Functions, Domains, Ranges and Optimization Previous Year Questions (PYQs) for CAT: 11+ Solved Questions with Step-by-Step Solutions

Solve 11+ Functions, Domains, Ranges and Optimization previous year questions for CAT with answers and detailed solutions. Free sample questions below.

Chapter Roadmap: Functions, Domains, Ranges and Optimization

Chapter roadmap

Functions, Domains, Ranges and Optimization

🔁 Function Definitions and Composition

Learn function notation, substitution, composition, piecewise definitions, and functional equations.

5 direct CAT PYQs | selected topic

🧱 Domain and Range of Rational Functions

Find allowed inputs and possible outputs, especially when denominators and composite functions create restrictions.

3 direct CAT PYQs

🎯 Optimization and Minimum-Maximum Values

Use algebraic structure, quadratics, and constraints to find best possible values.

5 direct CAT PYQs

By the end, you should read functions as input-output rules, not as random formulas.

Topic Hero: Function Definitions and Composition

Algebra → Functions, Domains, Ranges and Optimization → Topic 1

f(x)

Input goes in. Output comes out.

Function Definitions and Composition

CAT function questions are often just substitution puzzles wearing formal notation.

✅ Understand

f (x)

as a rule

✅ Evaluate composite functions like

f (f (x))

✅ Decode functional equations by substitution

✅ Handle piecewise and integer-defined functions

Functions, Domains, Ranges and Optimization: Solved Questions with Step-by-Step Explanations (5 Problems)

Question 1 · Quantitative Ability MCQ

For all real values of x, the range of the function $f (x) = \frac{x ^{2} + 2 x + 4}{2 x ^{2} + 4 x + 9}$ is:

A.
$[\frac{4}{9}, \frac{8}{9}]$
B.
$[\frac{3}{7}, \frac{8}{9})$
C.
$(\frac{3}{7}, \frac{1}{2})$
D.
$[\frac{3}{7}, \frac{1}{2})$

Correct Answer:

Step-by-Step Solution

Key idea: This is a rational function range question where the numerator and denominator quadratics are proportional. This signals that algebraic substitution is much faster than the discriminant method.

Step 1: Spot the proportional terms and substitute.

Let $u = x^{2} + 2 x + 4$ .

Notice that the denominator is $2 x^{2} + 4 x + 9 = 2 (x^{2} + 2 x + 4) + 1 = 2 u + 1$ .

The function simplifies to $y = \frac{u}{2 u + 1}$ .

Step 2: Find the valid range of $u$ .

Complete the square for $u$ : $u = (x^{2} + 2 x + 1) + 3 = (x + 1)^{2} + 3$ .

Since $(x + 1)^{2} \geq 0$ for all real $x$ , the minimum value of $u$ is $3$ . Thus, $u \geq 3$ .

Step 3: Analyze the behavior of $y$ with respect to $u$ .

Rewrite $y$ to make the limits obvious:

$y = \frac{1}{2} (\frac{2 u}{2 u + 1}) = \frac{1}{2} (\frac{2 u + 1 - 1}{2 u + 1}) = \frac{1}{2} (1 - \frac{1}{2 u + 1})$ .

Step 4: Map the domain of $u$ to the range of $y$ .

Since $u \geq 3$ , we have $2 u + 1 \geq 7$ .

Therefore, $0 < \frac{1}{2 u + 1} \leq \frac{1}{7}$ .

Multiplying by $- 1$ reverses the inequalities: $- \frac{1}{7} \leq - \frac{1}{2 u + 1} < 0$ .

Adding $1$ : $1 - \frac{1}{7} \leq 1 - \frac{1}{2 u + 1} < 1 ⟹ \frac{6}{7} \leq 1 - \frac{1}{2 u + 1} < 1$ .

Finally, multiply by $\frac{1}{2}$ : $\frac{3}{7} \leq \frac{1}{2} (1 - \frac{1}{2 u + 1}) < \frac{1}{2}$ .

Step 5: Write the final interval.

The range is $[\frac{3}{7}, \frac{1}{2})$ .

