

Step 1 — Let the roots be in G.P.

Assume the three roots are $\displaystyle \frac{1}{r},\; 1,\; r$ (we scale the middle term to $1$ because product will be convenient).

Step 2 — Use the product of roots

Product of roots $= \dfrac{1}{r}\cdot 1 \cdot r = 1$. This matches the constant term $(-1)^{3}\cdot(-1)=1$, so scaling is consistent (no contradiction).

Step 3 — Sum of roots gives $k$

$\displaystyle \alpha+\beta+\gamma = \frac{1}{r} + 1 + r = k$

Step 4 — Sum of pairwise products also equals $k$ (consistency check)

$\displaystyle \alpha\beta+\beta\gamma+\gamma\alpha = \frac{1}{r}\cdot 1 + 1\cdot r + \frac{1}{r}\cdot r = \frac{1}{r} + r + 1 = k$

Thus

$\boxed{\,k = r + 1 + \dfrac{1}{r}\,}$, where $r\in\mathbb{R}\setminus\{0\}$ and the three numbers $\dfrac{1}{r},1,r$ are real.

Step 5 — Find the range of real $k$ (so that all three roots are real)

Consider the function $f(r)=r+1+\dfrac{1}{r}$ for real $r\ne0$. Analyze for $r>0$ and $r<0$:< /p>

For $r>0$: by AM-GM, $r + \dfrac{1}{r} \ge 2$, so $f(r)\ge 1+2=3$, equality at $r=1$.
For $r<0$: put $r=-s$ with $s>0$, then $f(r)=1 -\big(s+\dfrac{1}{s}\big)\le 1-2=-1$, equality at $s=1$ (i.e. $r=-1$).

Therefore the set of real values attained by $k$ is

$\boxed{\,k\in (-\infty,-1] \;\cup\; [3,\infty)\, }$

Final: $k = r + 1 + \dfrac{1}{r}$ for some real $r\ne0$, and the real values of $k$ are exactly $(-\infty,-1]\cup[3,\infty)$.

(Verification example: for $r=1$ we get $k=3$ and polynomial $x^{3}-3x^{2}+3x-1=(x-1)^3$; for $r=2$ we get $k=2+1+\tfrac12=3.5$ and roots $\tfrac12,1,2$.)

Quadratic Equation

A quadratic equation in one variable x is an equation of the form:

$ ax^2 + bx + c = 0 $

where $ a, b, c $ are real coefficients and $ a \neq 0 $. The highest power of $ x $ is $ 2 $, hence it is called a quadratic equation.

Standard Form:

$ ax^2 + bx + c = 0 $

Here,
• $ a $ = coefficient of $ x^2 $
• $ b $ = coefficient of $ x $
• $ c $ = constant term

Roots of the Quadratic Equation:

$ x = \dfrac{ -b \pm \sqrt{b^2 - 4ac} }{2a} $

The values of $ x $ that satisfy the equation are called its roots or solutions.

Nature of Roots:

Let $ D = b^2 - 4ac $ be the discriminant.

If $ D > 0 $, the roots are real and distinct.
If $ D = 0 $, the roots are real and equal.
If $ D < 0 $, the roots are complex conjugates.

Example:

Solve $ 2x^2 + 3x - 2 = 0 $

$ x = \dfrac{ -3 \pm \sqrt{3^2 - 4(2)(-2)} }{2(2)} = \dfrac{ -3 \pm \sqrt{25} }{4} $
$ \Rightarrow x = \dfrac{1}{2},\ -2 $

Forms of Quadratic Equations

A quadratic equation can be expressed in several equivalent forms. Each form reveals different properties of the equation or graph of the quadratic function.

1. Standard Form:

$ y = ax^2 + bx + c $

This is the most common representation of a quadratic equation. It clearly shows the coefficients and constant term.

2. Factored Form:

$ y = a(x - r_1)(x - r_2) $

In this form, $ r_1 $ and $ r_2 $ are the roots of the equation, i.e., the values of $ x $ where $ y = 0 $.

3. Vertex Form:

$ y = a(x - h)^2 + k $

This form shows the vertex of the parabola at the point $(h, k)$.
It is useful for analyzing the maximum or minimum value of the quadratic function.

Conversion Between Forms:

Standard → Vertex form: by completing the square.
Standard → Factored form: by factoring or using the quadratic formula.

Example:

Convert $ y = 2x^2 - 4x + 1 $ into vertex form.

$ y = 2(x^2 - 2x) + 1 $
$ = 2[(x - 1)^2 - 1] + 1 $
$ = 2(x - 1)^2 - 1 $

So, the vertex form is $ y = 2(x - 1)^2 - 1 $ and the vertex is $(1, -1)$.

Graph of a Quadratic Equation

The graph of a quadratic equation $ y = ax^2 + bx + c $ is a parabola. The shape and orientation of the parabola depend on the sign and value of the coefficient $ a $.

1. Shape and Orientation:

If $ a > 0 $, the parabola opens upwards.
If $ a < 0 $, the parabola opens downwards.

$ \text{Parabola: } y = ax^2 + bx + c $

2. Vertex:

The vertex is the turning point of the parabola. For the equation $ y = ax^2 + bx + c $, the coordinates of the vertex are:

$ \left( -\dfrac{b}{2a},\ \dfrac{4ac - b^2}{4a} \right) $

• If $ a > 0 $, the vertex gives the minimum value of $ y $.
• If $ a < 0 $, the vertex gives the maximum value of $ y $.

3. Axis of Symmetry:

The parabola is symmetric about a vertical line passing through the vertex called the axis of symmetry.

$ x = -\dfrac{b}{2a} $

4. Y-Intercept:

The point where the parabola cuts the $ y $-axis is called the y-intercept. It occurs when $ x = 0 $.

$ y = c $

5. X-Intercepts (Roots):

The points where the parabola cuts the $ x $-axis correspond to the roots of the equation, obtained by solving $ ax^2 + bx + c = 0 $.

$ x = \dfrac{ -b \pm \sqrt{b^2 - 4ac} }{2a} $

Example:

For $ y = x^2 - 4x + 3 $:

$ a = 1,\ b = -4,\ c = 3 $
Vertex: $ (2, -1) $
Axis of Symmetry: $ x = 2 $
Y-Intercept: $ (0, 3) $
X-Intercepts: $ x = 1,\ 3 $

Relationship Between Coefficients and Roots

Let the roots of the quadratic equation $ ax^2 + bx + c = 0 $ be $ \alpha $ and $ \beta $. These roots satisfy the equation for both values of $ x $.

1. Derivation:

Since $ \alpha $ and $ \beta $ are roots,

$ a(x - \alpha)(x - \beta) = 0 $

Expanding this expression:

$ a(x^2 - (\alpha + \beta)x + \alpha\beta) = 0 $

Comparing with $ ax^2 + bx + c = 0 $, we get:

$ \alpha + \beta = -\dfrac{b}{a} $
$ \alpha \beta = \dfrac{c}{a} $

2. Relationship:

Sum of roots = $ \alpha + \beta = -\dfrac{b}{a} $
Product of roots = $ \alpha\beta = \dfrac{c}{a} $

3. Forming a Quadratic Equation from Given Roots:

If the roots of a quadratic equation are known to be $ \alpha $ and $ \beta $, then the equation can be written as:

$ x^2 - (\alpha + \beta)x + \alpha\beta = 0 $

Example 1:

If the roots of a quadratic equation are $ 2 $ and $ 3 $, find the equation.

$ \alpha + \beta = 2 + 3 = 5 $
$ \alpha\beta = 2 \times 3 = 6 $
Therefore, $ x^2 - 5x + 6 = 0 $

Example 2:

Find the sum and product of the roots of $ 2x^2 - 3x + 5 = 0 $.

$ \alpha + \beta = -\dfrac{-3}{2} = \dfrac{3}{2} $
$ \alpha\beta = \dfrac{5}{2} $

Nature of Roots and Discriminant

The discriminant of a quadratic equation $ ax^2 + bx + c = 0 $ is given by:

$ D = b^2 - 4ac $

The discriminant helps determine the nature of the roots of the quadratic equation.

Cases Based on Discriminant:

If $ D > 0 $, the equation has two distinct real roots.
If $ D = 0 $, the equation has two equal real roots (a repeated root).
If $ D < 0 $, the equation has two complex conjugate roots.

Example 1:

Find the nature of roots of $ x^2 - 5x + 6 = 0 $.

$ D = (-5)^2 - 4(1)(6) = 25 - 24 = 1 > 0 $
Therefore, roots are real and distinct.

Example 2:

Find the nature of roots of $ 2x^2 - 4x + 2 = 0 $.

$ D = (-4)^2 - 4(2)(2) = 16 - 16 = 0 $
Therefore, roots are real and equal.

Example 3:

Find the nature of roots of $ x^2 + x + 1 = 0 $.

$ D = 1^2 - 4(1)(1) = 1 - 4 = -3 < 0 $
Therefore, roots are complex conjugates.

Forming Quadratic Equations with Given Conditions

Quadratic equations can be constructed if certain properties of the roots are known, such as the sum, product, or actual values of roots.

1. If Roots Are Given:

If the roots of the quadratic equation are $ \alpha $ and $ \beta $, the equation can be written as:

$ x^2 - (\alpha + \beta)x + \alpha\beta = 0 $

2. If Sum and Product of Roots Are Given:

Let $ S = \alpha + \beta $ and $ P = \alpha \beta $, then the quadratic equation is:

$ x^2 - Sx + P = 0 $

3. If Roots Have a Certain Relationship:

Suppose the roots are in a specific ratio or differ by a certain number, say $ m $ and $ n $. First, express the roots in terms of a variable, then use sum and product to form the equation.

Example 1:

Form a quadratic equation whose roots are $ 3 $ and $ -2 $.

$ x^2 - (3 + (-2))x + (3 \cdot -2) = 0 $
$ x^2 - x - 6 = 0 $

Example 2:

Form a quadratic equation whose roots are in the ratio $ 2:3 $ and sum $ 10 $.

Let roots be $ 2k $ and $ 3k $.
$ 2k + 3k = 5k = 10 \Rightarrow k = 2 $
Roots: $ 4 $ and $ 6 $
Equation: $ x^2 - (4+6)x + (4 \cdot 6) = 0 \Rightarrow x^2 - 10x + 24 = 0 $

Solve: $$ x^2 - 7x + 12 = 0 $$

Solution (step-by-step):

Step 1 — Identify coefficients

Here, $a=1$, $b=-7$, $c=12$.

Step 2 — Use quadratic formula

$ x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \dfrac{-(-7) \pm \sqrt{(-7)^2 - 4(1)(12)}}{2(1)} $

Step 3 — Simplify

$ x = \dfrac{7 \pm \sqrt{49 - 48}}{2} = \dfrac{7 \pm 1}{2} $

Step 4 — Find the roots

$ x = 4, \; 3 $

Verification: $(x-4)(x-3) = x^2 - 7x + 12$ ✅

Determine all real values of $k$ for which the cubic $$ x^3 - kx^2 + kx - 1 = 0 $$ has three real roots in geometric progression (G.P.).

Solution (step-by-step):

Step 1 — Let roots be in G.P.

Assume the roots are $\dfrac{1}{r}, 1, r$ for some real $r \neq 0$.

Step 2 — Sum of roots

$\dfrac{1}{r} + 1 + r = k$

Step 3 — Pairwise sum of products (consistency check)

$\dfrac{1}{r}\cdot 1 + 1\cdot r + \dfrac{1}{r}\cdot r = \dfrac{1}{r} + r + 1 = k$

Step 4 — Determine range of $k$ for real roots

$f(r) = r + 1 + \dfrac{1}{r}$
$r > 0 → r + 1/r ≥ 2 → f(r) ≥ 3 (equality at r=1)$
$ r < 0 → r + 1/r + 1 ≤ -1 (equality at r=-1)$

Step 5 — Conclusion

$\boxed{k \in (-\infty, -1] \cup [3, \infty)}$

Verification: For r=1, k=3 and roots are (1,1,1); for r=2, k=3.5 and roots are (1/2,1,2).

Find a quadratic equation whose roots are in the ratio $2:3$ and whose sum is $25$.

Solution (step-by-step):

Step 1 — Let roots be in ratio 2:3

Assume roots are $2k$ and $3k$ for some real $k$.

Step 2 — Use sum of roots

$ 2k + 3k = 5k = 25 \;\Rightarrow\; k = 5$

Step 3 — Determine the roots

$ \text{Roots: } 2k = 10, \; 3k = 15$

Step 4 — Form the quadratic equation

$x^2 - (\text{sum of roots})x + (\text{product of roots}) = 0$
$x^2 - (10 + 15)x + (10 \cdot 15) = 0 $
$x^2 - 25x + 150 = 0$

Verification: Sum = 10+15=25 ✅, Product = 10*15=150 ✅

Find a quadratic equation whose roots differ by $3$ and whose sum is $7$.

Solution (step-by-step):

Step 1 — Let roots differ by 3

Assume roots are $ \alpha $ and $ \alpha + 3 $.

