NDA Relations and Functions previous year questions with solutions

Answer: (b) IV only

Method. a function from A to B gives each element of A exactly one image

I: 3 has two images (2 and 6) → not a function.
II: 3 has two images (4 and 2) → not a function.
III: 5 has no image → not a function from A.
IV: 1 → 6, 3 → 2, 5 → 2; each element of A has exactly one image → a function.
Only IV.

Answer: (a) I only

Method. definitions of domain, codomain and range of a relation

I: for R ⊆ A × B, B is the codomain. Correct.
II: range = set of second elements actually used, a subset of B; e.g. R = {(1, 2)} from {1} to {2, 3} has range {2} ≠ B. Wrong.
III: domain = set of first elements actually used, a subset of A; the same R from {1, 4} gives domain {1} ≠ A. Wrong.
Only I.

Answer: (d) 2n²

Method. n(P × Q) = n(P) n(Q)

n(X −⁠ Y) = n(X) − n(X ∩ Y) = 3n − n = 2n.
n(Y −⁠ X) = n(Y) − n(X ∩ Y) = 2n − n = n.
n((X −⁠ Y) × (Y −⁠ X)) = 2n × n = 2n².

Answer: (c) 15

Method. count pairs y by y

y = 0: |x| < 0 is impossible → 0 pairs.
y = 1: |x| < 1 → x = 0 → 1 pair.
y = 4: |x| < 4 → all 7 elements of A → 7 pairs.
y = 9: |x| < 9 → 7 pairs.
Total = 0 + 1 + 7 + 7 = 15.

Answer: (d) 100

Method. (A × B) ∩ (B × A) = (A ∩ B) × (A ∩ B)

(x, y) is in both A × B and B × A exactly when x ∈ A ∩ B and y ∈ A ∩ B.
So the common part is (A ∩ B) × (A ∩ B), with 10 × 10 = 100 elements.

Answer: (c) Both 1 and 2

Method. n(A × A) = n(A)²

n(A)² = 16, so n(A) = 4.
(0, 2) ∈ A × A gives 0, 2 ∈ A; (1, 3) ∈ A × A gives 1, 3 ∈ A.
These are four distinct elements, so A = {0, 1, 2, 3}: statement 1 is correct.
3 ∈ A and 2 ∈ A, so (3, 2) ∈ A × A: statement 2 is correct.

Answer: (b) 2 only

Method. list the pairs with y = 4x − 13 inside A

y = 4x − 13 is an integer only when x = 1, 4, 7, 10, 13, 16, 19.
These give y = 1, 5, 9, 13, 17, 21, 25; y must be ≤ 20, so x = 16 and 19 drop out.
R = {(1, 1), (4, 5), (7, 9), (10, 13), (13, 17)}.
Domain = {1, 4, 7, 10, 13}: statement 1 (which includes 16) is wrong.
Range = {1, 5, 9, 13, 17}: statement 2 is correct.
Codomain is A = {1, …, 20} ≠ range: statement 3 is wrong.

Answer: (a) 1 only

Method. a piecewise rule is a function only if overlapping pieces agree at the common point

Both rules cover x = 2 (in f) and x = 4 (in g) twice.
f at x = 2: x³ = 8 and 4x = 8 — the same value, so f is a function.
g at x = 4: x² = 16 but 3x = 12 — two images of 4, so g is not a function.
Only statement 1 is correct.

Answer: (d) Neither 1 nor 2

Method. a function needs a single image for each input

1. Input 0 arises from (x, y) = (0, 1) and (0, 2), giving images 0 − 1 = -1 and 0 − 2 = -2. One input, two images, so f is not a function.
2. Input 4 arises from (x, y) = (1, 4) and (2, 2), giving images 5 and 4. Again not a function.
Neither statement is correct.

Answer: (b) 2 and 3 only

Method. a function is a special kind of relation; a relation is a subset of a Cartesian product

Statement 1: every function is a relation, but not every relation is a function (an element may be related to two images). Statement 1 states the reverse, so it is false.
Statement 2: a relation from A to B is, by definition, a subset of A × B. True.
Statement 3: a relation in A is a relation from A to A, a subset of A × A. True.

