Functions, Domains, Ranges and Optimization Practice Questions for CAT: 127+ Solved Questions with Step-by-Step Solutions

Solve 127+ Functions, Domains, Ranges and Optimization practice 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 NAT

Let $f (x) = \frac{x ^{2} - x + 1}{x ^{2} + x + 1}$ for all real $x$ . If $M$ and $m$ are the maximum and minimum values of $f (x)$ respectively, what is the value of $3 (M + m)$ ?

Correct Answer:

Step-by-Step Solution

Key idea: This is an optimization via discriminant question. When finding the range (max/min values) of a rational function where both numerator and denominator are quadratics, the most efficient method is to use the discriminant.

Step 1: Set the function equal to $y$ .

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

Step 2: Cross-multiply and rearrange into a standard quadratic equation in terms of $x$ .

$y (x^{2} + x + 1) = x^{2} - x + 1$

$y x^{2} + y x + y = x^{2} - x + 1$

$(y - 1) x^{2} + (y + 1) x + (y - 1) = 0$ .

Step 3: Apply the condition for real roots.

Since the domain is all real $x$ , for any valid output $y$ , there must exist at least one real $x$ that satisfies this equation. Therefore, the discriminant $Δ$ must be greater than or equal to zero.

$Δ = b^{2} - 4 a c \geq 0$

$(y + 1)^{2} - 4 (y - 1) (y - 1) \geq 0$ .

Step 4: Solve the inequality for $y$ .

Expand the terms:

$(y^{2} + 2 y + 1) - 4 (y^{2} - 2 y + 1) \geq 0$

$y^{2} + 2 y + 1 - 4 y^{2} + 8 y - 4 \geq 0$

$- 3 y^{2} + 10 y - 3 \geq 0$ .

Multiply by $- 1$ and flip the inequality sign:

$3 y^{2} - 10 y + 3 \leq 0$ .

Factor the quadratic:

$(3 y - 1) (y - 3) \leq 0$ .

The roots are $y = 1/3$ and $y = 3$ . Since the parabola opens upwards and is $\leq 0$ , the solution is the interval between the roots:

$1/3 \leq y \leq 3$ .

Step 5: Identify $M$ and $m$ and calculate the final answer.

The maximum value $M = 3$ .

The minimum value $m = 1/3$ .

The question asks for $3 (M + m)$ :

$3 (3 + 1/3) = 3 (10/3) = 10$ .

Answer: 10

Question 2 · Quantitative Ability NAT

How many integer values are in the range of the function $f (x) = \frac{8 x - 16}{x ^{2} - 4 x + 5}$ for all real numbers $x$ ?

Correct Answer:

Step-by-Step Solution

Key idea: This is a linear over quadratic range question. The trigger is a rational function where the numerator is degree 1 and the denominator is degree 2. The most robust method is substitution to simplify the expression, followed by the discriminant method to find the exact bounds of the range.

Step 1: Simplify the function using substitution.

Notice the denominator: $x^{2} - 4 x + 5 = (x^{2} - 4 x + 4) + 1 = (x - 2)^{2} + 1$ .

Notice the numerator: $8 x - 16 = 8 (x - 2)$ .

Let $u = x - 2$ . As $x$ ranges over all real numbers, $u$ also ranges over all real numbers.

The function becomes $y = \frac{8 u}{u ^{2} + 1}$ .

Step 2: Set up the quadratic equation in terms of $u$ .

Cross-multiply to get $y (u^{2} + 1) = 8 u$ .

$y u^{2} - 8 u + y = 0$ .

Step 3: Apply the discriminant condition for real $u$ .

Case A: If $y = 0$ , the equation becomes $- 8 u = 0 ⟹ u = 0$ . Since $u = 0$ is a real number, $y = 0$ is in the range.

Case B: If $y \neq = 0$ , this is a quadratic equation in $u$ . For real solutions to exist, the discriminant $D$ must be non-negative.

$D = B^{2} - 4 A C = (- 8)^{2} - 4 (y) (y) \geq 0$ .

$64 - 4 y^{2} \geq 0$ .