Step 2 — Use sum of roots

$\alpha + (\alpha + 3) = 2\alpha + 3 = 7 \;\Rightarrow\; 2\alpha = 4 \;\Rightarrow\; \alpha = 2$

Step 3 — Determine the roots

$\alpha = 2, \;\alpha + 3 = 5 $
$\text{Roots: } 2, 5$

Step 4 — Form the quadratic equation

$x^2 - (\text{sum of roots})x + (\text{product of roots}) = 0$
$ x^2 - (2+5)x + (2*5) = 0 $
$x^2 - 7x + 10 = 0$

Verification: Roots differ by 3 ✅, Sum = 2+5=7 ✅, Product = 2*5=10 ✅

Find a quadratic equation whose roots $ \alpha $ and $ \beta $ satisfy $$ \alpha + \beta = 5 \quad \text{and} \quad \alpha^2 + \beta^2 = 13. $$

Solution (step-by-step):

Step 1 — Express product using sum-of-squares

$\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$ $\Rightarrow 13 = 5^2 - 2\alpha\beta \;\Rightarrow 13 = 25 - 2\alpha\beta$

Step 2 — Solve for product of roots

$2\alpha\beta = 25 - 13 = 12 \;\Rightarrow\; \alpha\beta = 6$

Step 3 — Form the quadratic equation

$ x^2 - (\text{sum of roots})x + (\text{product of roots}) = 0 $
$x^2 - 5x + 6 = 0$

Verification: Sum = 2+3=5 ✅, Sum of squares = 2^2 + 3^2 = 4+9=13 ✅

Find a quadratic equation whose roots $\alpha$ and $\beta$ satisfy $$ \frac{1}{\alpha} + \frac{1}{\beta} = 5 \quad \text{and} \quad \alpha + \beta = 6. $$

Solution (step-by-step):

Step 1 — Express sum of reciprocals in terms of sum and product

$ \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} = 5 \;\Rightarrow\; \frac{6}{\alpha\beta} = 5$

Step 2 — Solve for product of roots

$ \alpha\beta = \frac{6}{5}$

Step 3 — Form the quadratic equation

$x^2 - (\text{sum of roots})x + (\text{product of roots}) = 0 $
$x^2 - 6x + \frac{6}{5} = 0$
$5x^2 - 30x + 6 = 0 \quad (\text{clear denominators})$

Verification: Sum = 6 ✅, Product = 6/5 ✅, Sum of reciprocals = 6 / (6/5) = 5 ✅

Definition

A complex number $z$ is written as: $z = a + ib$, where:

$a$ → Real part of $z$ ($\mathcal{R}(z)$)

$b$ → Imaginary part of $z$ ($\mathcal{I}(z)$)

Example 1

Consider $z = 5 + 2i$:

$\mathcal{R}(z) = 5$

$\mathcal{I}(z) = 2$

Example 2

Consider $z = -3 + 7i$:

$\mathcal{R}(z) = -3$

$\mathcal{I}(z) = 7$

1. Cartesian / Rectangular Form

$z = x + iy$

$x = \mathcal{R}(z)$ → Real part

$y = \mathcal{I}(z)$ → Imaginary part

Example: $z = 3 + 4i → x = 3, y = 4$

Operations easy in Cartesian form: addition, subtraction, conjugate.

2. Polar Form

$z = r(\cos\theta + i\sin\theta)$

$r = |z| = \sqrt{x^2 + y^2}$ → modulus

$\theta = \arg(z) = \tan^{-1}(y/x)$ → argument

Use: Multiplication, division, powers, roots.

Example: $1 + i = \sqrt{2}(\cos 45^\circ + i \sin 45^\circ)$

3. Euler / Exponential Form

$z = r e^{i\theta}$ (Euler’s formula: $e^{i\theta} = \cos\theta + i\sin\theta$)

Multiplication: $z_1 z_2 = r_1 r_2 e^{i(\theta_1 + \theta_2)}$

Division: $z_1 / z_2 = (r_1/r_2) e^{i(\theta_1 - \theta_2)}$

Powers: $z^n = r^n e^{in\theta}$

4. Trigonometric / Sin-Cos Form

Essentially same as polar form: $z = r(\cos\theta + i\sin\theta)$

Often used in De Moivre’s Theorem.

5. Modulus-Argument Form

|z| = r, \arg(z) = \theta

Can write: $z = |z|(\cos \arg(z) + i \sin \arg(z))$

6. Conjugate Form

$z = x + iy, \bar{z} = x - iy$

Useful for division, real part extraction:

$\mathcal{R}(z) = \frac{z + \bar{z}}{2}$

$\mathcal{I}(z) = \frac{z - \bar{z}}{2i}$

7. Geometric / Vector Form

Consider $z = x + iy$ as a vector in Argand plane: $\vec{z} = \langle x, y \rangle$

Modulus → magnitude of vector, Argument → angle with positive x-axis

JEE often uses this for loci, perpendicularity, collinearity.

8. Matrix Form (Advanced / Optional)

$z = x + iy \equiv \begin{bmatrix} x & -y \\ y & x \end{bmatrix}$

Multiplication of complex numbers ↔ matrix multiplication. Rare in JEE, mostly for Advanced derivations.

$z_1 = x_1 + iy_1, z_2 = x_2 + iy_2$ are equal if $x_1 = x_2 \text{ and } y_1 = y_2$

Example: $3 + 4i = 3 + 4i$ → True

$z_1 + z_2 = (x_1 + x_2) + i(y_1 + y_2)$

Example: $(2 + 3i) + (1 + 4i) = 3 + 7i$

$z_1 - z_2 = (x_1 - x_2) + i(y_1 - y_2)$

Example: $(5 + 6i) - (2 + 3i) = 3 + 3i$

$z_1 z_2 = (x_1 x_2 - y_1 y_2) + i(x_1 y_2 + x_2 y_1)$

Example: $(1 + 2i)(3 + 4i) = -5 + 10i$

$\frac{z_1}{z_2} = \frac{(x_1 x_2 + y_1 y_2) + i(y_1 x_2 - x_1 y_2)}{x_2^2 + y_2^2}, z_2 \neq 0$

Example: $\frac{1 + 2i}{1 + i} = 1.5 + 0.5i$

$\bar{z} = x - iy$

Properties:

$\overline{z_1 + z_2} = \bar{z_1} + \bar{z_2}$
$\overline{z_1 z_2} = \bar{z_1} \bar{z_2}$
$\overline{\bar{z}} = z$

$\arg(z) = \theta$ → angle with positive x-axis

Properties:

$\arg(z_1 z_2) = \arg(z_1) + \arg(z_2)$
$\arg(z_1 / z_2) = \arg(z_1) - \arg(z_2)$
$\arg(\bar{z}) = -\arg(z)$

Find the roots of the quadratic equation in complex numbers: $z^2 - (3 + 4i)z + (5 + 6i) = 0$

Solution:

Use quadratic formula: $z = \frac{(3+4i) \pm \sqrt{(3+4i)^2 - 4(5+6i)}}{2}$

Step 1: Compute discriminant: $(3+4i)^2 - 4(5+6i) = (9 + 24i - 16) - (20 + 24i) = -7 + 0i$

Step 2: Square root of discriminant: $\sqrt{-7} = i\sqrt{7}$

Step 3: Apply formula: $z = \frac{3+4i \pm i\sqrt{7}}{2}$

Step 4: Final roots: $z_1 = \frac{3 + 4i + i\sqrt{7}}{2} = \frac{3}{2} + i\frac{4+\sqrt{7}}{2}, \quad z_2 = \frac{3 + 4i - i\sqrt{7}}{2} = \frac{3}{2} + i\frac{4-\sqrt{7}}{2}$

Find the modulus and argument of the complex number:

$z = -1 + \sqrt{3} i$

Step 1: Modulus:

$|z| = \sqrt{(-1)^2 + (\sqrt{3})^2} = 2$

Step 2: Argument:

$\theta = \tan^{-1}(y/x) = \tan^{-1}(\sqrt{3}/-1)$

Step 3: Quadrant check: x < 0, y> 0 → second quadrant:

$\theta = 180^\circ - 60^\circ = 120^\circ = \frac{2 \pi}{3}$

Final Answer:

$|z| = 2, \quad \arg(z) = \frac{2\pi}{3}$

Find the locus of the complex number $z = x + iy$ satisfying $|z - 1| = 2$.

Solution:

Step 1: Substitute $z = x + iy$: $|(x + iy) - 1| = 2$

Step 2: Simplify: $|(x - 1) + iy| = 2 \implies \sqrt{(x-1)^2 + y^2} = 2$

Step 3: Square both sides: $(x - 1)^2 + y^2 = 4$

Step 4: Interpretation: This represents a circle with center at $(1,0)$ and radius $2$ in the Argand plane.

Answer: Circle, center = (1,0), radius = 2

Solve the quadratic equation $z^2 - (3 - 4i)z + (5 - 12i) = 0$.

Solution:

Step 1: Use quadratic formula: $z = \frac{(3-4i) \pm \sqrt{(3-4i)^2 - 4(5-12i)}}{2}$

Step 2: Compute discriminant: $(3-4i)^2 - 4(5-12i) = (9 - 24i + 16i^2) - (20 - 48i) = (-7 - 24i) - (20 - 48i) = -27 + 24i$

Step 3: Compute square root of discriminant: Let $\sqrt{-27 + 24i} = a + bi$. Solve $a^2 - b^2 = -27, 2ab = 24 \implies a = 3, b = 4$ So, $\sqrt{-27 + 24i} = 3 + 4i$

Step 4: Roots: $z = \frac{(3-4i) \pm (3+4i)}{2}$

Step 5: Compute each root: $z_1 = \frac{(3-4i) + (3+4i)}{2} = \frac{6}{2} = 3$

$z_2 = \frac{(3-4i) - (3+4i)}{2} = \frac{-8i}{2} = -4i$

Answer: $z = 3, \; z = -4i$

If $z + \bar{z} = 6$ and $z\bar{z} = 10$, find $z$.

Solution:

Step 1: Let $z = x + iy$. Then $z + \bar{z} = 2x = 6 \implies x = 3$.

Step 2: $z \bar{z} = x^2 + y^2 = 10 \implies 3^2 + y^2 = 10 \implies y^2 = 1 \implies y = \pm 1$.

Answer: $z = 3 + i, \; 3 - i$

If $z_1 = 2(\cos 30^\circ + i \sin 30^\circ)$ and $z_2 = 3(\cos 45^\circ + i \sin 45^\circ)$, find $z_1 z_2$ in polar form.

Solution:

Step 1: Modulus of product: $|z_1 z_2| = |z_1||z_2| = 2 \cdot 3 = 6$.

Step 2: Argument of product: $\arg(z_1 z_2) = 30^\circ + 45^\circ = 75^\circ$.

Answer: $z_1 z_2 = 6(\cos 75^\circ + i \sin 75^\circ)$

Points $z_1 = 0$, $z_2 = 2$, and $z_3$ form an equilateral triangle. Find $z_3$.

Solution:

Step 1: Formula for rotation: $z_3 = z_1 + (z_2 - z_1)e^{i 60^\circ} = 0 + 2 e^{i\pi/3} = 2(\cos 60^\circ + i \sin 60^\circ)$

Step 2: Convert to rectangular form: $z_3 = 2(1/2 + i\sqrt{3}/2) = 1 + i\sqrt{3}$

Answer: $z_3 = 1 + i\sqrt{3}$

Solve for complex number $z$ given $|z - 1| = |z + 1|$.

Solution:

Step 1: Let $z = x + iy$, then equation becomes: $| (x+iy) - 1 | = | (x+iy) + 1 |$

Step 2: Simplify modulus: $\sqrt{(x-1)^2 + y^2} = \sqrt{(x+1)^2 + y^2}$

Step 3: Square both sides: $(x-1)^2 + y^2 = (x+1)^2 + y^2$

Step 4: Simplify equation: $x^2 - 2x + 1 + y^2 = x^2 + 2x + 1 + y^2 \implies -2x = 2x \implies x = 0$

Step 5: Conclusion: $x = 0 \implies z = iy, \text{ any real } y$

Answer: $z = iy, \; y \in \mathbb{R}$

Solve for $z$ if $z + \bar{z} = 4$ and $z \bar{z} = 5$.

Solution:

Step 1: Let $z = x + iy \implies \bar{z} = x - iy$.

Step 2: First equation $z + \bar{z} = 4 \implies (x + iy) + (x - iy) = 4 \implies 2x = 4 \implies x = 2$.

Step 3: Second equation $z \bar{z} = 5 \implies (x+iy)(x-iy) = x^2 + y^2 = 5$.

Step 4: Substitute $x = 2$: $2^2 + y^2 = 5 \implies 4 + y^2 = 5 \implies y^2 = 1 \implies y = \pm 1$.

Step 5: Write all solutions: $z = 2 + i$ or $z = 2 - i$.

Answer: $z = 2 + i, \; 2 - i$

Definition of Logarithms

A logarithm is the inverse operation of exponentiation. In other words, it tells us the power to which a certain number, called the base, must be raised to obtain another number.

1. Basic Definition:

$ \log_b a = c \iff b^c = a $

Where:

$b > 0$ and $b \neq 1$ is the base
$a > 0$ is the argument or the number whose logarithm is being found
$c$ is the logarithm itself, i.e., the exponent to which $b$ must be raised to obtain $a$

2. Example:

$ \log_2 8 = 3 \quad \text{because} \quad 2^3 = 8 $

This means 2 must be raised to the power 3 to get 8.

3. Inverse Relation with Exponentials:

Logarithms and exponentials are inverse functions. If $y = b^x$, then $\log_b y = x$. Conversely, if $y = \log_b x$, then $b^y = x$.

4. Conditions for Logarithms to Exist:

The base $b$ must be positive and not equal to 1: $b > 0$, $b \neq 1$
The argument $a$ must be positive: $a > 0$

Basic Properties and Laws of Logarithms

Logarithms follow specific rules and properties which make it easier to simplify expressions and solve equations involving logarithms.

1. Logarithm of 1 and the Base:

$ \log_b 1 = 0, \quad \log_b b = 1 $

These come directly from the definition, because $b^0 = 1$ and $b^1 = b$.

2. Product Rule:

$ \log_b(xy) = \log_b x + \log_b y $

The logarithm of a product is the sum of the logarithms.