Answer: (b) 3

Method. list natural-number solutions of a linear equation

2x = 20 − 3y must be even and positive, so y is even and 3y < 20.
y = 2 ⇒ x = 7; y = 4 ⇒ x = 4; y = 6 ⇒ x = 1.
R = {(7, 2), (4, 4), (1, 6)}: 3 elements.

Answer: (d) S is not a function

Method. a function from A gives each x exactly one image

x² + y² = 1 gives y = ±⁠√(1⁠ −⁠ x²).
For x = 0, both (0, 1) and (0, -1) are in S.
So the element 0 of A has two images, and S is not a function.

Answer: (b) transitive only

Method. test reflexivity with a negative number; transitivity via y ≥ 0

Reading xRy as |x| ≤ y.
Reflexive needs |x| ≤ x for all real x; x = -1 gives 1 ≤ -1, false. Not reflexive.
Transitive: if |x| ≤ y and |y| ≤ z, then y ≥ |x| ≥ 0 so |y| = y, and |x| ≤ y ≤ z, i.e. xRz. Transitive.
R is transitive only.

Answer: (d) Neither I nor II

Method. test each property with a counterexample

Symmetric: (8, 2) ∈ R since 8 = 2³, but (2, 8) ∉ R since 2 ≠ 8³. Not symmetric: I is wrong.
Transitive: (512, 8) ∈ R (512 = 8³) and (8, 2) ∈ R, but (512, 2) ∉ R since 2³ = 8 ≠ 512. Not transitive: II is wrong.
Neither statement is correct.

Answer: (d) I, II and III

Method. check each property against the definitions

I: (a, a) ∈ P and (a, a) ∈ Q for every a ∈ A, so (a, a) ∈ P ∩ Q: true.
II: (a, b) ∈ P ∪ Q lies in P or in Q; that relation is symmetric, so (b, a) lies in it too, hence in P ∪ Q: true.
III: (a, b), (b, c) ∈ P ∩ Q give (a, c) ∈ P(P transitive) and (a, c) ∈ Q (Q transitive), so (a, c) ∈ P ∩ Q: true.
All three are correct.

Answer: (d) R is an equivalence relation

Method. for x, y in (-1, 1) the condition |x + y| < 2 always holds

For x, y ∈ (-1, 1): -2 < x + y < 2, so |x + y| < 2 for every pair.
Hence R = (-1, 1) × (-1, 1), the universal relation.
The universal relation is reflexive, symmetric and transitive, so R is an equivalence relation.

Answer: (d) 1, 2 and 3

Method. x² = y³ in ℕ forces x = t³, y = t²; test each property

1. (x, x) ∈ R needs x² = x³, true only for x = 1; e.g. (2, 2) ∉ R. Not correct.
2. (8, 4) ∈ R since 64 = 64, but (4, 8) ∉ R since 16 ≠ 512. Not correct.
3. (512, 64) ∈ R (512² = 2¹⁸ = 64³) and (64, 16) ∈ R (64² = 2¹² = 16³), but (512, 16) ∉ R since 2¹⁸ ≠ 2¹². Not correct.
All three are not correct.

Answer: (a) 1 only

Method. factorise the condition: xRy ⇔ x = y or x = 4y

x² - 5xy + 4y² = (x − y)(x − 4y), so xRy ⇔ x = y or x = 4y.
Reflexive: x = x always, so xRx. True.
Symmetric: 4R1 (4 = 4 × 1), but 1R4 fails (1 ≠ 4 and 1 ≠ 16). False.
Transitive: 16R4 and 4R1, but 16R1 fails (16 ≠ 1 and 16 ≠ 4). False.
Only statement 1 is correct.

Answer: (d) 1, 2 and 3

Method. R^-1 = {(b, a): (a, b) ∈ R} inherits each property

1. R reflexive ⇒ (a, a) ∈ R for all a ⇒ (a, a) ∈ R^-1. True.
2. R symmetric ⇒ R^-1 = R, which is symmetric. True.
3. (a, b), (b, c) ∈ R^-1 ⇒ (c, b), (b, a) ∈ R ⇒ (c, a) ∈ R (R transitive) ⇒ (a, c) ∈ R^-1. True.