$4 y^{2} \leq 64 ⟹ y^{2} \leq 16$ .

Taking the square root gives $- 4 \leq y \leq 4$ .

Step 4: Combine and count the integers.

The full range is the closed interval $[- 4, 4]$ .

The integer values in this interval are: $- 4, - 3, - 2, - 1, 0, 1, 2, 3, 4$ .

Counting these gives exactly 9 integers.

Answer: 9

Question 3 · Quantitative Ability NAT

Let $f (x) = \frac{x ^{2} + a x + 1}{x ^{2} - x + 1}$ for all real numbers $x$ . If the range of $f (x)$ is exactly the closed interval $[\frac{1}{3}, 3]$ , find the value of $a^{2} + a$ .

Correct Answer:

Step-by-Step Solution

Key idea: This is a reverse engineering range question, recognisable because the range is given and you must find the unknown coefficient in the function. The trigger is using the discriminant method to generate a quadratic inequality in $y$ , then matching its roots to the given range endpoints.

Step 1: Set $y = f (x)$ and cross-multiply.

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

$y (x^{2} - x + 1) = x^{2} + a x + 1$

$(y - 1) x^{2} - (y + a) x + (y - 1) = 0$

Step 2: Check the $y = 1$ case.

If $y = 1$ , the equation becomes $- (1 + a) x = 0$ . For this to have a real solution, we just need $x = 0$ (which works as long as $a \neq = - 1$ , but even if $a = - 1$ , $0 = 0$ means all $x$ work, which would make the range just ${1}$ , not $[1/3, 3]$ ). So $y = 1$ is in the range.

Step 3: Apply the discriminant for $y \neq = 1$ .

For real $x$ , we need $Δ \geq 0$ :

$Δ = [- (y + a)]^{2} - 4 (y - 1) (y - 1) \geq 0$

$(y + a)^{2} - 4 (y - 1)^{2} \geq 0$

Expand both squares:

$(y^{2} + 2 a y + a^{2}) - 4 (y^{2} - 2 y + 1) \geq 0$

$- 3 y^{2} + (2 a + 8) y + (a^{2} - 4) \geq 0$

Multiply by $- 1$ (flip sign):

$3 y^{2} - (2 a + 8) y - (a^{2} - 4) \leq 0$

Step 4: Match the roots to the given range.

We are given that the range is $[1/3, 3]$ . This means the roots of the quadratic $3 y^{2} - (2 a + 8) y - (a^{2} - 4) = 0$ must be exactly $1/3$ and $3$ .

Using Vieta's formulas:

Sum of roots: $\frac{1}{3} + 3 = \frac{10}{3}$

From the equation: Sum $= \frac{2 a + 8}{3}$

Equating them: $\frac{2 a + 8}{3} = \frac{10}{3} ⟹ 2 a + 8 = 10 ⟹ 2 a = 2 ⟹ a = 1$ .

Step 5: Verify with the product of roots.

Product of roots: $\frac{1}{3} \times 3 = 1$

From the equation: Product $= \frac{- ( a ^{2} - 4 )}{3}$

Equating them: $\frac{- ( a ^{2} - 4 )}{3} = 1 ⟹ - (a^{2} - 4) = 3 ⟹ a^{2} - 4 = - 3 ⟹ a^{2} = 1 ⟹ a = \pm 1$ .

The only value that satisfies BOTH sum and product conditions is $a = 1$ .

Step 6: Calculate the final answer.

$a^{2} + a = 1^{2} + 1 = 2$

Answer: 2

Common trap: Only using the product of roots to find $a = \pm 1$ and forgetting to check the sum of roots, which would leave ambiguity. Both Vieta's conditions must be satisfied simultaneously.

Question 4 · Quantitative Ability NAT

Let $f (x) = (x^{2} - 5 x - 2) (x^{2} - 5 x + 2)$ . What is the sum of all real solutions to the equation $f (x) + 4 = 6$ ?

Correct Answer:

Step-by-Step Solution

Key idea: This is a hidden perfect square question. The function is a product of two expressions that share a common core, and adding a constant turns it into a perfect square.

Step 1: Identify the common core and substitute.

