Mathematical Functions for Placement Aptitude

Aptitude rounds at TCS NQT, AMCAT, and eLitmus test three core things about functions: domain and range, absolute value equations, and function composition.

Trigonometry and calculus are not on the list. Placement aptitude sections focus on algebra-level function work. Knowing what the test actually covers lets you skip a semester’s worth of content and spend time on the patterns that actually appear.

What Aptitude Tests Actually Ask

Functions appear in the quantitative ability sections of most campus placement tests. The testable scope is narrower than a textbook chapter:

Function type	Common question pattern
Absolute value (modular)	Solve |x - a| = b; find x satisfying |x| < c or |x| > c
Linear	Evaluate f(x) = ax + b at a given x; find x when f(x) = 0
Quadratic	Find roots; check discriminant; locate the vertex
Exponential	Growth or decay word problems; compare two a^x values
Logarithmic	Apply product, quotient, or power rules to simplify
Composite	Evaluate f(g(x)) for a specific value; note order vs. g(f(x))

The table above is the syllabus. Series, limits, derivatives, and trig identities belong to a different exam.

Domain and Range

The domain of a function is the complete set of x values for which the function produces a valid output. The range is the complete set of possible output values.

Three domain-restriction rules appear in almost every aptitude test involving functions:

• Square roots: the expression under the root must be greater than or equal to zero.
• Fractions: the denominator cannot equal zero.
• Logarithms: the argument must be strictly greater than zero.

Worked examples, derived from first principles:

• f(x) = sqrt(x - 4)

  • Step 1: Set x - 4 to be 0 or greater.
  • Step 2: x must be 4 or greater.
  • Domain: all real x where x is at least 4.

• f(x) = 1 / (x - 2)

  • Step 1: Denominator x - 2 cannot equal zero.
  • Step 2: x cannot equal 2.
  • Domain: all real numbers except x = 2.

• f(x) = log(x + 3)

  • Step 1: Argument x + 3 must be strictly greater than zero.
  • Step 2: x must be greater than -3.
  • Domain: all real x where x is greater than -3.

Practice sets built around these domain patterns are available at IndiaBix’s functions and graphs section.

Absolute Value (Modular) Functions

The absolute value |x| gives the non-negative distance of x from zero. When x is zero or positive, |x| = x. When x is negative, |x| = -x. So |3| = 3 and |-3| = 3.

Two question patterns dominate in placement tests.

Solving Absolute Value Equations

Example: Solve |x - 3| = 5

• Case 1: x - 3 = 5, so x = 8.
• Case 2: x - 3 = -5, so x = -2.
• Both values satisfy the original equation. Answer: x = 8 or x = -2.
• Verify: |8 - 3| = |5| = 5 ✓ and |-2 - 3| = |-5| = 5 ✓.

Solving Absolute Value Inequalities

Example: Solve |2x - 1| < 7

• Step 1: Rewrite as -7 < 2x - 1 < 7.
• Step 2: Add 1 throughout: -6 < 2x < 8.
• Step 3: Divide by 2: -3 < x < 4.
• Answer: x lies strictly between -3 and 4.

Rule: |expression| < c converts to -c < expression < c. |expression| > c converts to expression < -c OR expression > c.

Linear and Quadratic Functions

Linear Functions

A linear function takes the form f(x) = ax + b. Aptitude questions usually ask you to evaluate the function at a given x, or find x when f(x) = 0.

• Example: f(x) = 3x - 6. Find x when f(x) = 0.
  • Step 1: Set 3x - 6 = 0.
  • Step 2: 3x = 6, so x = 2.

Linear functions also model rate relationships. For instance, time and work problems often express a worker’s output as a linear function of hours.

Quadratic Functions

A quadratic function takes the form f(x) = ax² + bx + c. Three things placement tests ask:

1. Roots — the values of x where f(x) = 0.
2. Discriminant (b² - 4ac) — positive means two distinct real roots, zero means one repeated root, negative means no real roots.
3. Vertex — the minimum (when a > 0) or maximum (when a < 0) point.