Answer: (c) transitive only

Method. log to a base below 1 is decreasing, so xRy means x < y

With a = 12, log_a t is a decreasing function, so log_a x > log_a y ⇔ x < y.
Reflexive: x < x is false, so not reflexive.
Symmetric: 1 < 2 but not 2 < 1, so not symmetric.
Transitive: x < y and y < z ⇒ x < z, so transitive.
Hence R is transitive only.

Answer: (d) R is neither reflexive nor transitive, but symmetric

Method. check each property against the listed pairs

Reflexive: (4, 4) ∉ R though 4 ∈ X, so R is not reflexive.
Symmetric: (1, 2), (2, 1) and (2, 3), (3, 2) both appear; the rest are (a, a). So R is symmetric.
Transitive: (1, 2) ∈ R and (2, 3) ∈ R but (1, 3) ∉ R, so R is not transitive.
So R is symmetric but neither reflexive nor transitive.

Answer: (d) R is reflexive, but neither symmetric nor transitive

Method. factor the quadratic: x R y ⇔ x = y or x = 3y

x² - 4xy + 3y² = (x − y)(x − 3y), so x R y ⇔ x = y or x = 3y.
Reflexive: x = x, so x R x.
Not symmetric: 3 R 1 (3 = 3.1) but 1 R 3 fails (1 ≠ 3, 1 ≠ 9).
Not transitive: 9 R 3 and 3 R 1, but 9 R 1 fails (9 ≠ 1, 9 ≠ 3).
So R is reflexive, but neither symmetric nor transitive.

Answer: (c) equivalence relation

Method. check reflexive, symmetric, transitive; equivalently a − b = c − d

a + d = b + c ⇔ a − b = c − d.
Reflexive: a − b = a − b, so (a, b) R (a, b).
Symmetric: a − b = c − d ⇒ c − d = a − b.
Transitive: a − b = c − d and c − d = e − f ⇒ a − b = e − f.
So R is an equivalence relation.

Answer: (b) The relation is not reflexive but it is symmetric and transitive

Method. test each property of 'born in Delhi on the same day'

Reflexive needs x R x for every x ∈ S, i.e. every Delhi resident was born in Delhi. A resident born elsewhere is not related to himself, so R is not reflexive.
Symmetric: if x and y were born in Delhi on the same day, so were y and x.
Transitive: x, y born in Delhi on the same day and y, z born in Delhi on the same day ⇒ x, z born in Delhi on the same day.
So R is symmetric and transitive but not reflexive.

Answer: (b) (0, ∞) - {1}

Method. change of base: log_x e = 1/⁠ln x

log_x e = ln eln x = 1ln x.
Need x > 0 (base positive) and ln x ≠ 0, i.e. x ≠ 1.
Domain = (0, ∞) - {1}.

Answer: (b) (-∞, ∞)

Method. a cube root is defined for every real number

(x − 1)² is defined for all real x, and ∛ of any real number is real.
So f(x) = 1 - ∛((x − 1)²) is defined for every x: domain = (-∞, ∞).

Answer: (b) a maximum at x = 1

Method. ∛((x − 1)²) ≥ 0 with equality only at x = 1

∛((x − 1)²) ≥ 0 for all x, and = 0 only at x = 1.
So f(x) = 1 - ∛((x − 1)²) ≤ 1 = f(1) for all x, with f(x) < 1 for x ≠ 1.
f has a maximum (value 1) at x = 1 (a cusp, where f' does not exist, but the maximum stands).

Answer: (b) [0.5, ∞)

Method. bound 1 − cos x and take the reciprocal

x ≠ 2nπ, so cos x ≠ 1 and -1 ≤ cos x < 1; hence 0 < 1 − cos x ≤ 2.
Taking reciprocals reverses the order: y = 11 − cos x ≥ 12, with y = 1/⁠2 at x = (2n + 1)π.
As cos x → 1, 1 − cos x → 0⁺ and y → ∞. Range = [0.5, ∞).