3. Quotient Rule:

$ \log_b\left(\dfrac{x}{y}\right) = \log_b x - \log_b y $

The logarithm of a quotient is the difference of the logarithms.

4. Power Rule:

$ \log_b(x^n) = n \log_b x $

The logarithm of a number raised to a power is equal to the exponent times the logarithm of the number.

5. Change of Base Formula:

$ \log_b a = \dfrac{\log_k a}{\log_k b} $

This allows converting logarithms from one base to another, commonly used with base 10 or base $e$.

6. Reciprocal Base Formula:

$ \log_{1/b} x = -\log_b x $

Flipping the base to its reciprocal changes the sign of the logarithm.

Logarithmic Equations

A logarithmic equation is an equation in which the unknown variable appears inside a logarithm. Solving these equations usually involves applying the laws of logarithms and ensuring that all arguments of logarithms remain positive.

1. General Form:

$ \log_b f(x) = \log_b g(x) \quad \Rightarrow \quad f(x) = g(x), \quad f(x) > 0, g(x) > 0 $

For real solutions, the arguments of the logarithms must be positive, and the bases must satisfy $b>0$, $b \neq 1$.

2. Example 1:

Solve for $x$:

$ \log_2(x-1) = 3 $

Solution (step-by-step):

Step 1 — Convert to exponential form:

$ 2^3 = x - 1 $

Step 2 — Solve for x:

$ x - 1 = 8 \quad \Rightarrow \quad x = 9 $

Step 3 — Check domain condition:

Argument must be positive: $x - 1 > 0 \;\Rightarrow\; 9 - 1 > 0$ ✅ Valid.

Final Answer: $x = 9$

Note: Always check the domain of the logarithm. Any solution that makes the argument non-positive is invalid.

Example 2: Logarithmic Equation with Variables on Both Sides

Solve for $x$:

$ \log_3(x+2) - \log_3(x-1) = 1 $

Solution (step-by-step):

Step 1 — Apply the Quotient Rule:

$\log_3\left(\frac{x+2}{x-1}\right) = 1$

Step 2 — Convert to exponential form:

$\frac{x+2}{x-1} = 3^1 = 3$

Step 3 — Solve for x:

$x + 2 = 3(x - 1) \;\Rightarrow\; x + 2 = 3x - 3 \;\Rightarrow\; 2 + 3 = 3x - x \;\Rightarrow\; 5 = 2x \;\Rightarrow\; x = \frac{5}{2}$

Step 4 — Check domain conditions:

Arguments must be positive: $x + 2 > 0 \;\Rightarrow\; \frac{5}{2}+2 >0$ ✅, $x - 1 > 0 \;\Rightarrow\; \frac{5}{2}-1 = \frac{3}{2}>0$ ✅

Final Answer: $x = \dfrac{5}{2}$

Example 5: Logarithmic Equation with Unknown Inside and Outside

Solve for $$x$$:

$$ \log(x) + x = 3 $$

Solution (step-by-step):

Step 1 — Consider substitution:

Let $$y = \log(x) \implies x = 10^y$$. Substituting gives:

$$ y + 10^y = 3 $$

Step 2 — Solve for $$y$$:

This is a transcendental equation and can be solved using inspection for small integer solutions. Try $$y = 1$$: $$1 + 10^1 = 1 + 10 = 11 \neq 3$$ Try $$y = 0$$: $$0 + 10^0 = 1 \neq 3$$ Try $$y = \log(2) \approx 0.301$$: $$0.301 + 2 \approx 2.301 \neq 3$$ Try $$y = \log(2) +?$$ ...

Note: Equations like $$\log(x)+x=3$$ usually require **graphical or numerical methods** (trial/error or Newton-Raphson) for exact solution. For JEE-level, an approximate solution is often acceptable.

Solve for $$x$$: $$\log_2(x-1) + \log_2(x-3) = 3$$

Solution (step-by-step):

Step 1 — Apply the Product Rule of Logarithms

$$\log_2\big((x-1)(x-3)\big) = 3$$

Step 2 — Convert to exponential form

$$(x-1)(x-3) = 2^3 = 8$$

Step 3 — Expand and form a quadratic equation

$$x^2 - 4x + 3 = 8 \quad \Rightarrow \quad x^2 - 4x - 5 = 0$$

Step 4 — Solve the quadratic

$$x = \frac{4 \pm \sqrt{16 + 20}}{2} = \frac{4 \pm \sqrt{36}}{2} = \frac{4 \pm 6}{2}$$
$$x = 5 \quad \text{or} \quad x = -1$$

Step 5 — Check domain restrictions

Arguments must be positive: $$x-1>0 \implies x>1$$, $$x-3>0 \implies x>3$$ So only $$x = 5$$ is valid.

Final Answer: $$x = 5$$

Solve for $$x$$: $$\log_2(x) + \log_4(x-2) = 3$$

Solution (step-by-step):

Step 1 — Convert all logarithms to the same base

$$\log_4(x-2) = \frac{\log_2(x-2)}{\log_2 4} = \frac{\log_2(x-2)}{2}$$ $$\Rightarrow \log_2(x) + \frac{1}{2}\log_2(x-2) = 3$$

Step 2 — Multiply through by 2

$$2\log_2(x) + \log_2(x-2) = 6$$

Step 3 — Apply the Product Rule

$$\log_2(x^2) + \log_2(x-2) = \log_2(x^2(x-2)) = 6$$

Step 4 — Convert to exponential form

$$x^2(x-2) = 2^6 = 64$$ $$x^3 - 2x^2 - 64 = 0$$

Step 5 — Solve the cubic equation

Try factorization: $$(x-4)(x^2 + 2x + 16) = 0$$ Only real solution: $$x = 4$$

Step 6 — Check domain conditions

Arguments must be positive: $$x>0 \quad \text{and} \quad x-2>0 \implies x>2$$ Valid solution: $$x = 4$$ ✅

Final Answer: $$x = 4$$

Solve for $$x$$: $$\log_3(x+1) + \log_9(x-2) = 2$$

Solution (step-by-step):

Step 1 — Express all logs in base 3

$$\log_9(x-2) = \frac{\log_3(x-2)}{\log_3 9} = \frac{\log_3(x-2)}{2}$$ $$\Rightarrow \log_3(x+1) + \frac{1}{2}\log_3(x-2) = 2$$

Step 2 — Multiply both sides by 2

$$2\log_3(x+1) + \log_3(x-2) = 4$$

Step 3 — Apply the Product Rule of Logarithms

$$\log_3((x+1)^2) + \log_3(x-2) = \log_3((x+1)^2(x-2)) = 4$$

Step 4 — Convert to exponential form

$$(x+1)^2(x-2) = 3^4 = 81$$

Step 5 — Expand and form a cubic equation

$$(x^2 + 2x + 1)(x-2) = x^3 - 2x^2 + 2x^2 - 4x + x - 2 = x^3 - x - 2 = 81$$ $$x^3 - x - 83 = 0$$

Step 6 — Solve the cubic equation (approximation)

This cubic does not factor nicely; approximate solution using trial or graphing: $$x \approx 4.4$$

Step 7 — Check domain restrictions

Arguments must be positive: $$x+1>0 \implies x>-1$$, $$x-2>0 \implies x>2$$ Approximate solution $$x \approx 4.4$$ satisfies both ✅

Final Answer (approximate): $$x \approx 4.4$$

Solve for $$x$$: $$\log_2(x^2 - 5x + 6) = 3$$

Solution (step-by-step):

Step 1 — Convert to exponential form

$$x^2 - 5x + 6 = 2^3 = 8$$

Step 2 — Form the quadratic equation

$$x^2 - 5x + 6 - 8 = 0 \quad \Rightarrow \quad x^2 - 5x - 2 = 0$$

Step 3 — Solve the quadratic

$$x = \frac{5 \pm \sqrt{25 + 8}}{2} = \frac{5 \pm \sqrt{33}}{2}$$

Step 4 — Check domain restrictions

Argument of log must be positive: $$x^2 - 5x + 6 > 0$$ Factor: $$(x-2)(x-3) > 0 \implies x < 2 \text{ or } x> 3$$ Only $$x = \frac{5 + \sqrt{33}}{2} \approx 5.372$$ satisfies $$x>3$$

Final Answer: $$x = \frac{5 + \sqrt{33}}{2}$$

Solve for $$x$$: $$x + \log_2(x-1) = 5$$

Solution (step-by-step):

Step 1 — Domain restriction

Argument of log must be positive: $$x-1 > 0 \implies x > 1$$

Step 2 — Substitution

Let $$y = x - 1 \implies x = y + 1$$. Then the equation becomes:

$$y + 1 + \log_2(y) = 5 \quad \Rightarrow \quad y + \log_2(y) = 4$$

Step 3 — Solve using trial/iteration

Since this is a transcendental equation, try integer powers of 2:

Try $$y = 2$$: $$2 + \log_2(2) = 2 + 1 = 3 \neq 4$$
Try $$y = 4$$: $$4 + \log_2(4) = 4 + 2 = 6 \neq 4$$
Try $$y = 3$$: $$3 + \log_2(3) \approx 3 + 1.585 = 4.585 \approx 4?$$

Using closer approximation, the solution is around $$y \approx 2.415 \implies x \approx 3.415$$

Step 4 — Final answer

Approximate solution: $$x \approx 3.415$$

Step 1: Concept of Limit

Consider a function

$ f(x) $

defined near $ x = a $ (except possibly at $ a $ itself).

We want to understand what value $ f(x) $ approaches as $ x $ gets closer and closer to $ a $.

Step 2: Informal Definition

The limit of $ f(x) $ as $ x \to a $ is $ L $ if:

$ \displaystyle \lim_{x \to a} f(x) = L $

meaning $ f(x) $ can be made arbitrarily close to $ L $ by taking $ x $ sufficiently close to $ a $ (but $ x \neq a $).

Step 3: Left-hand and Right-hand Limits

Left-hand limit:

$ \displaystyle \lim_{x \to a^-} f(x) = L $

Right-hand limit:

$ \displaystyle \lim_{x \to a^+} f(x) = L $

The limit exists if:

$ \displaystyle \lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) = L $

Step 4: Rigorous ε-δ Definition

The limit of $ f(x) $ as $ x \to a $ is $ L $ means:

$$ \displaystyle \lim_{x \to a} f(x) = L \iff \forall \varepsilon > 0, \; \exists \delta > 0 :\\ 0 < |x - a| < \delta \implies |f(x) - L| < \varepsilon $$

In the rigorous ε-δ definition of a limit: ε (epsilon) represents how close we want f(x) to be to L, and δ (delta) represents how close x should be to a (excluding a itself). No matter how small ε is, we can always find a δ that ensures f(x) stays within ε of L.

Step 5: Important Notes

1. Function need not be defined at $ x = a $.

2. Limits can be finite or infinite.

3. One-sided limits are useful for studying discontinuities.

4. JEE problems often involve algebraic manipulation to evaluate limits.

Here are some examples

Step 1: Simple Limit Example

Evaluate: $ \lim_{x \to 2} (3x + 1) $

$ \lim_{x \to 2} (3x + 1) = ? $

Solution: Substitute $ x = 2 $

$ 3(2) + 1 = 7 $

Answer: 7

Step 2: Limit Leading to 0/0

Evaluate: $ \lim_{x \to 2} \frac{x^2 - 4}{x - 2} $

$ \frac{x^2 - 4}{x - 2} $

Solution: Factor numerator:

$ x^2 - 4 = (x-2)(x+2) $

Cancel (x-2):

$ \lim_{x \to 2} (x + 2) = 4 $

Answer: 4

Step 3: One-Sided Limits

Function:

$ f(x) = \begin{cases} x^2 & x < 1 \\ 2x + 1 & x \ge 1 \end{cases} $

Left-hand limit: $ \lim_{x \to 1^-} f(x) = 1^2 = 1 $

Right-hand limit: $ \lim_{x \to 1^+} f(x) = 2(1) + 1 = 3 $

Since LHL ≠ RHL, overall limit does not exist.

Step 4: Limit at Infinity

Evaluate: $ \lim_{x \to \infty} \frac{3x^2 + 5}{2x^2 - x} $

Divide numerator & denominator by $ x^2 $:

$ \frac{3 + 5/x^2}{2 - 1/x} \to \frac{3}{2} $

Answer: 3/2

Step 5: Standard Tricky Limit

Evaluate: $ \lim_{x \to 0} \frac{\sin x}{x} $

Answer: 1 (well-known JEE standard limit)

Standard Limits in Calculus

Step 1: Definition of a Limit

The limit of a function $ f(x) $ as $ x \to a $ is the value that $ f(x) $ approaches as $ x $ gets arbitrarily close to $ a $ (but $ x \neq a $).

$ \displaystyle \lim_{x \to a} f(x) = L $

Step 2: Some Standard Limits

$ \displaystyle \lim_{x \to 0} \frac{\sin x}{x} = 1 $
$ \displaystyle \lim_{x \to 0} \frac{1 - \cos x}{x} = 0 $
$ \displaystyle \lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2} $
$ \displaystyle \lim_{x \to 0} \frac{\tan x}{x} = 1 $
$ \displaystyle \lim_{x \to 0} (1 + x)^{1/x} = e $
$ \displaystyle \lim_{x \to \infty} \left(1 + \frac{1}{x}\right)^x = e $

Step 3: Examples

Example 1: $ \displaystyle \lim_{x \to 0} \frac{\sin 5x}{x} $

$ \displaystyle \lim_{x \to 0} \frac{\sin 5x}{x} = \lim_{x \to 0} 5 \cdot \frac{\sin 5x}{5x} = 5 $

Example 2: $ \displaystyle \lim_{x \to 0} \frac{1 - \cos 3x}{x^2} $

$ \displaystyle \lim_{x \to 0} \frac{1 - \cos 3x}{x^2} = \lim_{x \to 0} \frac{9(1 - \cos 3x)}{(3x)^2} = \frac{9}{2} $

Step 4: Important Notes

Always look for standard limits before applying L'Hôpital's Rule.
Multiplying and dividing by constants can simplify limits.
Many JEE questions combine these limits with algebraic manipulation.
Trigonometric limits often require using small angle approximations: $ \sin x \approx x, \tan x \approx x, \cos x \approx 1 - \frac{x^2}{2} $ as $ x \to 0 $.

