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FUNCTIONS: EAPCET PYQs (2020-2023)

Topic Wise Compilation

Expert’s Note: With 15 years of experience guiding EAPCET aspirants, I’ve compiled this topic-wise collection to maximize your revision efficiency. Mastering Functions is about understanding definitions and practicing their application.

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TOPIC 1: DOMAIN OF A FUNCTION

The set of all possible input values (x) for which the function is defined.

Key Concepts:

· √(f(x)) requires f(x) ≥ 0
· 1/f(x) requires f(x) ≠ 0
· log(f(x)) requires f(x) > 0
· For sin⁻¹(f(x)) and cos⁻¹(f(x)): -1 ≤ f(x) ≤ 1

PYQs with Solutions:

1. [2020 Shift-1] Find the domain of f(x) = \cos^{-1} \left( \frac{x – 3}{2} \right) – \log_{10} (4 – x)

Solution:

· For cos⁻¹: -1 ≤ (x-3)/2 ≤ 1 → 1 ≤ x ≤ 5
· For log: 4 – x > 0 → x < 4
· Combining: [1, 4)

2. [2020 Shift-2] The domain of \sqrt{|x| – x} is:

Solution:
√requires |x| – x ≥ 0 → |x| ≥ x
This is true for all x≤ 0
Domain:(-∞, 0]

3. [2021 Shift-1] Domain of f(x) = \sec^{-1} (3x – 4) + \tanh^{-1} \left( \frac{x+3}{5} \right)

Solution:

· For sec⁻¹: |3x-4| ≥ 1
· For tanh⁻¹: -1 < (x+3)/5 < 1
· Solving both: (-8, 1] ∪ [5/3, 2)

4. [2021 Shift-1] Domain of f(x) = \sqrt{\log_{10} \left( \frac{5x – x^2}{4} \right)}

Solution:

1. (5x – x²)/4 > 0 → 0 < x < 5
2. log₁₀((5x – x²)/4) ≥ 0 → (5x – x²)/4 ≥ 1 → x² – 5x + 4 ≤ 0 → 1 ≤ x ≤ 4

· Combining: [1, 4]

5. [2022 Shift-1] Domain of f(x) = \frac{\log_2 (x + 3)}{\sqrt{x^2 + 3x + 2}}

Solution:

· Numerator: x + 3 > 0 → x > -3
· Denominator: x² + 3x + 2 > 0 → (x+1)(x+2) > 0 → x ∈ (-∞, -2) ∪ (-1, ∞)
· Combining: (-3, -2) ∪ (-1, ∞)

6. [2022 Shift-2] Domain of f(x) = \frac{\sqrt{2 – x} + \sqrt{1 + x}}{\sqrt{x + 3}}

Solution:

· 2 – x ≥ 0 → x ≤ 2
· 1 + x ≥ 0 → x ≥ -1
· x + 3 > 0 → x > -3
· Combining: [-1, 2]

7. [2023 Shift-1] Domain of f(x) = \frac{\log_{0.5}(x – 3)}{\sqrt{x – 1}}

Solution:

· x – 3 > 0 → x > 3
· x – 1 > 0 → x > 1
· Domain: (3, ∞)

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TOPIC 2: RANGE OF A FUNCTION

The set of all possible output values (f(x)) obtained.

Key Concepts:

· For quadratic functions, complete the square
· For rational functions, express x in terms of y
· For expressions involving √, find min/max values

PYQs with Solutions:

8. [2020 Shift-2] If f : R \rightarrow R is defined by f(x) = \frac{x^6}{x^6 + 2020} , find range

Solution:
Let y= x⁶/(x⁶+2020)
Solve for x⁶:x⁶ = 2020y/(1-y)
Since x⁶≥ 0, we require 2020y/(1-y) ≥ 0
Range:[0, 1)

9. [2021 Shift-1] Range of f(x) = x^2 + \frac{1}{x^2 + 1}

Solution:
Let t= x² + 1, so t ≥ 1
f(x)= (t – 1) + 1/t = t + 1/t – 1
For t≥ 1, minimum at t=1: 1 + 1 – 1 = 1
As t→∞,f(x)→∞
Range:[1, ∞)