Answer: [\frac{3}{7}, \frac{1}{2})

Question 2 · Quantitative Ability MCQ

For real values of $x$ , the range of the function $f (x) = \frac{2 x - 3}{2 x ^{2} + 4 x - 6}$ is

A.
$(- \infty, \frac{1}{8}] \cup [1, \infty)$
B.
$(- \infty, \frac{1}{4}] \cup [1, \infty)$
C.
$(- \infty, \frac{1}{8}] \cup [\frac{1}{2}, \infty)$
D.
$(- \infty, \frac{1}{4}] \cup [\frac{1}{2}, \infty)$

Correct Answer:

Step-by-Step Solution

Key idea: This is a linear-over-quadratic range question. It is recognisable because the function has a degree-1 numerator and a degree-2 denominator, and the question asks for all possible output values.

Why this method applies: To find the range, treat the output as $y$ . A value $y$ is in the range exactly when the equation $y = f (x)$ has at least one real solution $x$ that is allowed by the original denominator.

Step 1: Put $y = f (x)$ .

$y = \frac{2 x - 3}{2 x ^{2} + 4 x - 6}$

Step 2: Cross-multiply to get an equation in $x$ .

$y (2 x^{2} + 4 x - 6) = 2 x - 3$

$2 y x^{2} + (4 y - 2) x + (3 - 6 y) = 0$

For $y \neq = 0$ , this is a quadratic in $x$ . For $y = 0$ , the original equation gives $2 x - 3 = 0$ , so $x = \frac{3}{2}$ , and $y = 0$ is certainly possible.

Step 3: Require real $x$ . For the quadratic in $x$ , the discriminant must be non-negative.

$D = (4 y - 2)^{2} - 4 (2 y) (3 - 6 y)$

Expand carefully:

$D = 16 y^{2} - 16 y + 4 - 24 y + 48 y^{2}$

$D = 64 y^{2} - 40 y + 4 = 4 (16 y^{2} - 10 y + 1)$

Step 4: Solve $D \geq 0$ .

$16 y^{2} - 10 y + 1 \geq 0$

Factor:

$16 y^{2} - 10 y + 1 = 16 (y - \frac{1}{8}) (y - \frac{1}{2})$

A positive quadratic is non-negative outside its roots, so

$y \leq \frac{1}{8} or y \geq \frac{1}{2}$

Step 5: Check endpoints and domain restrictions. The original denominator is zero at $x = - 3$ and $x = 1$ . Neither of these can satisfy the cross-multiplied equation, so no extra output value has to be removed. Also:

$y = \frac{1}{8} \Rightarrow x = 3$

and

$y = \frac{1}{2} \Rightarrow x = 0$

both are valid inputs.

Answer: $(- \infty, \frac{1}{8}] \cup [\frac{1}{2}, \infty)$ , which is option C.

Question 3 · Quantitative Ability MCQ

For any non-zero real number x, let $f (x) + 2 f (\frac{1}{x}) = 3 x$ . Then, the sum of all possible values of x for which $f (x) = 3$ , is

A.
3
B.
-2
C.
-3
D.
2

Correct Answer:

Step-by-Step Solution

This is a functional equation question — you can tell because $f$ appears at two related inputs, $x$ and $1/ x$ , connected by one equation. The trigger is seeing $f (x)$ and $f (1/ x)$ together: this always means you should substitute $x \to 1/ x$ to generate a second equation, then solve the two equations simultaneously for $f (x)$ .

Step 1 — Write the given equation.

$f (x) + 2 f (\frac{1}{x}) = 3 x ...(1)$

Step 2 — Substitute $x \to \frac{1}{x}$ to get a second equation.

$f (\frac{1}{x}) + 2 f (x) = \frac{3}{x} ...(2)$

Step 3 — Eliminate $f (1/ x)$ . From (1): $f (1/ x) = \frac{3 x - f ( x )}{2}$ . Substitute into (2):

$\frac{3 x - f ( x )}{2} + 2 f (x) = \frac{3}{x}$

Step 4 — Multiply through by 2 to clear the fraction.

$3 x - f (x) + 4 f (x) = \frac{6}{x}$

$3 x + 3 f (x) = \frac{6}{x}$

Step 5 — Solve for $f (x)$ .

$3 f (x) = \frac{6}{x} - 3 x ⟹ f (x) = \frac{2}{x} - x$

Step 6 — Set $f (x) = 3$ and solve for x.