Techniques to Evaluate Limits

Step 1: Direct Substitution

If the function is continuous at the point, directly substitute the value of $ x $ to find the limit.

Example:

$ \displaystyle \lim_{x \to 2} (3x^2 + 5) = 3(2)^2 + 5 = 17 $

Step 2: Factorization

If direct substitution gives an indeterminate form like $ \frac{0}{0} $, try factoring numerator and denominator.

Example:

$ \displaystyle \lim_{x \to 3} \frac{x^2 - 9}{x - 3} = \lim_{x \to 3} \frac{(x-3)(x+3)}{x-3} = \lim_{x \to 3} (x+3) = 6 $

Step 3: Rationalization

Multiply by the conjugate if the limit involves square roots to remove indeterminate forms.

Example:

$ \displaystyle \lim_{x \to 4} \frac{\sqrt{x} - 2}{x-4} = \lim_{x \to 4} \frac{\sqrt{x}-2}{x-4} \cdot \frac{\sqrt{x}+2}{\sqrt{x}+2} = \lim_{x \to 4} \frac{x-4}{(x-4)(\sqrt{x}+2)} = \frac{1}{4} $

Step 4: Using Standard Limits

Apply well-known standard limits such as:

$ \displaystyle \lim_{x \to 0} \frac{\sin x}{x} = 1 $
$ \displaystyle \lim_{x \to 0} (1+x)^{1/x} = e $
$ \displaystyle \lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2} $

Step 5: L'Hôpital's Rule

For indeterminate forms like $ \frac{0}{0} $ or $ \frac{\infty}{\infty} $, take derivatives of numerator and denominator:

Example:

$ \displaystyle \lim_{x \to 0} \frac{\sin x}{x} = \lim_{x \to 0} \frac{\cos x}{1} = 1 $

Step 6: Algebraic Manipulation

Sometimes rearranging terms or splitting fractions helps simplify limits.

Example:

$ \displaystyle \lim_{x \to 0} \frac{x^2 + 3x}{x} = \lim_{x \to 0} (x + 3) = 3 $

Step 7: Important Notes

Always check if the limit can be evaluated directly before using advanced methods.
Factorization and rationalization are the most commonly used techniques in JEE problems.
Keep standard limits handy—they save a lot of computation time.
Check for one-sided limits in case of piecewise functions.

Definition of Integration

Integration is the inverse process of differentiation. It is a fundamental operation in calculus that allows us to find:

The original function from its derivative.
The area under a curve represented by a function.
The accumulation of quantities over an interval, such as distance from velocity or total growth from a rate of change.

If $$F(x)$$ is a function such that $$\frac{d}{dx}F(x) = f(x),$$ then $$F(x)$$ is called an antiderivative or indefinite integral of $$f(x)$$.

$$ \int f(x)\,dx = F(x) + C $$

where:

$$f(x)$$ is called the integrand.
$$dx$$ indicates the variable of integration.
$$C$$ is the constant of integration, representing the fact that there are infinitely many antiderivatives differing by a constant.

Example: If $$f(x) = 2x$$, then

$$ \int 2x \, dx = x^2 + C $$

Integration can also be interpreted geometrically as the area under the curve $$y = f(x)$$ between two points $$x = a$$ and $$x = b$$:

$$ \int_a^b f(x) \, dx = \text{Area under the curve from } x=a \text{ to } x=b $$

Note: Integration is widely used in physics, engineering, and other fields to calculate quantities such as displacement, area, volume, work, and probability.

Types of Integration

Integration can be broadly classified into the following types, depending on the limits and the method used:

1. Indefinite Integral:

Represents the general antiderivative of a function without specifying the limits of integration. It includes an arbitrary constant $$C$$.

$$ \int f(x)\,dx = F(x) + C $$

Example: $$ \int 3x^2 \, dx = x^3 + C $$

2. Definite Integral:

Represents the net area under the curve of a function between two limits $$x = a$$ and $$x = b$$. The result is a number, not a function.

$$ \int_a^b f(x)\,dx = F(b) - F(a) $$

Example: $$ \int_0^2 2x \, dx = [x^2]_0^2 = 4 - 0 = 4 $$

3. Methods of Integration:

Integration by Substitution: Used when a part of the integrand is a derivative of another part.
Integration by Parts: Based on the formula $$\int u \, dv = uv - \int v \, du$$
Partial Fraction Decomposition: Used for rational functions to break them into simpler fractions before integrating.
Trigonometric Substitution: Useful when the integrand contains square roots of quadratic expressions.

Note: Mastery of these types and methods is essential for solving complex JEE calculus problems, including area, volume, and physics applications.

Properties and Formulas of Integration

Integration follows several important properties and formulas that simplify the process of evaluating integrals:

1. Linearity:

$$ \int [f(x) + g(x)] \, dx = \int f(x) \, dx + \int g(x) \, dx $$
$$ \int k \, f(x) \, dx = k \int f(x) \, dx $$ where $$k$$ is a constant

2. Integration of a sum/difference:

The integral of a sum or difference of functions equals the sum or difference of their integrals:

$$ \int [f(x) \pm g(x)] \, dx = \int f(x) \, dx \pm \int g(x) \, dx $$

3. Properties of definite integrals:

$$ \int_a^a f(x) \, dx = 0 $$
$$ \int_a^b f(x) \, dx = - \int_b^a f(x) \, dx $$
$$ \int_a^b [f(x) + g(x)] \, dx = \int_a^b f(x) \, dx + \int_a^b g(x) \, dx $$
Splitting limits: $$ \int_a^b f(x) \, dx = \int_a^c f(x) \, dx + \int_c^b f(x) \, dx $$

4. Standard Integrals:

$$ \int x^n \, dx = \frac{x^{n+1}}{n+1} + C, \quad n \neq -1 $$
$$ \int \frac{dx}{x} = \ln|x| + C $$
$$ \int e^x \, dx = e^x + C $$
$$ \int a^x \, dx = \frac{a^x}{\ln a} + C, \quad a>0, a\neq1 $$
$$ \int \sin x \, dx = -\cos x + C $$
$$ \int \cos x \, dx = \sin x + C $$
$$ \int \sec^2 x \, dx = \tan x + C $$
$$ \int \csc^2 x \, dx = -\cot x + C $$
$$ \int \sec x \tan x \, dx = \sec x + C $$
$$ \int \csc x \cot x \, dx = -\csc x + C $$

Note: Remembering these properties and standard formulas is crucial for solving JEE-level integration problems efficiently.

Important Concepts in Integration

1. Definite vs Indefinite Integrals:

Indefinite Integral: $$ \int f(x)\,dx = F(x) + C $$ It gives the general antiderivative of $$f(x)$$.
Definite Integral: $$ \int_a^b f(x)\,dx = F(b) - F(a) $$ It represents the net area under the curve between $$x=a$$ and $$x=b$$.

2. Geometric Interpretation:

The definite integral $$ \int_a^b f(x)\,dx $$ represents the **signed area** between the curve $$y=f(x)$$ and the x-axis. Positive area is above the x-axis, negative below.

3. Fundamental Theorem of Calculus:

If $$F(x)$$ is an antiderivative of $$f(x)$$, then $$ \int_a^b f(x) \, dx = F(b) - F(a) $$
It connects **differentiation and integration**, showing they are inverse operations.

5. Applications of Integration:

Area under curves: $$\int_a^b f(x) \, dx$$
Area between two curves: $$\int_a^b [f(x)-g(x)] \, dx$$
Volume of solids of revolution: $$ V = \pi \int_a^b [f(x)]^2 \, dx $$
Work done by a variable force: $$ W = \int F(x) \, dx $$
Average value of a function: $$ f_{\text{avg}} = \frac{1}{b-a} \int_a^b f(x)\, dx $$

6. Important Tips:

1. Always check **domain restrictions** before integrating functions like $$\frac{1}{x}, \sqrt{x}, \ln x$$.
2. Factorize polynomials first if possible.
3. Look for substitution that converts complex functions into standard forms.
4. Use symmetry properties for definite integrals: if $$f(x)$$ is even or odd over $$[-a,a]$$.

Trigonometric Ratios and Identities

Trigonometric ratios relate the angles of a right-angled triangle to the ratios of its sides. For a right triangle with angle $\theta$, hypotenuse $h$, opposite side $o$, and adjacent side $a$:

$$ \sin \theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{o}{h} $$
$$ \cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{a}{h} $$
$$ \tan \theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{o}{a} $$
$$ \csc \theta = \frac{1}{\sin \theta} = \frac{h}{o} $$
$$ \sec \theta = \frac{1}{\cos \theta} = \frac{h}{a} $$
$$ \cot \theta = \frac{1}{\tan \theta} = \frac{a}{o} $$

Fundamental Trigonometric Identities:

Pythagorean Identity: $$ \sin^2 \theta + \cos^2 \theta = 1 $$
Derived identities:
$$ 1 + \tan^2 \theta = \sec^2 \theta $$ $$ 1 + \cot^2 \theta = \csc^2 \theta $$
Reciprocal relations: $$ \csc \theta = \frac{1}{\sin \theta},\; \sec \theta = \frac{1}{\cos \theta},\; \cot \theta = \frac{1}{\tan \theta} $$
Quotient relations: $$ \tan \theta = \frac{\sin \theta}{\cos \theta},\; \cot \theta = \frac{\cos \theta}{\sin \theta} $$

Note: These basic ratios and identities are the foundation for solving all trigonometric equations and proving advanced identities in JEE. Memorizing them thoroughly will save time in problem-solving.

Definition and Concept of Trigonometric Ratios

Trigonometry is a branch of mathematics that studies the relationships between the sides and angles of triangles. The basic idea is to associate each angle with specific ratios of the sides of a right-angled triangle. These ratios are called trigonometric ratios.

Consider a right-angled triangle ABC, with:

Angle $ \theta $ at vertex A
Hypotenuse $ h $ (opposite the right angle)
Side opposite $ \theta $: $ o $
Side adjacent to $ \theta $: $ a $

Then, the six primary trigonometric ratios are defined as:

$ \sin \theta = \frac{o}{h} $ — sine: ratio of opposite side to hypotenuse
$ \cos \theta = \frac{a}{h} $ — cosine: ratio of adjacent side to hypotenuse
$ \tan \theta = \frac{o}{a} $ — tangent: ratio of opposite side to adjacent side
$ \csc \theta = \frac{h}{o} $ — cosecant: reciprocal of sine
$ \sec \theta = \frac{h}{a} $ — secant: reciprocal of cosine
$ \cot \theta = \frac{a}{o} $ — cotangent: reciprocal of tangent

Notes: 1. These ratios are defined only for angles in a right triangle (0° < θ < 90°).
2. Trigonometric ratios can also be extended to any angle using the unit circle.
3. These ratios are fundamental for solving triangle problems, proving identities, and solving trigonometric equations in JEE-level mathematics.

Extension of Trigonometric Ratios to Any Angle

Trigonometric ratios can be defined for **any angle** (not just acute angles) using the **unit circle**. A unit circle is a circle of radius 1 centered at the origin of the coordinate plane.

Consider a point $P(x, y)$ on the unit circle, corresponding to an angle $\theta$ measured from the positive x-axis:

$ \sin \theta = y $ — vertical coordinate
$ \cos \theta = x $ — horizontal coordinate
$ \tan \theta = \frac{y}{x} $, provided $ x \neq 0 $
$ \csc \theta = \frac{1}{y} $, provided $ y \neq 0 $
$ \sec \theta = \frac{1}{x} $, provided $ x \neq 0 $
$ \cot \theta = \frac{x}{y} $, provided $ y \neq 0 $

Notes: 1. This definition allows trigonometric functions to be used for **angles greater than 90°**, **negative angles**, and **angles beyond 360°**.
2. The **sign of the ratio** depends on the quadrant:
• Quadrant I: all positive • Quadrant II: sine positive, others negative • Quadrant III: tangent positive, others negative • Quadrant IV: cosine positive, others negative
3. This unit-circle approach is essential for proving **trigonometric identities** and solving **JEE-level equations**.

Fundamental Trigonometric Identities

1. Pythagorean Identities:

$\sin^2 \theta + \cos^2 \theta = 1$
$1 + \tan^2 \theta = \sec^2 \theta$
$1 + \cot^2 \theta = \csc^2 \theta$

2. Reciprocal Identities:

$\csc \theta = 1 / \sin \theta$
$\sec \theta = 1 / \cos \theta$
$\cot \theta = 1 / \tan \theta$

3. Quotient Identities:

$\tan \theta = \sin \theta / \cos \theta$
$\cot \theta = \cos \theta / \sin \theta$

4. Co-function Identities:

$\sin (90^\circ - \theta) = \cos \theta$
$\cos (90^\circ - \theta) = \sin \theta$
$\tan (90^\circ - \theta) = \cot \theta$
$\cot (90^\circ - \theta) = \tan \theta$
$\sec (90^\circ - \theta) = \csc \theta$
$\csc (90^\circ - \theta) = \sec \theta$

Notes: These identities form the **foundation of all trigonometric problem-solving** in JEE. Memorize them thoroughly for proving advanced identities and solving equations efficiently.