Answer: (b) -cot(x/⁠2) + c

Method. half-angle identity 1 − cos x = 2 sin²(x/⁠2)

y = 11 − cos x = 12 sin²(x/⁠2) = 12 cosec²(x/⁠2).
∫ 12 cosec²(x/⁠2) dx = 12 × (-cot(x/⁠2)) × 2 + c = -cot(x/⁠2) + c.

Answer: (c) (-∞, ∞)

Method. modulus is defined for every real number

|x − 3| is defined for every real x (it equals x − 3 for x ≥ 3 and 3 − x for x < 3).
Domain = (-∞, ∞).

Answer: (d) 9 square units

Method. area of the triangle cut off by y = 3 from the V-shaped graph

y = |x − 3| meets y = 3 where |x − 3| = 3, i.e. x = 0 and x = 6; the vertex of the V is (3, 0).
The bounded region is the triangle with vertices (0, 3), (6, 3), (3, 0): base 6, height 3.
Area = 12 × 6 × 3 = 9 square units.

Shortcut. ∫₀⁶ (3 − |x − 3|) dx = 18 − 9 = 9.

Answer: (c) 1

Method. fit a line through the given pairs

From (1, 1): p + q = 1. From (2, 4): 2p + q = 4, so p = 3, q = -2.
Check: f(3) = 7, f(4) = 10. (holds)
p + q = 3 − 2 = 1.

Shortcut. p + q = f(1) = 1 directly.

Answer: (a) I only

Method. one-one: distinct inputs give distinct outputs; onto: range equals codomain

f maps 1, 2, 3, 4 to 1, 4, 7, 10, which are all different, so f is one-one: I is correct.
Range = {1, 4, 7, 10} ≠ N (for example 2 has no pre-image), so f is not onto N: II is wrong.
Only I is correct.

Answer: (a) f(x) has some domain and g(x) has no domain

Method. the expression under a square root in a denominator must be > 0

f: need |x| − x > 0. For x ≥ 0, |x| − x = 0; for x < 0, |x| − x = -2x > 0. Domain of f = (-∞, 0).
g: need x − |x| > 0. For x ≥ 0 it is 0; for x < 0 it is 2x < 0. Never positive, so g has empty domain.
f has some domain, g has none.

Answer: (b) [π, π + 1]

Method. 0 ≤ sin² x ≤ 1

sin² x takes every value from 0 (at x = 0) to 1 (at x = π2).
Adding π: π ≤ π + sin² x ≤ π + 1.
Range = [π, π + 1].

Answer: (b) π

Method. sin² x = 1 − cos 2x2 has period π

f(x) = π + 1 − cos 2x2.
cos 2x has period 2π2 = π, and adding constants does not change the period.
Check: f(x + π) = π + sin²(x + π) = π + (-sin x)² = f(x); no smaller positive period works: a period T needs f(T) = f(0), i.e. sin² T = 0, so T is a multiple of π (in particular f(π2) = π + 1 ≠ f(0) = π).
Period = π.

Answer: (b) [-2, 2]

Method. both expressions under the square roots must be ≥ 0

2 − x ≥ 0 gives x ≤ 2; 2 + x ≥ 0 gives x ≥ -2.
Domain = [-2, 2] (the end points are allowed, since ⁠√0⁠ = 0).

Answer: (c) ⁠√8⁠

Method. square f and maximise 4 − x²

f(x)² = (2 − x) + (2 + x) + 2⁠√((2⁠ −⁠ x)(2 + x)) = 4 + 2⁠√(4⁠ −⁠ x²).
4 − x² is largest (= 4) at x = 0, so f² ≤ 4 + 2.2 = 8.
Greatest value = ⁠√8⁠ (at x = 0: f(0) = ⁠√2⁠ + ⁠√2⁠ = 2⁠√2⁠ = ⁠√8).⁠

Answer: (c) 6

Method. list the highest prime factor of each element

7 → 7, 8 → 2, 9 → 3, 10 → 5, 11 → 11, 12 → 3, 13 → 13, 14 → 7, 15 → 5, 16 → 2.
Range = {2, 3, 5, 7, 11, 13}: 6 elements.