Example: f(x) = x² - 5x + 6. Find the roots and minimum value.

• Step 1: Factorise. x² - 5x + 6 = (x - 2)(x - 3).
• Step 2: Roots are x = 2 and x = 3.
• Step 3: Vertex x-coordinate = -b / (2a) = 5 / 2 = 2.5.
• Step 4: f(2.5) = (2.5)² - 5(2.5) + 6 = 6.25 - 12.5 + 6 = -0.25.
• Minimum value of the function is -0.25, at x = 2.5.
• Discriminant check: b² - 4ac = 25 - 24 = 1, which is greater than zero, confirming two distinct real roots.

Exponential and Logarithmic Functions

Exponential Functions

An exponential function takes the form f(x) = a^x, where a is a positive constant not equal to 1. Growth problems use a greater than 1; decay problems use a between 0 and 1.

Example: A colony of bacteria doubles every 4 hours. Starting count is 500. What is the count after 12 hours?

• Step 1: Number of doubling periods in 12 hours = 12 / 4 = 3.
• Step 2: Count after 12 hours = 500 × 2³ = 500 × 8 = 4,000.
• Answer: 4,000 bacteria after 12 hours.

Logarithmic Functions

log_a(x) is the inverse of a^x. If a^y = x, then log_a(x) = y.

Three properties cover most placement log questions:

Property	Rule
Product	log(mn) = log(m) + log(n)
Quotient	log(m/n) = log(m) - log(n)
Power	log(m^n) = n × log(m)

Two useful constants: log_a(a) = 1 and log_a(1) = 0 for any valid base a.

Example: If log(2) = 0.301, find log(32).

• Step 1: Write 32 = 2⁵.
• Step 2: log(32) = log(2⁵) = 5 × log(2) = 5 × 0.301 = 1.505.

Example: Evaluate log₃(81).

• Step 1: Identify — 3 to what power gives 81?
• Step 2: 3⁴ = 81.
• Answer: log₃(81) = 4.

Khan Academy’s algebra and functions module covers the foundational theory behind these log properties if you want a deeper walkthrough between practice sessions.

Composite and Inverse Functions

Composite Functions

f(g(x)) means: apply g first to x, then apply f to the result. Order is not interchangeable.

Example: f(x) = 3x + 2 and g(x) = x² - 1. Find f(g(3)) and g(f(2)).

For f(g(3)):

• Step 1: g(3) = 3² - 1 = 9 - 1 = 8.
• Step 2: f(8) = 3(8) + 2 = 24 + 2 = 26.
• Answer: f(g(3)) = 26.

For g(f(2)) (reversed order):

• Step 1: f(2) = 3(2) + 2 = 8.
• Step 2: g(8) = 8² - 1 = 64 - 1 = 63.
• Answer: g(f(2)) = 63.

The two compositions give 26 and 63, confirming that order matters and cannot be swapped.

Inverse Functions

f⁻¹(x) reverses what f(x) does. To find it: write y = f(x), swap x and y, then solve for y.

Example: Find the inverse of f(x) = 2x + 6.

• Step 1: Write y = 2x + 6.
• Step 2: Swap x and y: x = 2y + 6.
• Step 3: Solve for y: y = (x - 6) / 2.
• f⁻¹(x) = (x - 6) / 2.
• Verify: f(f⁻¹(x)) = 2 × ((x - 6) / 2) + 6 = (x - 6) + 6 = x. ✓

For all the function types above, timed drills on each type individually build speed before you mix them. The campus placement evaluation test overview breaks down which aptitude modules carry the most weight on actual placement days, so you can allocate prep time accordingly. For a curated reading list across all quantitative aptitude topics, best books for placement preparation covers the options worth your money.

The log and exponential properties covered in this article are also the core mathematics behind how AI models score predictions. Log loss in classification tasks and the softmax transformation in multi-class models use the same rules. If you want to see those connections directly:

• TinkerLLM puts real LLM API calls in your hands for ₹299, and the mini-projects you build there are the kind of work that stands out when a technical interviewer asks what you have actually shipped.