Angle Sum and Difference Formulas

1. Sine Formulas:

$\sin(A + B) = \sin A \cos B + \cos A \sin B$
$\sin(A - B) = \sin A \cos B - \cos A \sin B$

2. Cosine Formulas:

$\cos(A + B) = \cos A \cos B - \sin A \sin B$
$\cos(A - B) = \cos A \cos B + \sin A \sin B$

3. Tangent Formulas:

$\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$
$\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$

Notes: These formulas are essential for solving trigonometric equations, proving identities, and calculating exact values of angles in JEE problems.

Double Angle and Half Angle Formulas

1. Double Angle Formulas:

$\sin 2A = 2 \sin A \cos A$
$\cos 2A = \cos^2 A - \sin^2 A = 2 \cos^2 A - 1 = 1 - 2 \sin^2 A$
$\tan 2A = \frac{2 \tan A}{1 - \tan^2 A}$

2. Half Angle Formulas:

$\sin \frac{A}{2} = \pm \sqrt{\frac{1 - \cos A}{2}}$
$\cos \frac{A}{2} = \pm \sqrt{\frac{1 + \cos A}{2}}$
$\tan \frac{A}{2} = \pm \sqrt{\frac{1 - \cos A}{1 + \cos A}} = \frac{\sin A}{1 + \cos A} = \frac{1 - \cos A}{\sin A}$

Notes: These formulas are extensively used in JEE-level trigonometric equations, simplifying expressions, and calculating exact values of angles.

Product-to-Sum and Sum-to-Product Formulas

1. Product-to-Sum Formulas:

$\sin A \sin B = \frac{1}{2} [\cos(A - B) - \cos(A + B)]$
$\cos A \cos B = \frac{1}{2} [\cos(A - B) + \cos(A + B)]$
$\sin A \cos B = \frac{1}{2} [\sin(A + B) + \sin(A - B)]$

2. Sum-to-Product Formulas:

$\sin A + \sin B = 2 \sin\frac{A + B}{2} \cos\frac{A - B}{2}$
$\sin A - \sin B = 2 \cos\frac{A + B}{2} \sin\frac{A - B}{2}$
$\cos A + \cos B = 2 \cos\frac{A + B}{2} \cos\frac{A - B}{2}$
$\cos A - \cos B = -2 \sin\frac{A + B}{2} \sin\frac{A - B}{2}$

Notes: These formulas are extremely useful for simplifying complex trigonometric expressions, evaluating integrals, and solving advanced JEE-level problems.

Important Trigonometric Values and Special Angles

1. Standard Angle Values (in degrees and radians):

$0^\circ (0)$: $\sin 0 = 0$, $\cos 0 = 1$, $\tan 0 = 0$
$30^\circ (\pi/6)$: $\sin 30^\circ = 1/2$, $\cos 30^\circ = \sqrt{3}/2$, $\tan 30^\circ = 1/\sqrt{3}$
$45^\circ (\pi/4)$: $\sin 45^\circ = \cos 45^\circ = 1/\sqrt{2}$, $\tan 45^\circ = 1$
$60^\circ (\pi/3)$: $\sin 60^\circ = \sqrt{3}/2$, $\cos 60^\circ = 1/2$, $\tan 60^\circ = \sqrt{3}$
$90^\circ (\pi/2)$: $\sin 90^\circ = 1$, $\cos 90^\circ = 0$, $\tan 90^\circ$ undefined

2. Negative and Quadrantal Angles:

$\sin(-\theta) = -\sin \theta$, $\cos(-\theta) = \cos \theta$, $\tan(-\theta) = -\tan \theta$
Quadrantal angles (0°, 90°, 180°, 270°) have sine or cosine equal to 0 or ±1

Notes: Memorizing these standard values is crucial for solving trigonometric equations, evaluating integrals, and quickly checking answers in JEE problems.

Trigonometric Equations and Their General Solutions

1. Sine Equations:

Equation: $\sin \theta = k$
General Solution: $\theta = \sin^{-1} k + 360^\circ n$ or $\theta = 180^\circ - \sin^{-1} k + 360^\circ n$, $n \in \mathbb{Z}$

2. Cosine Equations:

Equation: $\cos \theta = k$
General Solution: $\theta = \cos^{-1} k + 360^\circ n$ or $\theta = -\cos^{-1} k + 360^\circ n$, $n \in \mathbb{Z}$

3. Tangent Equations:

Equation: $\tan \theta = k$
General Solution: $\theta = \tan^{-1} k + 180^\circ n$, $n \in \mathbb{Z}$

Notes: Understanding these general solutions is critical for solving trigonometric equations in JEE, including multiple-angle equations, and for finding all possible solutions in a given interval.

Inverse Trigonometric Functions

Inverse trigonometric functions are the inverse functions of the trigonometric functions. They are used to obtain an angle from any of the angle's trigonometric ratios.

$$ \sin^{-1} x \quad \text{or} \quad \arcsin x $$
$$ \cos^{-1} x \quad \text{or} \quad \arccos x $$
$$ \tan^{-1} x \quad \text{or} \quad \arctan x $$
$$ \csc^{-1} x \quad \text{or} \quad \operatorname{arccsc} x $$
$$ \sec^{-1} x \quad \text{or} \quad \operatorname{arcsec} x $$
$$ \cot^{-1} x \quad \text{or} \quad \operatorname{arccot} x $$

Domain and Range of Inverse Trigonometric Functions:

$$ \sin^{-1} x : \text{Domain} = [-1, 1], \text{Range} = \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] $$
$$ \cos^{-1} x : \text{Domain} = [-1, 1], \text{Range} = [0, \pi] $$
$$ \tan^{-1} x : \text{Domain} = \mathbb{R}, \text{Range} = \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) $$
$$ \csc^{-1} x : \text{Domain} = (-\infty, -1] \cup [1, \infty), \text{Range} = \left[-\frac{\pi}{2}, 0\right) \cup \left(0, \frac{\pi}{2}\right] $$
$$ \sec^{-1} x : \text{Domain} = (-\infty, -1] \cup [1, \infty), \text{Range} = \left[0, \frac{\pi}{2}\right) \cup \left(\frac{\pi}{2}, \pi\right] $$
$$ \cot^{-1} x : \text{Domain} = \mathbb{R}, \text{Range} = (0, \pi) $$

Principal Values:

The inverse trigonometric functions are multi-valued. To make them single-valued, we restrict their ranges to specific intervals called principal value branches.

Note: Inverse trigonometric functions are crucial for solving equations involving trigonometric functions and appear frequently in JEE problems, especially in integration and differentiation.

Definition and Concept of Inverse Trigonometric Functions

Inverse trigonometric functions are the inverses of trigonometric functions. Since trigonometric functions are periodic, they are not one-to-one, so their inverses are defined only on restricted domains where they are one-to-one.

For example, the sine function:

Sine function: $y = \sin x$ is not one-to-one on its entire domain
We restrict the domain to $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ to make it one-to-one
Then we define the inverse: $x = \sin^{-1} y$ or $x = \arcsin y$

This means: If $y = \sin x$ with $x \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, then $x = \sin^{-1} y$.

Notes: 1. The notation $\sin^{-1} x$ does NOT mean $(\sin x)^{-1} = \frac{1}{\sin x}$.
2. The range of each inverse trigonometric function is carefully chosen to ensure it's single-valued.
3. These functions are essential for solving trigonometric equations and appear in calculus problems.

Graphs of Inverse Trigonometric Functions

The graphs of inverse trigonometric functions can be obtained by reflecting the graphs of the corresponding trigonometric functions about the line $y = x$, but only for the restricted domains.

Key Properties of Graphs:

$\sin^{-1} x$: Increasing function, passes through origin
$\cos^{-1} x$: Decreasing function, passes through $(1, 0)$ and $(-1, \pi)$
$\tan^{-1} x$: Increasing function, has horizontal asymptotes at $y = \pm \frac{\pi}{2}$
$\cot^{-1} x$: Decreasing function, has horizontal asymptotes at $y = 0$ and $y = \pi$
$\sec^{-1} x$ and $\csc^{-1} x$: Defined only for $|x| \geq 1$

Notes: Understanding these graphs helps visualize the domain and range restrictions and is useful for solving inequalities involving inverse trigonometric functions.

Properties of Inverse Trigonometric Functions

1. Basic Properties:

$\sin(\sin^{-1} x) = x$ for $x \in [-1, 1]$
$\sin^{-1}(\sin x) = x$ for $x \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$
$\cos(\cos^{-1} x) = x$ for $x \in [-1, 1]$
$\cos^{-1}(\cos x) = x$ for $x \in [0, \pi]$
$\tan(\tan^{-1} x) = x$ for $x \in \mathbb{R}$
$\tan^{-1}(\tan x) = x$ for $x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$

2. Properties with Negative Arguments:

$\sin^{-1}(-x) = -\sin^{-1} x$
$\cos^{-1}(-x) = \pi - \cos^{-1} x$
$\tan^{-1}(-x) = -\tan^{-1} x$
$\csc^{-1}(-x) = -\csc^{-1} x$
$\sec^{-1}(-x) = \pi - \sec^{-1} x$
$\cot^{-1}(-x) = \pi - \cot^{-1} x$

3. Reciprocal Relations:

$\sin^{-1}\left(\frac{1}{x}\right) = \csc^{-1} x$ for $|x| \geq 1$
$\cos^{-1}\left(\frac{1}{x}\right) = \sec^{-1} x$ for $|x| \geq 1$
$\tan^{-1}\left(\frac{1}{x}\right) = \cot^{-1} x$ for $x > 0$
$\tan^{-1}\left(\frac{1}{x}\right) = \cot^{-1} x - \pi$ for $x < 0$

Notes: These properties are essential for simplifying expressions and solving equations involving inverse trigonometric functions in JEE problems.

Sum and Difference Formulas

1. Addition Formulas:

$\sin^{-1} x + \sin^{-1} y = \sin^{-1}(x\sqrt{1-y^2} + y\sqrt{1-x^2})$ for $x^2+y^2 \leq 1$ or $xy < 0$
$\sin^{-1} x + \sin^{-1} y = \pi - \sin^{-1}(x\sqrt{1-y^2} + y\sqrt{1-x^2})$ for $x^2+y^2 > 1$ and $x,y > 0$
$\sin^{-1} x + \sin^{-1} y = -\pi - \sin^{-1}(x\sqrt{1-y^2} + y\sqrt{1-x^2})$ for $x^2+y^2 > 1$ and $x,y < 0$
$\cos^{-1} x + \cos^{-1} y = \cos^{-1}(xy - \sqrt{1-x^2}\sqrt{1-y^2})$ for $x+y \geq 0$
$\cos^{-1} x + \cos^{-1} y = 2\pi - \cos^{-1}(xy - \sqrt{1-x^2}\sqrt{1-y^2})$ for $x+y < 0$
$\tan^{-1} x + \tan^{-1} y = \tan^{-1}\left(\frac{x+y}{1-xy}\right)$ for $xy < 1$
$\tan^{-1} x + \tan^{-1} y = \pi + \tan^{-1}\left(\frac{x+y}{1-xy}\right)$ for $x > 0, y > 0, xy > 1$
$\tan^{-1} x + \tan^{-1} y = -\pi + \tan^{-1}\left(\frac{x+y}{1-xy}\right)$ for $x < 0, y < 0, xy> 1$

2. Subtraction Formulas:

$\sin^{-1} x - \sin^{-1} y = \sin^{-1}(x\sqrt{1-y^2} - y\sqrt{1-x^2})$
$\cos^{-1} x - \cos^{-1} y = \cos^{-1}(xy + \sqrt{1-x^2}\sqrt{1-y^2})$ for $x \leq y$
$\cos^{-1} x - \cos^{-1} y = -\cos^{-1}(xy + \sqrt{1-x^2}\sqrt{1-y^2})$ for $x > y$
$\tan^{-1} x - \tan^{-1} y = \tan^{-1}\left(\frac{x-y}{1+xy}\right)$ for $xy > -1$
$\tan^{-1} x - \tan^{-1} y = \pi + \tan^{-1}\left(\frac{x-y}{1+xy}\right)$ for $x > 0, y < 0, xy < -1$
$\tan^{-1} x - \tan^{-1} y = -\pi + \tan^{-1}\left(\frac{x-y}{1+xy}\right)$ for $x < 0, y> 0, xy < -1$

Notes: These formulas are crucial for simplifying complex expressions and solving equations involving sums and differences of inverse trigonometric functions in JEE problems.

Multiple Angle Formulas

1. Double Angle Formulas:

$2\sin^{-1} x = \sin^{-1}(2x\sqrt{1-x^2})$ for $|x| \leq \frac{1}{\sqrt{2}}$
$2\sin^{-1} x = \pi - \sin^{-1}(2x\sqrt{1-x^2})$ for $\frac{1}{\sqrt{2}} < x \leq 1$
$2\sin^{-1} x = -\pi - \sin^{-1}(2x\sqrt{1-x^2})$ for $-1 \leq x < -\frac{1}{\sqrt{2}}$
$2\cos^{-1} x = \cos^{-1}(2x^2 - 1)$ for $0 \leq x \leq 1$
$2\cos^{-1} x = 2\pi - \cos^{-1}(2x^2 - 1)$ for $-1 \leq x < 0$
$2\tan^{-1} x = \tan^{-1}\left(\frac{2x}{1-x^2}\right)$ for $|x| < 1$
$2\tan^{-1} x = \pi + \tan^{-1}\left(\frac{2x}{1-x^2}\right)$ for $x > 1$
$2\tan^{-1} x = -\pi + \tan^{-1}\left(\frac{2x}{1-x^2}\right)$ for $x < -1$

2. Triple Angle Formulas:

$3\sin^{-1} x = \sin^{-1}(3x - 4x^3)$ for $|x| \leq \frac{1}{2}$
$3\cos^{-1} x = \cos^{-1}(4x^3 - 3x)$ for $\frac{1}{2} \leq x \leq 1$
$3\tan^{-1} x = \tan^{-1}\left(\frac{3x - x^3}{1 - 3x^2}\right)$ for $|x| < \frac{1}{\sqrt{3}}$

Notes: These formulas are useful for solving equations where arguments of inverse trigonometric functions are multiples of each other, a common type of problem in JEE.