Let $t = x^{2} - 5 x$ . The function can be rewritten as:

$f (x) = (t - 2) (t + 2) = t^{2} - 4$ .

Step 2: Simplify the expression under the square root.

The equation is $f (x) + 4 = 6$ .

Substitute $f (x)$ :

$(t^{2} - 4) + 4 = 6$

$t^{2} = 6$ .

Step 3: Handle the square root correctly.

The square root of a square is the absolute value: $t^{2} = ∣ t ∣$ .

So, $∣ t ∣ = 6$ . This splits into two cases:

Case 1: $t = 6$

Case 2: $t = - 6$ .

Step 4: Solve each case for $x$ and check for real roots.

Case 1: $x^{2} - 5 x = 6 ⟹ x^{2} - 5 x - 6 = 0$ .

Factoring: $(x - 6) (x + 1) = 0 ⟹ x = 6, x = - 1$ . (Both are real).

Case 2: $x^{2} - 5 x = - 6 ⟹ x^{2} - 5 x + 6 = 0$ .

Factoring: $(x - 2) (x - 3) = 0 ⟹ x = 2, x = 3$ . (Both are real).

Step 5: Sum all the valid real solutions.

Sum $= 6 + (- 1) + 2 + 3 = 10$ .

Trap to avoid: Forgetting the absolute value when simplifying $t^{2}$ . If you just wrote $t = 6$ , you would only find the roots $6$ and $- 1$ , missing the roots from the negative case.

Answer: 10

Question 5 · Quantitative Ability MCQ

Let $f (x) = \frac{9 ^{x}}{9 ^{x} + 3}$ for all real numbers $x$ . What is the exact value of the sum $\sum_{k = 1}^{2023} f (\frac{k}{2024})$ ?

A.
1011
B.
1011.5
C.
1012
D.
2023

Correct Answer:

Step-by-Step Solution

Key idea: This is a symmetric function summation question. The trigger is a sum of a fractional exponential function evaluated at evenly spaced points between 0 and 1. You should test for the symmetry property $f (x) + f (1 - x)$ .

Step 1: Test the symmetry $f (x) + f (1 - x)$ .

$f (1 - x) = \frac{9 ^{1 - x}}{9 ^{1 - x} + 3} = \frac{\frac{9}{9 ^{x}}}{\frac{9}{9 ^{x}} + 3}$ .

Multiply numerator and denominator by $9^{x}$ :

$f (1 - x) = \frac{9}{9 + 3 \cdot 9 ^{x}} = \frac{3}{3 + 9 ^{x}}$ .

Now add $f (x)$ and $f (1 - x)$ :

$f (x) + f (1 - x) = \frac{9 ^{x}}{9 ^{x} + 3} + \frac{3}{9 ^{x} + 3} = \frac{9 ^{x} + 3}{9 ^{x} + 3} = 1$ .

Step 2: Pair the terms in the summation.

The sum is $S = f (\frac{1}{2024}) + f (\frac{2}{2024}) + \dots + f (\frac{2023}{2024})$ .

Pair the first and last terms: $f (\frac{1}{2024}) + f (\frac{2023}{2024}) = 1$ .

Pair the second and second-to-last: $f (\frac{2}{2024}) + f (\frac{2022}{2024}) = 1$ .

Step 3: Count the pairs and the middle term.

There are 2023 terms in total.

Number of pairs = $\frac{2023 - 1}{2} = 1011$ pairs.

The sum of these pairs is $1011 \times 1 = 1011$ .

There is exactly 1 unpaired middle term, which occurs when the numerator is half the denominator: $k = 1012$ .

Step 4: Evaluate the middle term.

$f (\frac{1012}{2024}) = f (\frac{1}{2}) = \frac{9 ^{0.5}}{9 ^{0.5} + 3} = \frac{3}{3 + 3} = \frac{3}{6} = 0.5$ .

Step 5: Calculate the final sum.

Total Sum = (Sum of pairs) + (Middle term) = $1011 + 0.5 = 1011.5$ .

Answer: B

Functions, Domains, Ranges and Optimization Practice Questions for CAT: 127+ 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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