10. [2022 Shift-1] Range of f(x) = \frac{x^2 + x + 1}{x}

Solution:
y= x + 1 + 1/x
Let z= x + 1/x, where |z| ≥ 2
So y= z + 1
Thus,y ≥ 3 or y ≤ -1
Range:(-∞, -1] ∪ [3, ∞)

11. [2023 Shift-1] Range of f(x) = |x – 2| + |x – 3|

Solution:
Represents sum of distances from x to 2 and 3
Minimum value= distance between 2 and 3 = 1
Increases without bound as x→±∞
Range:[1, ∞)

12. [2023 Shift-1] Range of f(x) = \sqrt{9 – x^2}

Solution:
Expression inside root:9 – x²
Maximum= 9 (at x=0), Minimum = 0 (at x=±3)
Range:[0, 3]

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TOPIC 3: TYPES OF FUNCTIONS

(Injective, Surjective, Bijective)

Definitions:

· Injective (One-one): Every domain element maps to unique codomain element
· Surjective (Onto): Every codomain element has at least one pre-image
· Bijective: Both one-one and onto

PYQs with Solutions:

13. [2020 Shift-2] f : R \rightarrow R defined by f(x) = \frac{x}{\sqrt{1 + x^2}} is:

Solution:

· Check one-one: f(a)=f(b) ⇒ a=b ✓
· Check onto: Solve y = x/√(1+x²)
For |y|=1, no solution ✗
· Injective but not surjective

14. [2021 Shift-1] Let S be finite set, then non-identity function f : S \to S can be:

Solution:
For finite sets,injective ⇔ surjective ⇔ bijective
Non-identity function can still be bijective
Bijective

15. [2022 Shift-2] If f : N \times N \to N defined by f((m,n)) = 2^{m-1}(2n-1) , then f is:

Solution:
Every natural number can be uniquely expressed as power of 2× odd number
This is abijection

16. [2023 Shift-1] f : R \to R defined as f(x) = x – [x] + 3 is:

Solution:
{x}= x – [x] is periodic with period 1
Adding constant doesn’t change periodicity
Periodic function with period 1

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TOPIC 4: FUNCTIONAL EQUATIONS

Equations where the unknown is a function.

PYQs with Solutions:

17. [2020 Shift-1] How many bijections f : Z \rightarrow Z such that f(x + y) = f(x) + f(y) ?

Solution:
Solutions:f(x) = kx
For bijection on Z:k = ±1
Twobijections: f(x)=x and f(x)=-x

18. [2020 Shift-2] How many functions f : Z \rightarrow Z such that f(x + y) = f(x) + f(y) ?

Solution:
Solutions:f(x) = kx
For any integer k,maps Z→Z
Infinitely many

19. [2020 Shift-1] Find \sum_{t=1}^{30} f(t) if f(x + y) = f(x) + f(y) and f(1) = 7

Solution:
f is linear:f(x)=kx
f(1)=7⇒ k=7, so f(x)=7x
∑f(t)from 1 to 30 = 7∑t = 7×(30×31)/2 = 3255

20. [2020 Shift-2] If (f(x))^2 = f(x^2) + f(1) , find f(x)

Solution:
Try f(x)= x + 1/x
f(x)²= x² + 2 + 1/x²
f(x²)+ f(1) = (x² + 1/x²) + 2
They are equal
f(x) = x + 1/x

21. [2021 Shift-2] For A_n = \{(n+1)k / k \in N\} , X = \bigcup_{n \in N} A_n , f : X \rightarrow N defined by f(x) = x is:

Solution:
X= set of composite numbers (and some primes)
f(x)=x is one-one but not onto(primes have no pre-image)
One-one but not onto

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TOPIC 5: INVERSE & COMPOSITION

PYQs with Solutions:

22. [2020 Shift-1] Find g(t) if f(t) = 3t – 2 and (g∘f)⁻¹(t) = t – 2

Solution:
Let h(t)= (g∘f)⁻¹(t) = t – 2
Then g∘f= h⁻¹, so h⁻¹(t) = t + 2
Thus(g∘f)(t) = t + 2
g(f(t))= t + 2
g(3t- 2) = t + 2
Let u= 3t – 2 ⇒ t = (u+2)/3
g(u)= (u+2)/3 + 2 = (u+8)/3
g(t) = (t+8)/3

23. [2021 Shift-2] Inverse of y = \frac{10^x – 10^{-x}}{10^x + 10^{-x}}

Solution:
y= (10²ˣ – 1)/(10²ˣ + 1)
Solve for 10²ˣ:10²ˣ = (1+y)/(1-y)
2x= log₁₀((1+y)/(1-y))
f⁻¹(x) = ½ log₁₀((1+x)/(1-x))

24. [2022 Shift-1] If f(x) = (x + 2)^2 – 2, x \geq -2 , find f⁻¹(x)

Solution:
y= (x+2)² – 2
(x+2)²= y + 2
Since x≥ -2, x+2 ≥ 0
x+2= √(y+2)
f⁻¹(x) = √(x+2) – 2

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TOPIC 6: SPECIAL FUNCTIONS & [x]

(Greatest Integer Function)

PYQs with Solutions:

25. [2020 Shift-1] Domain of f(x) = [(x)^2 – [x] – 2]^{-1/2}

Solution:
Expression inside(-1/2) power must be > 0
[x]²- [x] – 2 > 0
Let t=[x]:t² – t – 2 > 0 ⇒ (t-2)(t+1) > 0 ⇒ t < -1 or t > 2
So[x] < -1 or [x] > 2
Domain:R – [-1, 3)

26. [2020 Shift-2] For f(x) = \frac{\sin \pi [x]}{1 + [x]} + \frac{x}{2 + 3x} , find domain and range

Solution:

· Domain: 1+[x] ≠ 0 ⇒ [x] ≠ -1 ⇒ x ∉ [-1, 0)
Also 2+3x ≠ 0 ⇒ x ≠ -2/3
Domain: R – [-1, -2/3)
· First term = 0 (since sin(πn)=0)
· Function simplifies to x/(2+3x) on its domain
· Range: R – {1/3}

27. [2021 Shift-2] Let f(x) = [x] and g(x) = 3[x/3], find {x: f(x) = g(x)}

Solution:
We need[x] = 3[x/3]
Let x= 3q + r, where 0 ≤ r < 3
Then[x] = 3q + [r]
[x/3] = [q + r/3] = q + [r/3]
g(x)= 3(q + [r/3]) = 3q + 3[r/3]
For equality:[r] = 3[r/3]
This holds for r∈ [0,1)
{x ∈ R / 3k ≤ x < 3k+1, k ∈ Z}

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QUICK ANSWER KEY

1. [1, 4)
2. (-∞, 0]
3. (-8, 1] ∪ [5/3, 2)
4. [1, 4]
5. (-3, -2) ∪ (-1, ∞)
6. [-1, 2]
7. (3, ∞)
8. [0, 1)
9. [1, ∞)
10. (-∞, -1] ∪ [3, ∞)
11. [1, ∞)
12. [0, 3]
13. Injective but not surjective
14. Bijective
15. Bijective
16. Periodic with period 1
17. 2
18. Infinitely many
19. 3255
20. x + 1/x
21. One-one but not onto
22. g(t) = (t+8)/3
23. ½ log₁₀((1+x)/(1-x))
24. √(x+2) – 2
25. R – [-1, 3)
26. R – [-1, -2/3) & R – {1/3}
27. {x ∈ R / 3k ≤ x < 3k+1, k ∈ Z}

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Final Expert Advice:

· Practice these questions repeatedly
· Understand the logic behind each solution
· In exam, if stuck, try substituting values or checking options
· All the best for your EAPCET preparation!

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