Conversion Between Inverse Trigonometric Functions

Conversion Formulas:

$\sin^{-1} x = \cos^{-1}\sqrt{1-x^2} = \tan^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right)$ for $0 \leq x \leq 1$
$\sin^{-1} x = -\cos^{-1}\sqrt{1-x^2} = \tan^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right)$ for $-1 \leq x < 0$
$\cos^{-1} x = \sin^{-1}\sqrt{1-x^2} = \tan^{-1}\left(\frac{\sqrt{1-x^2}}{x}\right)$ for $0 < x \leq 1$
$\cos^{-1} x = \pi - \sin^{-1}\sqrt{1-x^2} = \pi + \tan^{-1}\left(\frac{\sqrt{1-x^2}}{x}\right)$ for $-1 \leq x < 0$
$\tan^{-1} x = \sin^{-1}\left(\frac{x}{\sqrt{1+x^2}}\right) = \cos^{-1}\left(\frac{1}{\sqrt{1+x^2}}\right)$ for $x \geq 0$
$\tan^{-1} x = -\sin^{-1}\left(\frac{x}{\sqrt{1+x^2}}\right) = \pi - \cos^{-1}\left(\frac{1}{\sqrt{1+x^2}}\right)$ for $x < 0$

Special Cases:

$\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}$ for $x \in [-1, 1]$
$\tan^{-1} x + \cot^{-1} x = \frac{\pi}{2}$ for $x \in \mathbb{R}$
$\sec^{-1} x + \csc^{-1} x = \frac{\pi}{2}$ for $|x| \geq 1$

Notes: These conversion formulas are extremely useful for simplifying complex expressions and solving equations where different inverse trigonometric functions appear together.

Important Identities and Formulas

Key Identities:

$\tan^{-1} x + \tan^{-1} y + \tan^{-1} z = \tan^{-1}\left(\frac{x+y+z-xyz}{1-xy-yz-zx}\right)$
$\sin^{-1} x \pm \sin^{-1} y = \sin^{-1}(x\sqrt{1-y^2} \pm y\sqrt{1-x^2})$
$\cos^{-1} x \pm \cos^{-1} y = \cos^{-1}(xy \mp \sqrt{1-x^2}\sqrt{1-y^2})$
$\tan^{-1} x \pm \tan^{-1} y = \tan^{-1}\left(\frac{x \pm y}{1 \mp xy}\right)$
$\sin^{-1} x = 2\tan^{-1}\left(\frac{x}{1+\sqrt{1-x^2}}\right)$
$\cos^{-1} x = 2\tan^{-1}\left(\frac{\sqrt{1-x^2}}{1+x}\right)$

Special Values:

$\sin^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{6}$
$\cos^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{3}$
$\tan^{-1}(1) = \frac{\pi}{4}$
$\sin^{-1}\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{3}$
$\cos^{-1}\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{6}$
$\tan^{-1}(\sqrt{3}) = \frac{\pi}{3}$

Notes: Memorizing these identities and special values is crucial for quickly solving JEE problems involving inverse trigonometric functions.

Solving Equations with Inverse Trigonometric Functions

To solve equations involving inverse trigonometric functions:

Apply the appropriate trigonometric function to both sides
Use algebraic techniques to solve the resulting equation
Check that the solutions lie within the domain of the original inverse trigonometric functions
Verify solutions by substituting back into the original equation

Common Equation Types:

Equations of the form: $\sin^{-1} x + \sin^{-1} y = \text{constant}$
Equations of the form: $\tan^{-1} x + \tan^{-1} y = \tan^{-1} z$
Equations involving multiple inverse trigonometric functions
Equations where the argument is an expression in $x$

Notes: Always verify that your solutions satisfy the domain restrictions of the inverse trigonometric functions. Extraneous solutions are common in these types of equations.

Height and Distance - Applications of Trigonometry

Height and Distance problems involve using trigonometric ratios to find unknown heights, distances, or angles in real-world scenarios. These are practical applications of trigonometry used in surveying, navigation, and engineering.

Basic Concepts and Terminology:

Line of Sight: The straight line from the eye of the observer to the point being viewed
Angle of Elevation: The angle between the horizontal and the line of sight when viewing an object above the horizontal
Angle of Depression: The angle between the horizontal and the line of sight when viewing an object below the horizontal
Horizontal Line: A line parallel to the ground level

Note: Height and Distance problems test the application of trigonometric concepts in practical situations. They frequently appear in JEE examinations and require strong visualization skills.

Angle of Elevation and Depression

Angle of Elevation:

When an observer looks at an object situated above the horizontal level, the angle between the line of sight and the horizontal line is called the angle of elevation.

If the angle of elevation is θ, then:

$$ \tan \theta = \frac{\text{Height of object}}{\text{Distance from object}} $$

Angle of Depression:

When an observer looks at an object situated below the horizontal level, the angle between the line of sight and the horizontal line is called the angle of depression.

If the angle of depression is θ, then:

$$ \tan \theta = \frac{\text{Depth of object}}{\text{Distance from object}} $$

Important: The angle of elevation and angle of depression between two points are equal when the points are at the same horizontal level.

Basic Formulas and Relationships

Key Trigonometric Relationships:

For angle of elevation θ:

$$ \sin \theta = \frac{\text{Height}}{\text{Hypotenuse}} $$

$$ \cos \theta = \frac{\text{Distance}}{\text{Hypotenuse}} $$

$$ \tan \theta = \frac{\text{Height}}{\text{Distance}} $$
For angle of depression θ:

$$ \sin \theta = \frac{\text{Depth}}{\text{Hypotenuse}} $$

$$ \cos \theta = \frac{\text{Distance}}{\text{Hypotenuse}} $$

$$ \tan \theta = \frac{\text{Depth}}{\text{Distance}} $$

Pythagorean Theorem Applications:

In right-angled triangles formed in height and distance problems:

$$ (\text{Hypotenuse})^2 = (\text{Height})^2 + (\text{Distance})^2 $$

Notes: Always identify the right-angled triangle in the problem and apply the appropriate trigonometric ratio based on the given information.

Single Object Problems

1. Basic Single Object Problems:

These involve finding the height or distance of a single object using one angle of elevation/depression.

Example:

A person observes the top of a tower at an angle of elevation of 30°. If the person is 50 m away from the tower, find the height of the tower.

$$ \tan 30° = \frac{h}{50} \Rightarrow h = 50 \times \frac{1}{\sqrt{3}} = \frac{50}{\sqrt{3}} \text{ m} $$

2. Problems with Observer Height:

When the observer's eye level is at some height above ground.

Example:

A 1.5 m tall person observes the top of a building at 45° elevation from 20 m away. Find building height.

$$ \tan 45° = \frac{h-1.5}{20} \Rightarrow h = 20 + 1.5 = 21.5 \text{ m} $$

Notes: Remember to account for the observer's height when it's given. The actual height of the object = calculated height + observer's height.

Two Position Problems

1. Two Positions on Same Side:

Observer moves toward or away from the object, measuring angles from two different positions.

Example:

From a point on the ground, the angles of elevation of bottom and top of a tower are 45° and 60° respectively. If the tower is 20 m high, find the distance of the point from the tower.

Let distance = x, then:

$$ \tan 45° = \frac{h}{x}, \quad \tan 60° = \frac{h+20}{x} $$

Solving: $$ x = \frac{20}{\sqrt{3}-1} \text{ m} $$

2. Two Positions on Opposite Sides:

Observer takes measurements from two points on opposite sides of the object.

Example:

From two points A and B on opposite sides of a tower, the angles of elevation are 30° and 45° respectively. If AB = 100 m, find tower height.

Let height = h, distances = x and 100-x

$$ \tan 30° = \frac{h}{x}, \quad \tan 45° = \frac{h}{100-x} $$

Solving: $$ h = \frac{100}{\sqrt{3}+1} \text{ m} $$

Notes: In two-position problems, create two equations and solve them simultaneously. Pay attention to whether the positions are on the same side or opposite sides.

Moving Observer Problems

1. Horizontal Movement Problems:

Observer moves horizontally toward or away from the object.

Example:

A person walking toward a tower observes the angle of elevation changes from 30° to 60° after walking 40 m. Find tower height.

Let initial distance = x, height = h

$$ \tan 30° = \frac{h}{x}, \quad \tan 60° = \frac{h}{x-40} $$

Solving: $$ h = 20\sqrt{3} \text{ m} $$

2. Vertical Movement Problems:

Observer moves vertically (e.g., in an elevator or climbing).

Example:

From the top of a 50 m building, the angle of depression of a car is 45°. After the observer climbs 10 m higher, the angle becomes 30°. Find the car's distance from building.

Let distance = x

$$ \tan 45° = \frac{50}{x}, \quad \tan 30° = \frac{60}{x} $$

Solving: $$ x = 50 \text{ m} $$

Notes: In moving observer problems, track the change in position carefully. The distance moved becomes a key parameter in setting up equations.

Two Objects Problems

1. Two Objects in Same Plane:

Finding relationship between heights/distances of two objects.

Example:

The angles of elevation of two towers from a point between them are 60° and 30°. If the towers are 100 m apart and one is taller than the other, find their heights.

Let distances be x and 100-x

$$ h_1 = x\tan 60°, \quad h_2 = (100-x)\tan 30° $$

2. Objects in Different Planes:

Objects at different heights or positions.

Example:

From the top of a tower, the angles of depression of two objects on the same side are 45° and 30°. If the objects are 100 m apart, find tower height.

Let height = h, distances = x and x+100

$$ \tan 45° = \frac{h}{x}, \quad \tan 30° = \frac{h}{x+100} $$

Notes: In two-object problems, you'll typically need to create equations for each object and find relationships between them.

Shadow Problems

Shadow Length Problems:

These involve the relationship between an object's height, its shadow length, and the sun's angle of elevation.

Example:

If the length of the shadow of a tower is increasing at 2 m/s when the sun's elevation is 45°, find the rate at which the angle of elevation is changing.

Let height = h, shadow length = x, angle = θ

$$ \tan \theta = \frac{h}{x} \Rightarrow \sec^2 \theta \cdot \frac{d\theta}{dt} = -\frac{h}{x^2} \cdot \frac{dx}{dt} $$

Rate of Change Problems:

These involve differentiation and related rates.

Example:

A man 2 m tall walks away from a lamp post 5 m high at 1.5 m/s. Find the rate at which his shadow length increases.

Using similar triangles and differentiation.

Notes: Shadow problems often involve similar triangles and may require calculus for rate of change problems.

Problem Solving Strategy

Systematic Approach to Height and Distance Problems:

Draw a clear diagram showing all given information
Identify right-angled triangles in the diagram
Label all known and unknown quantities
Choose appropriate trigonometric ratios based on given information
Set up equations using trigonometric relationships
Solve the equations systematically
Verify your answer makes sense in the context

Common Mistakes to Avoid:

Forgetting to account for observer's height
Confusing angle of elevation with angle of depression
Using wrong trigonometric ratios
Not converting angles to radians when required
Forgetting to include units in final answer

Final Tip: Practice drawing clear diagrams and visualizing the scenarios. This is the most important skill for solving height and distance problems efficiently in JEE examinations.

Straight Lines - Coordinate Geometry

Straight Lines are fundamental concepts in coordinate geometry that represent the simplest form of geometric figures. They play a crucial role in JEE Mathematics with applications in various domains.

Basic Forms of Straight Lines:

General Form: $$ Ax + By + C = 0 $$
Slope-Intercept Form: $$ y = mx + c $$
Point-Slope Form: $$ y - y_1 = m(x - x_1) $$
Two-Point Form: $$ \frac{y - y_1}{y_2 - y_1} = \frac{x - x_1}{x_2 - x_1} $$
Intercept Form: $$ \frac{x}{a} + \frac{y}{b} = 1 $$
Normal Form: $$ x\cos\alpha + y\sin\alpha = p $$

Note: Straight lines form the foundation for more complex coordinate geometry topics. Mastery of line equations and their properties is essential for JEE success.

Slope of a Line

The slope (m) of a line measures its steepness and direction.