Answer: (b) 1

Method. complete the square inside the log; log₁₀ is increasing

x² + 2x + 11 = (x + 1)² + 10 ≥ 10, with equality at x = -1.
Since log₁₀ is increasing, minimum f = log₁₀ 10 = 1.

Answer: (d) [0, 2]

Method. expression under the square root must be non-negative

Need 1 − (x − 1)² ≥ 0 ⇒ (x − 1)² ≤ 1 ⇒ -1 ≤ x − 1 ≤ 1 ⇒ 0 ≤ x ≤ 2.
The end points give ⁠√0⁠ = 0, so they are included: domain = [0, 2].

Answer: (b) [0, 2]

Method. range of sin x is [-1, 1]; transform it

-1 ≤ sin x ≤ 1, both ends attained.
Multiply by -1: -1 ≤ -sin x ≤ 1.
Add 1: 0 ≤ 1 − sin x ≤ 2.
Range = [0, 2] (closed: 0 at x = π2, 2 at x = -π2).

Answer: (a) (-∞, ∞)

Method. an exponential with positive base is defined for every real exponent

3^x is defined (and positive) for every real x, including x = 0 (3⁰ = 1) and negative x (3^-x = 13^x).
Domain = (-∞, ∞). (The range, not the domain, is (0, ∞).)

Answer: (d) [1, 3]

Method. range of cos x is [-1, 1]; shift it by 2

-1 ≤ cos x ≤ 1 for every x, with both ends attained (x = π and x = 0).
Adding 2: 1 ≤ 2 + cos x ≤ 3.
So the ordinate y takes exactly the values in [1, 3]; [0, 3] also contains them but is not the set of values taken.

Answer: (a) -8

Method. put x = 0 and evaluate f(2) + f(-2)

At x = 0 the expression is f(2) + f(-2).
f(2) = 2(2) − 2² = 0.
f(-2) = 2(-2) − (-2)² = -4 − 4 = -8.
Sum = 0 + (-8) = -8.

Answer: (b) The domain is {x ∈ R | x ≠ 1}, the range is {y ∈ R | y ≠ 0} and the graph intersects y-axis at (0, -1).

Method. domain, range and y-intercept of a rectangular hyperbola

Domain: x − 1 ≠ 0 ⇒ x ≠ 1.
Range: y = 1x − 1 ⇒ x = 1 + 1y, defined for every y ≠ 0, and 1x − 1 is never 0; so range = {y ∈ R | y ≠ 0}.
y-axis: x = 0 gives y = 1-1 = -1, so the graph meets it at (0, -1).
Option (a) is wrong because 0 is not in the range; (c) and (d) are wrong outright.

Answer: (c) 3x² - 2x + 5

Method. f(x) = ax² + bx + c with three conditions

f(0) = 5 ⇒ c = 5.
f(-1) = a − b + 5 = 10 ⇒ a − b = 5.
f(1) = a + b + 5 = 6 ⇒ a + b = 1.
So a = 3, b = -2, and f(x) = 3x² - 2x + 5.

Shortcut. Test f(1) = 6 in the options: (a) gives 8, (b) -4, (c) 6, (d) -2; only (c) fits (and its f(0) = 5, f(-1) = 10).

Answer: (c) 3

Method. perimeter/⁠area of a circle = 2/⁠r

f(r) = 2πrπr² = 2r.
f(1) + f(2) = 2 + 1 = 3.

Answer: (c) [2, 3]

Method. radicand ≥ 0

(2 − x)(x − 3) ≥ 0 ⇔ (x − 2)(x − 3) ≤ 0.
The product of (x − 2) and (x − 3) is ≤ 0 between the roots, endpoints included: 2 ≤ x ≤ 3.
Domain = [2, 3].

Answer: (d) [1, 4) ∪ (4, ∞)

Method. radicand ≥ 0 and denominator ≠ 0

⁠√(x⁠ −⁠ 1) is real only for x − 1 ≥ 0, i.e. x ≥ 1.
The denominator x − 4 must not be 0, so x ≠ 4.
Domain = [1, 4) ∪ (4, ∞).