$\frac{2}{x} - x = 3$

Multiply both sides by $x$ (note $x \neq = 0$ is given):

$2 - x^{2} = 3 x ⟹ x^{2} + 3 x - 2 = 0$

Step 7 — Use Vieta's formula for the sum of roots, without needing to find the roots individually:

$sum of roots = - \frac{b}{a} = - \frac{3}{1} = - 3$

Trap to avoid: A very common mistake is computing the product of the roots ( $c / a = - 2$ ) instead of the sum, or dropping a sign while rearranging $2 - x^{2} = 3 x$ into standard form.

Final answer: sum of all x = $- 3$ (option C).

Question 4 · Quantitative Ability NAT

Suppose f(x, y) is a real-valued function such that f(3x + 2y, 2x - 5y) = 19x, for all real numbers x and y. The value of x for which f(x, 2x) = 27, is

Correct Answer:

Step-by-Step Solution

This is a functional equation with linear substitution question, recognisable because f is defined through a linear transformation of its inputs.

Step 1: Understand the given functional equation.

We are told $f (3 x + 2 y, 2 x - 5 y) = 19 x$ for all real x and y.

Step 2: Invert the linear transformation.

Let $u = 3 x + 2 y$ and $v = 2 x - 5 y$ . We need to express x in terms of u and v.

Multiply the first equation by 5: $5 u = 15 x + 10 y$

Multiply the second by 2: $2 v = 4 x - 10 y$

Add them: $5 u + 2 v = 19 x$

Therefore: $x = \frac{5 u + 2 v}{19}$

Step 3: Find the general form of f.

Since $f (u, v) = 19 x = 19 \cdot \frac{5 u + 2 v}{19} = 5 u + 2 v$ :

$f (a, b) = 5 a + 2 b$

Step 4: Apply to the target.

We need $f (x, 2 x) = 27$ :

$f (x, 2 x) = 5 (x) + 2 (2 x) = 5 x + 4 x = 9 x$

$9 x = 27 ⟹ x = 3$

Answer: 3

Question 5 · Quantitative Ability NAT

Suppose for all integers x, there are two functions f and g such that $f (x) + f (x - 1) - 1 = 0$ and $g (x) = x^{2}$ . If $f (x^{2} - x) = 5$ , then the value of the sum $f (g (5)) + g (f (5))$ is

Correct Answer:

Step-by-Step Solution

This is a periodic functional equation with composition question, recognisable because f(x) + f(x-1) = 1 forces f to alternate between two values.

Step 1: Derive the period of f.

From $f (x) + f (x - 1) = 1$ :

$f (x) = 1 - f (x - 1)$

$f (x - 1) = 1 - f (x - 2)$

Substituting: $f (x) = 1 - (1 - f (x - 2)) = f (x - 2)$

So f has period 2: $f (x) = f (x - 2)$ for all integers x.

Step 2: Determine f(0) and f(1).

Since $x^{2} - x = x (x - 1)$ is always even (product of consecutive integers), and f has period 2:

$f (x^{2} - x) = f (0)$ for all integers x.

Given $f (x^{2} - x) = 5$ , we get $f (0) = 5$ .

From $f (0) + f (- 1) = 1$ : $f (- 1) = - 4$ . Since period is 2, $f (1) = f (- 1) = - 4$ .

Step 3: Compute f(g(5)).

$g (5) = 5^{2} = 25$ . Since 25 is odd, $f (25) = f (1) = - 4$ .

Step 4: Compute g(f(5)).

Since 5 is odd, $f (5) = f (1) = - 4$ .

$g (f (5)) = g (- 4) = (- 4)^{2} = 16$ .

Step 5: Add.

$f (g (5)) + g (f (5)) = - 4 + 16 = 12$

Answer: 12

Functions, Domains, Ranges and Optimization Previous Year Questions (PYQs) for CAT: 11+ Solved Questions with Step-by-Step Solutions

Chapter Roadmap: Functions, Domains, Ranges and Optimization

Functions, Domains, Ranges and Optimization

Topic Hero: Function Definitions and Composition

Function Definitions and Composition

Functions, Domains, Ranges and Optimization: Solved Questions with Step-by-Step Explanations (5 Problems)

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

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