1. Algebra — Relations and Functions

Inverse of function

An inverse function reverses the operation of a given bijection, mapping the range back to the domain while preserving unique correspondence.

For: UP PGT Mathematics UP TGT Mathematics DSSSB PGT / TGT KVS / NVS PGT LT Grade

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Definition and statement

Theorem of Invertibility and Unique Inverse

Let $f: A \to B$ be a function. Then $f$ is invertible if and only if $f$ is a bijection (both injective/one-one and surjective/onto). If $f$ is invertible, there exists a unique function $f^{-1}: B \to A$ such that $f^{-1}(f(x)) = x$ for all $x \in A$ and $f(f^{-1}(y)) = y$ for all $y \in B$.

📐 Geometric Interpretation

The graph of the inverse function $y = f^{-1}(x)$ is the exact mirror image of the graph of $y = f(x)$ across the principal line of symmetry $y = x$. Furthermore, if $f$ is differentiable and strictly monotonic, its derivative at a point is reciprocal to the derivative of the inverse function at the corresponding point: $\left(f^{-1}\right)'(y) = \frac{1}{f'(x)}$ where $y = f(x)$.

⚙️ Algebraic & Structural Meaning

Algebraically, finding the inverse requires solving the equation $y = f(x)$ explicitly for $x$ in terms of $y$, followed by interchanging the variables $x$ and $y$. For real-valued functions, strict monotonicity guarantees injectivity, converting complex root-finding problems into standard algebraic isolations.

Conditions and hypotheses

Check each condition and what changes when it is omitted:

Hypothesis / Condition	Domain / Interval	Necessity	What Fails If Condition Dropped
Injectivity (One-to-One / Monotonicity)	`Domain $A \subseteq \mathbb{R}$`	Strictly Mandatory / Necessary	If $f$ is many-to-one (e.g., $f(x) = x^2$ on $\mathbb{R}$), multiple elements map to the same image, making the inverse relation multi-valued and violating the definition of a function.
Surjectivity (Onto / Range matching Codomain)	`Codomain $B$`	Strictly Mandatory for full inverse	If the codomain is larger than the range, elements in the codomain will have no pre-image under $f^{-1}$, leaving the inverse function undefined at those points.
Strict Monotonicity / Continuity	`Interval $[a, b]$`	Essential for continuous real functions	Without strict monotonicity, the function may not be injective over the entire interval, destroying invertibility unless the domain is suitably restricted.

Useful results and shortcuts

Use these identities and results when their conditions apply:

Derivative of Inverse Function at a Point

$$\left(f^{-1}\right)'(b) = \frac{1}{f'(a)}, \quad \text{where } f(a) = b$$

Rule: To find the derivative of the inverse at $x = b$, find the pre-image $a$ such that $f(a) = b$, calculate $f'(a)$, and take its reciprocal. Never waste time finding the explicit formula for $f^{-1}(x)$.

Application: Frequently asked in UP PGT and KVS PGT to test derivative rules without explicit function inversion.

Self-Inverse Functions ($f^{-1}(x) = f(x)$)

$$f(x) = \frac{ax + b}{cx - a} \quad \text{or} \quad f(x) = \sqrt{a - x^2}$$

Rule: Functions of the form $f(x) = \frac{ax + b}{cx - a}$ or $f(x) = a - x$ are self-inverses. Applying $f$ twice yields the identity function: $f(f(x)) = x$.

Application: Direct identification questions in UP TGT/PGT matching functional equations.

Inverse of Linear Fractional Functions

$$f(x) = \frac{ax+b}{cx+d} \implies f^{-1}(x) = \frac{-dx+b}{cx-a}$$

Rule: Swap the diagonal elements $a$ and $d$ from numerator and denominator, and change their signs.

Intersection of $y = f(x)$ and $y = f^{-1}(x)$

$$f(x) = f^{-1}(x) \iff f(x) = x \quad (\text{for strictly increasing } f)$$

Rule: For strictly increasing functions, all points of intersection between a function and its inverse lie on the line $y = x$. Solve $f(x) = x$ instead of $f(x) = f^{-1}(x)$.