Answer: (c) [1, 5)

Method. complete the square and track the open ends of A

f(x) = (x − 2)² + 1.
On (1, 4) the minimum is f(2) = 1, attained since 2 ∈ (1, 4).
On (1, 2], f falls from values near f(1) = 2 down to 1; on [2, 4), f rises from 1 towards f(4) = 5, never reaching it.
So the range is [1, 5).

Answer: (a) [0, 1)

Method. write x²/⁠(1 + x²) = 1 − 1/⁠(1 + x²)

f(x) = 1 − 11 + x².
1 + x² ≥ 1, so 0 < 11 + x² ≤ 1, with the value 1 at x = 0.
Hence 0 ≤ f(x) < 1: f(0) = 0, and f(x) → 1 as |x| → ∞ but never reaches 1.
Every value in [0, 1) is attained (x = ⁠√(y1⁠ −⁠ y)), so the range is [0, 1).

Answer: (d) [-4, 6]

Method. a sin x + b cos x ranges over [-⁠√(a²⁠ + b²), ⁠√(a²⁠ + b²)]

4 sin x − 3 cos x = 5 sin(x − α), where cos α = 45, sin α = 35.
So 4 sin x − 3 cos x takes every value in [-5, 5].
f(x) = 4 sin x − 3 cos x + 1 takes every value in [-4, 6].
f is onto exactly when S is its range, so S = [-4, 6].

Answer: (a) (-∞, 0)

Method. the expression under the root in a denominator must be positive

Need |x| − x > 0.
For x ≥ 0: |x| − x = 0, not allowed.
For x < 0: |x| − x = -2x > 0, allowed.
Domain = (-∞, 0).

Answer: (d) [0, 1]

Method. AM ≥ GM bounds x²/⁠(1 + x⁴) to [0, 1/⁠2]

At x = 0 the value is 0; for x ≠ 0, x²1 + x⁴ = 1x² + 1x².
x² + 1x² ≥ 2 (AM ≥ GM), with equality at x = ±1, so 0 < value ≤ 12 for x ≠ 0.
Hence the expression takes every value in [0, 12], including 0 at x = 0.
(a), (b), (c) all exclude 0, the value at x = 0; only [0, 1] contains every value.

Answer: (c) Graph (c)

Method. simplify x/⁠x and mark the excluded point

For every x ≠ 0, xx = 1, and f is undefined at x = 0.
So the graph is the line y = 1 with the point (0, 1) removed.
Graph (c) shows y = 1 across the plane with an open circle at (0, 1). (a) has no gap at x = 0; (b) is the graph of x|x|.

Answer: (c) Both I and II

Method. f(x) = (ax + b)/⁠(cx − a) with a² + bc ≠ 0 is its own inverse

Range check: 3x + 25x − 3 = 35 ⇒ 15x + 10 = 15x − 9, impossible; so f maps R - {3/⁠5} into R - {3/⁠5}.
f(f(x)) = 3 · 3x + 25x − 3 + 25 · 3x + 25x − 3 − 3 = 9x + 6 + 10x − 615x + 10 − 15x + 9 = 19x19 = x.
So f ∘ f is the identity on R - {3/⁠5}: f is one-one and onto (bijective), and f^-1 = f.
Both I and II are correct.

Answer: (d) 9

Method. each element of A independently chooses an image in B

Each of the 2 elements of A can map to any of the 3 elements of B.
Number of mappings = 3 × 3 = 3² = 9.

Answer: (a) 1 only

Method. check one-one and onto on (-1, 1) for each function

f(x) = x|x| equals x² for x ≥ 0 and -x² for x < 0: it is strictly increasing and continuous.
As x runs over (-1, 1), f(x) runs over (-1, 1) exactly, so f is a bijection of A onto A.
g(x) = cos(πx) is even: g(-12) = g(12) = 0, so g is not one-one.
(Also g(0) = 1 ∉ A, so g does not even map A into A.)
Only 1 is bijective.

Answer: (c) 30

Method. onto maps to a 2-element set = all maps minus the 2 constant maps

Total functions A → B: 2⁵ = 32.
A function fails to be onto B only if all of A goes to 6 or all to 7: 2 such functions.
Onto functions = 32 − 2 = 30.