Formulas for Slope:

Using two points $(x_1, y_1)$ and $(x_2, y_2)$: $$ m = \frac{y_2 - y_1}{x_2 - x_1} $$
Angle with x-axis: $$ m = \tan\theta $$
From general equation $Ax + By + C = 0$: $$ m = -\frac{A}{B} $$
From intercept form: $$ m = -\frac{b}{a} $$

Special Cases:

Horizontal line: $m = 0$
Vertical line: $m = \infty$ (undefined)
Lines parallel to x-axis: $m = 0$
Lines parallel to y-axis: $m = \infty$

Angle Between Two Lines

For two lines with slopes $m_1$ and $m_2$, the angle between them is given by:

$$ \tan\theta = \left| \frac{m_1 - m_2}{1 + m_1m_2} \right| $$

Special Cases:

Parallel Lines: $m_1 = m_2$ ⇒ $\theta = 0^\circ$
Perpendicular Lines: $m_1m_2 = -1$ ⇒ $\theta = 90^\circ$
If lines are $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$:

Parallel if: $$ \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} $$

Perpendicular if: $$ a_1a_2 + b_1b_2 = 0 $$

Distance Formulas

1. Distance Between Two Points:

$$ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} $$

2. Distance from a Point to a Line:

Distance from point $(x_1, y_1)$ to line $Ax + By + C = 0$:

$$ d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} $$

3. Distance Between Parallel Lines:

Distance between $Ax + By + C_1 = 0$ and $Ax + By + C_2 = 0$:

$$ d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}} $$

Section Formulas

1. Internal Division:

Point dividing $(x_1, y_1)$ and $(x_2, y_2)$ in ratio m:n internally:

$$ P = \left( \frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n} \right) $$

2. External Division:

Point dividing $(x_1, y_1)$ and $(x_2, y_2)$ in ratio m:n externally:

$$ P = \left( \frac{mx_2 - nx_1}{m-n}, \frac{my_2 - ny_1}{m-n} \right) $$

3. Midpoint Formula:

Midpoint of $(x_1, y_1)$ and $(x_2, y_2)$:

$$ M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) $$

Family of Lines

1. Family Through Intersection of Two Lines:

Lines through intersection of $L_1=0$ and $L_2=0$:

$$ L_1 + \lambda L_2 = 0 $$

2. Family of Parallel Lines:

Lines parallel to $Ax + By + C = 0$:

$$ Ax + By + k = 0 $$

3. Family of Perpendicular Lines:

Lines perpendicular to $Ax + By + C = 0$:

$$ Bx - Ay + k = 0 $$

Concurrency of Lines

Condition for Concurrency:

Three lines $a_1x + b_1y + c_1 = 0$, $a_2x + b_2y + c_2 = 0$, $a_3x + b_3y + c_3 = 0$ are concurrent if:

$$ \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix} = 0 $$

Point of Concurrency:

Solve any two equations simultaneously to find the point of intersection.

Reflection and Image

1. Reflection of a Point:

Reflection of point $(x_1, y_1)$ about line $ax + by + c = 0$:

$$ \frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{-2(ax_1 + by_1 + c)}{a^2 + b^2} $$

2. Foot of Perpendicular:

Foot of perpendicular from $(x_1, y_1)$ to line $ax + by + c = 0$:

$$ \frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{-(ax_1 + by_1 + c)}{a^2 + b^2} $$

Pair of Straight Lines

General Second Degree Equation:

$$ ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0 $$

Conditions:

Represents pair of lines if: $$ \begin{vmatrix} a & h & g \\ h & b & f \\ g & f & c \end{vmatrix} = 0 $$
Angle between lines: $$ \tan\theta = \frac{2\sqrt{h^2 - ab}}{a + b} $$
Lines are perpendicular if: $a + b = 0$
Lines are parallel if: $h^2 = ab$

Important Results and Formulas

1. Area of Triangle:

Area of triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, $(x_3, y_3)$:

$$ \Delta = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| $$

2. Collinearity of Points:

Three points are collinear if:

$$ \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} = 0 $$

3. Centroid of Triangle:

$$ G = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$

Important: Practice all forms of line equations and their conversions. Straight lines frequently appear in JEE in combination with circles, parabolas, and other coordinate geometry topics.

Parabola - Conic Sections

A parabola is the set of all points in a plane that are equidistant from a fixed point (focus) and a fixed line (directrix). It is one of the four conic sections and has important applications in physics and engineering.

Standard Forms of Parabola:

Right Opening: $$ y^2 = 4ax $$
Left Opening: $$ y^2 = -4ax $$
Upward Opening: $$ x^2 = 4ay $$
Downward Opening: $$ x^2 = -4ay $$

Note: Parabola is a frequently tested topic in JEE with applications in coordinate geometry, calculus, and physics problems involving projectile motion.

Key Elements of Parabola

Important Elements (for y² = 4ax):

Focus: $S = (a, 0)$
Directrix: $x = -a$
Axis: x-axis ($y = 0$)
Vertex: $V = (0, 0)$
Length of Latus Rectum: $4a$
Endpoints of Latus Rectum: $L = (a, 2a)$, $L' = (a, -2a)$
Focal Distance: Distance from focus to any point on parabola

Parametric Coordinates:

For $y^2 = 4ax$: $$ x = at^2, \quad y = 2at $$
Parameter $t$ represents the slope of tangent at point $(at^2, 2at)$

Standard Equations and Properties

Equation	Focus	Directrix	Axis	Latus Rectum
$y^2 = 4ax$	$(a, 0)$	$x = -a$	$y = 0$	$4a$
$y^2 = -4ax$	$(-a, 0)$	$x = a$	$y = 0$	$4a$
$x^2 = 4ay$	$(0, a)$	$y = -a$	$x = 0$	$4a$
$x^2 = -4ay$	$(0, -a)$	$y = a$	$x = 0$	$4a$

Tangent to Parabola

Forms of Tangent:

Point Form: For $y^2 = 4ax$ at $(x_1, y_1)$: $$ yy_1 = 2a(x + x_1) $$
Parametric Form: For $y^2 = 4ax$ at $(at^2, 2at)$: $$ ty = x + at^2 $$
Slope Form: For $y^2 = 4ax$ with slope m: $$ y = mx + \frac{a}{m} $$

Condition of Tangency:

For $y^2 = 4ax$: Line $y = mx + c$ is tangent if $c = \frac{a}{m}$
For $x^2 = 4ay$: Line $y = mx + c$ is tangent if $c = -am^2$
Point of contact for $y = mx + \frac{a}{m}$ on $y^2 = 4ax$ is $\left(\frac{a}{m^2}, \frac{2a}{m}\right)$

Normal to Parabola

Forms of Normal:

Point Form: For $y^2 = 4ax$ at $(x_1, y_1)$: $$ y - y_1 = -\frac{y_1}{2a}(x - x_1) $$
Parametric Form: For $y^2 = 4ax$ at $(at^2, 2at)$: $$ y = tx - 2at - at^3 $$
Slope Form: For $y^2 = 4ax$ with slope m: $$ y = mx - 2am - am^3 $$

Important Properties:

Three normals can be drawn from a point to a parabola
The algebraic sum of slopes of three concurrent normals is zero
The algebraic sum of ordinates of feet of three concurrent normals is zero

Chord of Parabola

Forms of Chord:

Chord with given midpoint $(x_1, y_1)$: $$ T = S_1 $$ $$ yy_1 - 2a(x + x_1) = y_1^2 - 4ax_1 $$
Chord joining $P(t_1)$ and $Q(t_2)$: $$ (t_1 + t_2)y = 2x + 2at_1t_2 $$
Focal Chord: Chord passing through focus $$ t_1t_2 = -1 $$

Special Chords:

Focal Chord: Passes through focus, $t_1t_2 = -1$
Chord of Contact: From external point $(x_1, y_1)$: $$ yy_1 = 2a(x + x_1) $$
Equation of Chord with midpoint $(h,k)$: $$ T = S_1 $$

Director Circle and Diameter

Director Circle:

For parabola $y^2 = 4ax$, the director circle is:

$$ x + a = 0 $$

(The directrix itself serves as the director circle)

Diameter of Parabola:

Locus of midpoints of parallel chords of slope m:

$$ y = \frac{2a}{m} $$

This is a line parallel to the axis of the parabola

Reflection Property and Applications

Reflection Property:

Any ray parallel to the axis of parabola after reflection passes through the focus, and vice versa.

This property is used in:

Satellite dishes
Headlights of vehicles
Telescopes
Solar cookers

Parametric Coordinates for other forms:

$y^2 = -4ax$: $x = -at^2$, $y = 2at$
$x^2 = 4ay$: $x = 2at$, $y = at^2$
$x^2 = -4ay$: $x = 2at$, $y = -at^2$

Important Results and Formulas

1. Distance Formulas:

Focal distance of point $P(x_1, y_1)$ on $y^2 = 4ax$: $$ PS = x_1 + a $$
Distance of point from directrix = $x_1 + a$
For any point on parabola: Distance from focus = Distance from directrix

2. Chord Properties:

Length of focal chord with parameter t: $$ L = a\left(t + \frac{1}{t}\right)^2 $$
Length of chord with midpoint $(x_1, y_1)$: $$ L = \frac{2}{a}\sqrt{(y_1^2 - 4ax_1)(y_1^2 + 4a^2)} $$

3. Tangent Properties:

Angle between tangents at $t_1$ and $t_2$: $$ \tan\theta = \left|\frac{t_1 - t_2}{1 + t_1t_2}\right| $$
For perpendicular tangents: $t_1t_2 = -1$
Point of intersection of tangents at $t_1$ and $t_2$: $$ [at_1t_2, a(t_1 + t_2)] $$

Important: Practice all forms of parabola equations and their conversions. Focus on tangent, normal, and chord properties as they are frequently tested in JEE. Remember the reflection property for application-based problems.

Vectors - Mathematics

Vectors are mathematical quantities that have both magnitude and direction. They are fundamental in physics and engineering for representing physical quantities like force, velocity, and displacement.

Basic Definitions:

Scalar: Quantity with only magnitude (mass, temperature)
Vector: Quantity with both magnitude and direction (force, velocity)
Position Vector: $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$
Unit Vector: Vector with magnitude 1: $\hat{a} = \frac{\vec{a}}{|\vec{a}|}$
Null Vector: Vector with zero magnitude

Note: Vectors are crucial for JEE Mathematics and Physics. They form the foundation for 3D geometry, mechanics, and electromagnetic theory.

Vector Algebra Operations

1. Vector Addition and Subtraction:

Triangle Law: $\vec{a} + \vec{b} = \vec{c}$
Parallelogram Law: $\vec{a} + \vec{b} =$ diagonal
Component-wise: $$ \vec{a} \pm \vec{b} = (a_x \pm b_x)\hat{i} + (a_y \pm b_y)\hat{j} + (a_z \pm b_z)\hat{k} $$

2. Scalar Multiplication:

$$ k\vec{a} = ka_x\hat{i} + ka_y\hat{j} + ka_z\hat{k} $$

Magnitude: $|k\vec{a}| = |k||\vec{a}|$

3. Properties of Vector Addition:

Commutative: $\vec{a} + \vec{b} = \vec{b} + \vec{a}$
Associative: $(\vec{a} + \vec{b}) + \vec{c} = \vec{a} + (\vec{b} + \vec{c})$
Additive Identity: $\vec{a} + \vec{0} = \vec{a}$
Additive Inverse: $\vec{a} + (-\vec{a}) = \vec{0}$

Dot Product (Scalar Product)

Definition:

$$ \vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos\theta $$

where $\theta$ is the angle between $\vec{a}$ and $\vec{b}$

Component Form:

$$ \vec{a} \cdot \vec{b} = a_xb_x + a_yb_y + a_zb_z $$

Properties of Dot Product:

Commutative: $\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}$
Distributive: $\vec{a} \cdot (\vec{b} + \vec{c}) = \vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{c}$
$\vec{a} \cdot \vec{a} = |\vec{a}|^2$
$\vec{a} \cdot \vec{b} = 0$ if $\vec{a} \perp \vec{b}$ (and neither is null vector)
Angle between vectors: $\cos\theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}$

Applications:

Work done: $W = \vec{F} \cdot \vec{d}$
Projection of $\vec{a}$ on $\vec{b}$: $\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}$
Component of $\vec{a}$ along $\vec{b}$: $\left(\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2}\right)\vec{b}$

Cross Product (Vector Product)

Definition:

$$ \vec{a} \times \vec{b} = |\vec{a}||\vec{b}|\sin\theta\ \hat{n} $$

where $\hat{n}$ is unit vector perpendicular to both $\vec{a}$ and $\vec{b}$ (right-hand rule)

Component Form:

$$ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a_x & a_y & a_z \\ b_x & b_y & b_z \end{vmatrix} $$

Properties of Cross Product:

Anti-commutative: $\vec{a} \times \vec{b} = -\vec{b} \times \vec{a}$
Distributive: $\vec{a} \times (\vec{b} + \vec{c}) = \vec{a} \times \vec{b} + \vec{a} \times \vec{c}$
$\vec{a} \times \vec{a} = \vec{0}$
$\vec{a} \times \vec{b} = \vec{0}$ if $\vec{a} \parallel \vec{b}$ (and neither is null vector)
Magnitude: $|\vec{a} \times \vec{b}| = |\vec{a}||\vec{b}|\sin\theta$
Area of parallelogram = $|\vec{a} \times \vec{b}|$
Area of triangle = $\frac{1}{2}|\vec{a} \times \vec{b}|$

Scalar Triple Product

Definition:

$$ [\vec{a}\ \vec{b}\ \vec{c}] = \vec{a} \cdot (\vec{b} \times \vec{c}) $$

Component Form:

$$ [\vec{a}\ \vec{b}\ \vec{c}] = \begin{vmatrix} a_x & a_y & a_z \\ b_x & b_y & b_z \\ c_x & c_y & c_z \end{vmatrix} $$

Properties:

Cyclic property: $[\vec{a}\ \vec{b}\ \vec{c}] = [\vec{b}\ \vec{c}\ \vec{a}] = [\vec{c}\ \vec{a}\ \vec{b}]$
Anti-cyclic: $[\vec{a}\ \vec{b}\ \vec{c}] = -[\vec{a}\ \vec{c}\ \vec{b}]$
If any two vectors are equal: $[\vec{a}\ \vec{b}\ \vec{c}] = 0$
If vectors are coplanar: $[\vec{a}\ \vec{b}\ \vec{c}] = 0$
Volume of parallelepiped = $|[\vec{a}\ \vec{b}\ \vec{c}]|$
Volume of tetrahedron = $\frac{1}{6}|[\vec{a}\ \vec{b}\ \vec{c}]|$