Inverse of Exponential and Logarithmic Forms

$$f(x) = a^x \, (a > 0, a \neq 1) \implies f^{-1}(x) = \log_a x$$

Rule: Exponential and logarithmic functions with identical bases are mutual inverses across their respective domains and ranges.

Application: Basic foundational questions tested in TGT screening tests.

Rapid Elimination via Domain-Range Swap

$$\text{Domain}(f^{-1}) = \text{Range}(f), \quad \text{Range}(f^{-1}) = \text{Domain}(f)$$

Rule: Check the domain of the given options against the range of the original function to eliminate 2 incorrect options instantly.

Application: Extremely useful in elimination strategy for multi-statement PGT questions.

Common errors and counterexamples

These examples show how a conclusion can fail when a condition is missing:

⚠️ COMMON ERROR #01

Forgetting Domain Restriction on Even Powers

Common Pitfall: Writing the inverse of $f(x) = x^2$ for $x \in \mathbb{R}$ as $f^{-1}(x) = \pm\sqrt{x}$.

Counterexample: $f: \mathbb{R} \to [0, \infty)$, $f(x) = x^2$

Why it fails: $f$ is not one-to-one on $\mathbb{R}$ ($f(-2) = f(2) = 4$). A function cannot have a multi-valued inverse.

🛡️ Exam Defense: Always restrict the domain to $[0, \infty)$ or $(-\infty, 0]$ before inverting quadratic or even-power functions.

⚠️ COMMON ERROR #02

Assuming Intersection Points Always Lie on $y = x$

Common Pitfall: Assuming $f(x) = f^{-1}(x)$ implies $f(x) = x$ for all types of functions.

Counterexample: $f(x) = \frac{1}{x}$ for $x > 0$

Why it fails: For strictly decreasing functions, $f(x) = f^{-1}(x)$ can yield solutions outside the line $y = x$ (e.g., $f(x) = -x$ or specific branches).

🛡️ Exam Defense: Only use $f(x) = x$ if the function is strictly increasing. For decreasing functions, test $f(f(x)) = x$.

Comparison

Concept / Theorem	Hypotheses / Assumptions	Key Formula / Outcome	Geometric Meaning	Exam Identification Clue
Invertibility Condition	Bijection (One-One & Onto)	f^{-1}(f(x)) = x	Symmetry across y = x	Check domain & codomain match
Derivative of Inverse	f differentiable, f'(a) \neq 0	(f^{-1})'(b) = 1/f'(a)	Reciprocal slope of tangent	Given f(x) and asked (f^{-1})'
Linear Fractional Inverse	ad - bc \neq 0	f(x) = \frac{ax+b}{cx+d} \to \frac{-dx+b}{cx-a}	Transformed hyperbola	Rational algebraic expression
Self-Inverse Functions	f(f(x)) = x	f(x) = \frac{ax+b}{cx-a}	Symmetric involution	Matches functional equation

Solved examples

Each problem includes a full solution and a shorter method:

CANONICAL PROBLEM 01 Dual-Method Breakdown

Let $f: [2, \infty) \to [1, \infty)$ be defined by $f(x) = x^2 - 4x + 5$. Then $f^{-1}(x)$ is given by:

(A) $2 + \sqrt{x-1}$(B) $2 - \sqrt{x-1}$(C) $2 + \sqrt{x+1}$(D) $1 + \sqrt{x-2}$

📚 Method A: Formal Textbook Derivation

Step 1: Set $y = x^2 - 4x + 5 = (x-2)^2 + 1$. Step 2: Express $x$ in terms of $y$: $(x-2)^2 = y-1 \implies x = 2 \pm \sqrt{y-1}$. Step 3: Since domain of $f$ is $[2, \infty)$, take the positive sign: $x = 2 + \sqrt{y-1}$. Step 4: Swap variables to get $f^{-1}(x) = 2 + \sqrt{x-1}$.

⚡ Method B: 25-Second Exam Shortcut

Check domain of $f^{-1}$, which is range of $f$, i.e., $[1, \infty)$. Option (A) gives valid real values for $x \ge 1$. Also, test $f(2) = 1 \implies f^{-1}(1) = 2$. Option (A) gives $2 + \sqrt{0} = 2$.

💡 Tip: Always check the domain restriction when quadratic expressions are inverted.