Answer: (d) A = R ∖ {-53} and B = R ∖ {23}

Method. domain excludes the zero of the denominator; range excludes the ratio of leading coefficients

Domain: 3x + 5 ≠ 0 ⇒ A = R ∖ {-53}.
Solve y = 2x + 33x + 5 for x: 3xy + 5y = 2x + 3 ⇒ x = 3 − 5y3y − 2, which exists for every y ≠ 23.
y = 23 is never taken (it would need 6x + 10 = 6x + 9). So for onto, B = R ∖ {23}.

Answer: (c) Both 1 and 2

Method. one-one and onto tests for x ↦ x + 1 on ℤ and on ℕ

Both maps are one-one: x + 1 = y + 1 ⇒ x = y.
On ℤ: any y ∈ ℤ has preimage y − 1 ∈ ℤ, so f is onto. Statement 1 is correct.
On ℕ: y = 1 would need x = 0 ∉ ℕ, so 1 has no preimage and f is not onto. Statement 2 is correct.

Answer: (d) 2(1 + x)1 − x

Method. put y = f(x) and solve for x

y = x − 2x + 2 ⇒ xy + 2y = x − 2 ⇒ x(1 − y) = 2 + 2y ⇒ x = 2(1 + y)1 − y.
So f^-1(x) = 2(1 + x)1 − x.
Check: x = 0 gives f^-1(0) = 2, and f(2) = 0.

Answer: (a) x = y^{1/⁠(ln 5)}, y > 0

Method. rewrite 5^{ln x} as a power of x and solve for x

y = 5^{ln x} = e^{ln x · ln 5} = x^{ln 5}, defined for x > 0, so y > 0.
Raise both sides to 1ln 5: x = y^{1/⁠(ln 5)}, y > 0.

Shortcut. Take ln: ln y = ln x · ln 5, so ln x = ln yln 5.

Answer: (a) [0, π] and [-1, 1]

Method. a bijection needs cos to be monotonic on X and Y to be exactly its range

On [0, π], cos x decreases strictly from 1 to -1, so it is one-one there.
Its range on [0, π] is [-1, 1], so with Y = [-1, 1] it is onto.
(b) fails: cos(-x) = cos x on [-π2, π2], and the range is only [0, 1].
(c) fails: cos 0 = 1 ∉ (-1, 1). (d) fails: cos π = -1 ∉ [0, 1].

Answer: (d) one-one and onto

Method. write f − g piecewise and test one-one and onto

(f − g)(x) = x − 0 = x if x is rational; 0 − x = -x if x is irrational.
One-one: rational x gives a rational value and irrational x gives an irrational value, so equal values come from the same type; then x₁ = x₂ (or -x₁ = -x₂).
Onto: a rational y is the image of y; an irrational y is the image of the irrational -y.
So f − g is one-one and onto.

Answer: (a) 0

Method. f(0) = 0, so 0 is a fixed point

f(0) = 10⁰ - 10⁰10⁰ + 10⁰ = 1 − 12 = 0.
Then f(f(0)) = f(0) = 0, and each further application again gives 0.
f ∘ f ∘ f ∘ f ∘ f(0) = 0.

Answer: (d) 12 log₁₀(1 + x1 − x)

Method. solve y = f(x) for x

Multiply numerator and denominator by 10^x: y = 10^2x - 110^2x + 1.
y(10^2x + 1) = 10^2x - 1 ⇒ 10^2x(1 − y) = 1 + y ⇒ 10^2x = 1 + y1 − y.
x = 12 log₁₀(1 + y1 − y); so f^-1(x) = 12 log₁₀(1 + x1 − x), -1 < x < 1.

Answer: (a) 7

Method. equate the constant terms of f(g(x)) and g(f(x))

f(g(x)) = 4(kx + 2) + 1 = 4kx + 9.
g(f(x)) = k(4x + 1) + 2 = 4kx + k + 2.
Equality for all x: 9 = k + 2, so k = 7.