Vector Triple Product

Definition:

$$ \vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c} $$

Important Results:

$\vec{a} \times (\vec{b} \times \vec{c}) \neq (\vec{a} \times \vec{b}) \times \vec{c}$ (Not associative)
Lagrange's Identity: $$ (\vec{a} \times \vec{b}) \cdot (\vec{c} \times \vec{d}) = (\vec{a} \cdot \vec{c})(\vec{b} \cdot \vec{d}) - (\vec{a} \cdot \vec{d})(\vec{b} \cdot \vec{c}) $$
Jacobi Identity: $$ \vec{a} \times (\vec{b} \times \vec{c}) + \vec{b} \times (\vec{c} \times \vec{a}) + \vec{c} \times (\vec{a} \times \vec{b}) = \vec{0} $$

Position Vectors and Geometry

1. Section Formula:

Internal division (m:n): $$ \vec{r} = \frac{m\vec{b} + n\vec{a}}{m + n} $$
External division (m:n): $$ \vec{r} = \frac{m\vec{b} - n\vec{a}}{m - n} $$
Midpoint: $$ \vec{r} = \frac{\vec{a} + \vec{b}}{2} $$

2. Centroid of Triangle:

$$ \vec{G} = \frac{\vec{a} + \vec{b} + \vec{c}}{3} $$

3. Important Points in Triangle:

Incenter: $\vec{I} = \frac{a\vec{A} + b\vec{B} + c\vec{C}}{a + b + c}$
Circumcenter: $\vec{O} = \frac{\sin 2A\ \vec{A} + \sin 2B\ \vec{B} + \sin 2C\ \vec{C}}{\sin 2A + \sin 2B + \sin 2C}$
Orthocenter: $\vec{H} = \frac{\tan A\ \vec{A} + \tan B\ \vec{B} + \tan C\ \vec{C}}{\tan A + \tan B + \tan C}$

Line and Plane in Vector Form

1. Equation of Line:

Vector form: $\vec{r} = \vec{a} + \lambda\vec{b}$
Parametric form: $x = a_1 + \lambda b_1$, $y = a_2 + \lambda b_2$, $z = a_3 + \lambda b_3$
Cartesian form: $\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}$
Through two points: $\vec{r} = \vec{a} + \lambda(\vec{b} - \vec{a})$

2. Equation of Plane:

Normal form: $\vec{r} \cdot \hat{n} = d$
Through point with normal: $(\vec{r} - \vec{a}) \cdot \vec{n} = 0$
Through three points: $[\vec{r} - \vec{a}\ \vec{b} - \vec{a}\ \vec{c} - \vec{a}] = 0$
Parametric form: $\vec{r} = \vec{a} + \lambda\vec{b} + \mu\vec{c}$

Important Vector Identities

Basic Vector Identities:

$(\vec{a} + \vec{b})^2 = |\vec{a}|^2 + |\vec{b}|^2 + 2\vec{a} \cdot \vec{b}$
$(\vec{a} - \vec{b})^2 = |\vec{a}|^2 + |\vec{b}|^2 - 2\vec{a} \cdot \vec{b}$
$(\vec{a} + \vec{b}) \cdot (\vec{a} - \vec{b}) = |\vec{a}|^2 - |\vec{b}|^2$
$|\vec{a} + \vec{b}|^2 + |\vec{a} - \vec{b}|^2 = 2(|\vec{a}|^2 + |\vec{b}|^2)$ (Parallelogram Law)
$|\vec{a} + \vec{b}|^2 - |\vec{a} - \vec{b}|^2 = 4\vec{a} \cdot \vec{b}$

Advanced Vector Identities:

$\vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c}$
$(\vec{a} \times \vec{b}) \times \vec{c} = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{b} \cdot \vec{c})\vec{a}$
$(\vec{a} \times \vec{b}) \cdot (\vec{c} \times \vec{d}) = (\vec{a} \cdot \vec{c})(\vec{b} \cdot \vec{d}) - (\vec{a} \cdot \vec{d})(\vec{b} \cdot \vec{c})$
$\vec{a} \times (\vec{b} \times \vec{c}) + \vec{b} \times (\vec{c} \times \vec{a}) + \vec{c} \times (\vec{a} \times \vec{b}) = \vec{0}$

Important: Practice all vector operations and identities thoroughly. Vectors are extensively used in 3D geometry, mechanics, and electromagnetic theory in JEE. Focus on geometric interpretations of dot and cross products.

Probability - Mathematics

Probability is the branch of mathematics that deals with the study of random events and the likelihood of their occurrence. It quantifies uncertainty and is fundamental in statistics, data science, and decision-making.

Basic Definitions:

Random Experiment: An experiment with uncertain outcomes
Sample Space (S): Set of all possible outcomes
Event (E): Subset of sample space
Probability of Event E: $P(E) = \frac{\text{Favorable outcomes}}{\text{Total outcomes}}$
Sure Event: $P(S) = 1$
Impossible Event: $P(\phi) = 0$

Note: Probability is a high-weightage topic in JEE with applications in combinatorics, statistics, and real-world problem solving.

Types of Events

Classification of Events:

Equally Likely Events: Events with same probability
Mutually Exclusive Events: $A \cap B = \phi$
Exhaustive Events: $A_1 \cup A_2 \cup \dots \cup A_n = S$
Independent Events: $P(A \cap B) = P(A)P(B)$
Complementary Events: $A$ and $A'$ where $A \cup A' = S$

Algebra of Events:

$A \cup B$: Either A or B or both occur
$A \cap B$: Both A and B occur
$A'$ or $A^c$: A does not occur
$A - B$: A occurs but B does not
$A \Delta B$: Exactly one of A or B occurs

Basic Probability Theorems

1. Addition Theorem:

For any two events A and B: $$ P(A \cup B) = P(A) + P(B) - P(A \cap B) $$
For mutually exclusive events: $$ P(A \cup B) = P(A) + P(B) $$
For three events A, B, C: $$ P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(B \cap C) - P(C \cap A) + P(A \cap B \cap C) $$

2. Complementary Probability:

$$ P(A') = 1 - P(A) $$

3. Boole's Inequality:

$$ P\left(\bigcup_{i=1}^n A_i\right) \leq \sum_{i=1}^n P(A_i) $$

Conditional Probability

Definition:

$$ P(A|B) = \frac{P(A \cap B)}{P(B)},\quad \text{where } P(B) > 0 $$

Probability of A given that B has occurred

Properties of Conditional Probability:

$0 \leq P(A|B) \leq 1$
$P(S|B) = 1$
$P(A_1 \cup A_2|B) = P(A_1|B) + P(A_2|B) - P(A_1 \cap A_2|B)$
$P(A'|B) = 1 - P(A|B)$

Multiplication Theorem:

$$ P(A \cap B) = P(A) \cdot P(B|A) = P(B) \cdot P(A|B) $$

For n events: $$ P(A_1 \cap A_2 \cap \dots \cap A_n) = P(A_1) \cdot P(A_2|A_1) \cdot P(A_3|A_1 \cap A_2) \cdots P(A_n|A_1 \cap A_2 \cap \dots \cap A_{n-1}) $$

Independent Events

Definition:

Two events A and B are independent if:

$$ P(A \cap B) = P(A) \cdot P(B) $$

This implies $P(A|B) = P(A)$ and $P(B|A) = P(B)$

Properties of Independent Events:

If A and B are independent, then:
- A and B' are independent
- A' and B are independent
- A' and B' are independent
Three events A, B, C are mutually independent if:
- $P(A \cap B) = P(A)P(B)$
- $P(B \cap C) = P(B)P(C)$
- $P(C \cap A) = P(C)P(A)$
- $P(A \cap B \cap C) = P(A)P(B)P(C)$

Pairwise vs Mutual Independence:

Pairwise independence does not imply mutual independence. Three events can be pairwise independent but not mutually independent.

Total Probability Theorem

Theorem:

Let $B_1, B_2, \dots, B_n$ be mutually exclusive and exhaustive events, then for any event A:

$$ P(A) = \sum_{i=1}^n P(B_i) \cdot P(A|B_i) $$

Bayes' Theorem:

$$ P(B_i|A) = \frac{P(B_i) \cdot P(A|B_i)}{\sum_{j=1}^n P(B_j) \cdot P(A|B_j)} $$

Where $B_1, B_2, \dots, B_n$ are mutually exclusive and exhaustive events.

Applications:

Medical diagnosis
Quality control
Spam filtering
Risk assessment

Random Variables and Probability Distributions

Random Variable:

A function that assigns a real number to each outcome in sample space.

Discrete Random Variable: Takes finite or countable values
Continuous Random Variable: Takes uncountable values

Probability Distribution:

For discrete RV X with values $x_1, x_2, \dots, x_n$:
- $P(X = x_i) = p_i$
- $\sum p_i = 1$
- $0 \leq p_i \leq 1$
For continuous RV X:
- Probability density function $f(x)$
- $\int_{-\infty}^{\infty} f(x)dx = 1$
- $P(a \leq X \leq b) = \int_a^b f(x)dx$

Mathematical Expectation

Expectation (Mean):

For discrete RV: $$ E(X) = \sum x_i p_i $$
For continuous RV: $$ E(X) = \int_{-\infty}^{\infty} x f(x)dx $$
Properties:
- $E(aX + b) = aE(X) + b$
- $E(X + Y) = E(X) + E(Y)$
- If X and Y independent: $E(XY) = E(X)E(Y)$

Variance:

$$ Var(X) = E(X^2) - [E(X)]^2 $$

Properties:

$Var(aX + b) = a^2 Var(X)$
$Var(X + Y) = Var(X) + Var(Y) + 2Cov(X,Y)$
If X and Y independent: $Var(X + Y) = Var(X) + Var(Y)$

Standard Deviation:

$$ \sigma = \sqrt{Var(X)} $$

Important Probability Distributions

1. Binomial Distribution:

For n independent Bernoulli trials with success probability p:

$$ P(X = r) = ^nC_r p^r q^{n-r},\quad q = 1-p $$

$E(X) = np$
$Var(X) = npq$
Mode: Integer between $(n+1)p - 1$ and $(n+1)p$

2. Poisson Distribution:

$$ P(X = r) = \frac{e^{-\lambda} \lambda^r}{r!} $$

$E(X) = \lambda$
$Var(X) = \lambda$
Used for rare events

3. Normal Distribution:

$$ f(x) = \frac{1}{\sigma\sqrt{2\pi}} e^{-\frac{(x-\mu)^2}{2\sigma^2}} $$

$E(X) = \mu$
$Var(X) = \sigma^2$
Bell-shaped curve

Important Results and Formulas

1. Combinatorial Probability:

Probability = $\frac{\text{Favorable cases}}{\text{Total cases}}$
Use permutations and combinations for counting
For equally likely outcomes

2. Odds in Favor and Against:

Odds in favor = $\frac{P(E)}{P(E')}$
Odds against = $\frac{P(E')}{P(E)}$
If odds in favor are a:b, then $P(E) = \frac{a}{a+b}$

3. Important Inequalities:

$P(A \cap B) \leq \min(P(A), P(B))$
$P(A \cup B) \geq \max(P(A), P(B))$
$|P(A) - P(B)| \leq P(A \Delta B)$

Important: Practice problems on conditional probability, Bayes' theorem, and probability distributions extensively. These are frequently tested in JEE. Understand the difference between independent and mutually exclusive events clearly.

Select a topic from above to view complete study material with theory, formulas, and solved examples.

JEE Mathematics Complete Preparation

Definition of Polynomial

Degree of a Polynomial

Types of Polynomials

Roots of a Polynomial

Theorems of Polynomials

Relations Between Coefficients and Roots of a Polynomial

Quadratic Equation

Forms of Quadratic Equations

Graph of a Quadratic Equation

Relationship Between Coefficients and Roots

Nature of Roots and Discriminant

Forming Quadratic Equations with Given Conditions

Binomial Theorem

General Expansion

Properties & Applications

JEE Problem & Solution

Sequences & Series

AP, GP, HP

JEE Example

Definition

Example 1

Example 2

1. Cartesian / Rectangular Form

2. Polar Form

3. Euler / Exponential Form

4. Trigonometric / Sin-Cos Form

5. Modulus-Argument Form

6. Conjugate Form

7. Geometric / Vector Form

8. Matrix Form (Advanced / Optional)

Solution:

Solution:

Solution:

Solution:

Solution:

Solution:

Solution:

Solution:

Matrices & Determinants

Key Concepts

JEE Example

Definition of Logarithms

Basic Properties and Laws of Logarithms

Logarithmic Equations

Example 2: Logarithmic Equation with Variables on Both Sides

Example 5: Logarithmic Equation with Unknown Inside and Outside

Theory of Equations

Key Points

JEE Example

Here are some examples

Standard Limits in Calculus

Techniques to Evaluate Limits

Definition of Integration

Types of Integration

Properties and Formulas of Integration

Important Concepts in Integration

Continuity & Differentiability

Definitions

JEE Example

Application of Derivatives

Maxima & Minima

JEE Example

Definite Integrals

Properties

JEE Example

Differential Equations

Basics

JEE Example

Area under Curve

Formula

JEE Example

Trigonometric Ratios and Identities

Definition and Concept of Trigonometric Ratios

Extension of Trigonometric Ratios to Any Angle

Fundamental Trigonometric Identities

Angle Sum and Difference Formulas

Double Angle and Half Angle Formulas

Product-to-Sum and Sum-to-Product Formulas

Important Trigonometric Values and Special Angles