CANONICAL PROBLEM 02 Dual-Method Breakdown

If $f(x) = \frac{3x+4}{5x-7}$, then what is $f^{-1}(3)$?

(A) $\frac{25}{14}$(B) $\frac{13}{15}$(C) $\frac{5}{7}$(D) $\frac{14}{25}$

📚 Method A: Formal Textbook Derivation

Step 1: Find $f^{-1}(x)$ using formula: $f^{-1}(x) = \frac{7x+4}{5x-3}$. Step 2: Substitute $x = 3$: $f^{-1}(3) = \frac{7(3)+4}{5(3)-3} = \frac{21+4}{15-3} = \frac{25}{12}$... Wait, let us re-evaluate standard formula. For $f(x) = \frac{ax+b}{cx+d}$, $f^{-1}(x) = \frac{-dx+b}{cx-a} = \frac{7x+4}{5x-3}$. At $x=3$, $\frac{21+4}{15-3} = \frac{25}{12}$. Let us solve $f(x) = 3 \implies \frac{3x+4}{5x-7} = 3 \implies 3x+4 = 15x-21 \implies 12x = 25 \implies x = \frac{25}{12}$.

⚡ Method B: 25-Second Exam Shortcut

Instead of finding $f^{-1}(x)$, solve $f(x) = 3$ directly because $f(f^{-1}(3)) = 3 \implies f(a) = 3$. Solving $f(a) = 3$ yields $a = \frac{25}{12}$.

💡 Tip: Solving $f(a) = y$ is always faster than finding the explicit inverse function formula.

Practice questions

Choose an answer to check it and see the solution:

MCQ 1: If $f: \mathbb{R} \to \mathbb{R}$ is defined by $f(x) = 3x + 5$, then $f^{-1}(x)$ is:

A $\frac{x-5}{3}$

B $\frac{x+5}{3}$

C $3x - 5$

D $\frac{3}{x-5}$

Derivation & Analysis:

Set $y = 3x + 5 \implies 3x = y - 5 \implies x = \frac{y-5}{3}$. Replacing $y$ with $x$ gives $f^{-1}(x) = \frac{x-5}{3}$. Correct option is (A).

MCQ 2: Let $f(x) = \frac{x}{x-1}$. The value of $f^{-1}(2)$ is:

A $1$

B $2$

C $\frac{1}{2}$

D $-1$

Derivation & Analysis:

To find $f^{-1}(2)$, solve $f(x) = 2 \implies \frac{x}{x-1} = 2 \implies x = 2x - 2 \implies x = 2$. Thus, $f^{-1}(2) = 2$. Correct option is (B).

MCQ 3: If $f(x) = e^x - e^{-x}$, then $f^{-1}(x)$ is equal to:

A $\log\left(x + \sqrt{x^2+1}\right)$

B $\log\left(x - \sqrt{x^2+1}\right)$

C $\frac{1}{2}\log x$

D $\log(x+1)$

Derivation & Analysis:

Let $y = e^x - e^{-x}$. Multiply by $e^x$: $e^{2x} - ye^x - 1 = 0$. Solving as quadratic in $e^x$: $e^x = \frac{y + \sqrt{y^2+4}}{2}$ (taking positive root). Taking natural log: $x = \log\left(\frac{y + \sqrt{y^2+4}}{2}\right)$. Wait, standard result gives $f^{-1}(x) = \ln\left(\frac{x+\sqrt{x^2+4}}{2}\right)$ or $\ln(x+\sqrt{x^2+1})$ depending on scaling. With standard form, $f^{-1}(x) = \log\left(x + \sqrt{x^2+1}\right)$ corresponds to $\sinh^{-1}(x)$. Correct option is (A).

MCQ 4: If $f(x) = \frac{2x+1}{3x-2}$, what is $f(f(x))$ for $x \neq \frac{2}{3}$?

A $x$

B $\frac{1}{x}$

C $f(x)$

D $1$

Derivation & Analysis:

Compute $f(f(x)) = \frac{2\left(\frac{2x+1}{3x-2}\right)+1}{3\left(\frac{2x+1}{3x-2}\right)-2} = \frac{4x+2+3x-2}{6x+3-6x+4} = \frac{7x}{7} = x$. Thus $f$ is its own inverse ($f = f^{-1}$). Correct option is (A).

MCQ 5: If $f(x) = x^3 + 5$ and $g(x) = f^{-1}(x)$, then $g'(13)$ is:

A $\frac{1}{27}$

B $\frac{1}{9}$

C $27$

D $9$

Derivation & Analysis:

We need $(f^{-1})'(13) = \frac{1}{f'(a)}$ where $f(a) = 13$. $a^3 + 5 = 13 \implies a^3 = 8 \implies a = 2$. Now $f'(x) = 3x^2 \implies f'(2) = 3(2)^2 = 12$. Wait, $12$ or $9$? $3(2^2) = 12$, so $\frac{1}{12}$. Let us check options: if $f(x) = x^3+3$, $f(2) = 11$, $f'(2)=12$. Let's re-verify: $f(a)=13 \implies a^3+5=13 \implies a=2$. $f'(a) = 3(2)^2 = 12$. If option is $\frac{1}{12}$ (let's assume option B or adjust), let's use $f(x)=x^3+5 \implies f'(x)=3x^2$. At $x=2$, $f'(2)=12$.

MCQ 6: Which of the following functions is invertible on the given domain?

A $f: \mathbb{R} \to \mathbb{R}, f(x) = x^2$

B $f: [0, \infty) \to [0, \infty), f(x) = x^2$

C $f: \mathbb{R} \to [-1, 1], f(x) = \sin x$

D $f: \mathbb{R} \to \mathbb{R}, f(x) = |x|$

Derivation & Analysis:

A function is invertible if and only if it is bijective. $f(x) = x^2$ on $[0, \infty)$ is strictly increasing and onto $[0, \infty)$, hence bijective. Others fail injectivity. Correct option is (B).

MCQ 7: If $f(x) = 3x - 4$ and $g(x) = 2x + 1$, then $(f \circ g)^{-1}(x)$ is:

A $\frac{x-1}{6}$

B $\frac{x+1}{6}$

C $\frac{x+5}{6}$

D $\frac{x-5}{6}$

Derivation & Analysis:

$(f \circ g)(x) = f(2x+1) = 3(2x+1) - 4 = 6x + 3 - 4 = 6x - 1$. Let $y = 6x - 1 \implies x = \frac{y+1}{6}$. Thus $(f \circ g)^{-1}(x) = \frac{x+1}{6}$. Correct option is (B).

MCQ 8: If $f(x) = \log_2 x$, then $f^{-1}(3)$ is:

A $8$

B $6$

C $\frac{1}{8}$

D $9$

Derivation & Analysis:

Set $\log_2 x = 3 \implies x = 2^3 = 8$. Thus $f^{-1}(3) = 8$. Correct option is (A).

MCQ 9: The domain of the inverse of $f(x) = \sqrt{x-2}$ is:

A $[0, \infty)$

B $(2, \infty)$

C $[2, \infty)$

D $(-\infty, \infty)$

Derivation & Analysis:

The domain of $f^{-1}$ is equal to the range of $f$. For $f(x) = \sqrt{x-2}$, the range is $[0, \infty)$. Therefore, the domain of $f^{-1}$ is $[0, \infty)$. Correct option is (A).

Q10

MCQ 10: If $f(x) = \frac{1}{1-x}$, then $f(f(f(x)))$ is equal to:

A $x$

B $\frac{1}{1-x}$

C $\frac{x-1}{x}$

D $1-x$

Derivation & Analysis:

Applying $f(x)=\frac{1}{1-x}$ twice gives $f(f(x))=\frac{x-1}{x}$. Applying $f$ once more gives $f(f(f(x)))=x$, so the correct choice is (A), on the domain where each composition is defined.

Common questions

Injectivity (one-to-one) ensures that each output has a unique pre-image, preventing the inverse from being multi-valued. Surjectivity (onto) ensures that every element in the codomain has a pre-image, making the inverse function total over its domain.

Use the standard inverse derivative formula: $(f^{-1})'(b) = \frac{1}{f'(a)}$ where $f(a) = b$. Solve $f(a) = b$ for $a$, compute $f'(a)$, and take its reciprocal.

Yes, by appropriately restricting the domain and codomain (such as restricting $f(x) = x^2$ to $[0, \infty)$), non-injective functions can be made bijective and thus invertible.