Answer: (c) cos² x

Method. match √ and cos² so that both composites come out right

Try f(x) = cos² x and g(x) = ⁠√x.⁠
f(g(x)) = cos²(⁠√x)⁠ (holds).
g(f(x)) = ⁠√(cos²⁠ x) = |cos x| (holds).
So f(x) = cos² x.
(No other option works: e.g. f = cos x would need cos(g(x)) = cos² ⁠√x,⁠ which none of the listed g gives.)

Answer: (a) ⁠√x⁠

Method. from the previous item, f(x) = cos² x and g(x) = ⁠√x⁠

With g(x) = ⁠√x⁠ and f(x) = cos² x: f(g(x)) = cos² ⁠√x⁠ and g(f(x)) = ⁠√(cos²⁠ x) = |cos x|, both as given.
So g(x) = ⁠√x.⁠

Answer: (d) f(d) + g(b) = 2d

Method. equate the constant terms of f(g(x)) and g(f(x))

f(g(x)) = a(cx + d) − b = acx + ad − b.
g(f(x)) = c(ax − b) + d = acx − bc + d.
Equality gives ad − b = d − bc.
Now f(d) + g(b) = (ad − b) + (bc + d) = (d − bc) + bc + d = 2d.
(Option a would need ad − b = bc + d, i.e. bc = 0, which is not forced.)

Answer: (a) x³ - 1x³

Method. write x³ - 1x³ in terms of t = x − 1x

(x − 1x)³ = x³ - 1x³ − 3(x − 1x), so x³ - 1x³ = t³ + 3t.
Hence f(t) = t³ + 3t, i.e. f(x) = x³ + 3x.
f(x) − 3x = x³.
g[f(x) − 3x] = g(x³) = x³ - 1x³.

Answer: (d) 6x

Method. differentiate f(x) = x³ + 3x twice

From the previous item, f(x) = x³ + 3x.
f'(x) = 3x² + 3; f''(x) = 6x.

Answer: (b) ⁠√(x⁠ + 1) - ∜(x + 1) + 1

Method. substitute t = f(x) = x² - 1 and write x in terms of t

g(f(x)) = g(x² - 1) = x - ⁠√x⁠ + 1 (x ≥ 0, since ⁠√x⁠ appears).
Let t = x² - 1, so x = ⁠√(t⁠ + 1) and ⁠√x⁠ = ∜(t + 1).
g(t) = ⁠√(t⁠ + 1) - ∜(t + 1) + 1.
So g(x) = ⁠√(x⁠ + 1) - ∜(x + 1) + 1.

Answer: (c) 3

Method. evaluate g(x) = ⁠√(x⁠ + 1) - ∜(x + 1) + 1 at x = 15

g(15) = ⁠√16⁠ - ∜16 + 1 = 4 − 2 + 1 = 3.

Shortcut. g(15) = g(f(4)) since f(4) = 15, so g(15) = 4 - ⁠√4⁠ + 1 = 3.

Answer: (d) -3

Method. (fg)(x) means the product f(x)·g(x)

(fg)(1) = f(1)·g(1).
f(1) = 1² + 2 = 3; g(1) = 2(1) − 3 = -1.
(fg)(1) = 3 × (-1) = -3.
(The composite f∘g would give f(g(1)) = f(-1) = 3; the printed fg is the product.)

(fg)(1) is the product f(1) g(1), the standard notation (NCERT writes the composite as gof or g∘f), which gives −⁠3. Read as the composite f(g(1)), it would give 3, option (a).

Answer: (a) 7

Method. equate fog and gof and compare constant terms

fog(x) = 4(kx + 2) + 1 = 4kx + 9.
gof(x) = k(4x + 1) + 2 = 4kx + k + 2.
The x terms agree; equal constants give 9 = k + 2, so k = 7.

Answer: (a) -1

Method. -1 is a fixed point of f

f(-1) = 4(-1) + 3 = -1.
So f(f(-1)) = f(-1) = -1 and f(f(f(-1))) = -1.

Answer: (a) 0

Method. composition worked from the inside out: f first, then g, then h

[ho(gof)](x) = h(g(f(x))).
f(⁠√π2)⁠ = (⁠√π2)²⁠ = π4.
g(π4) = tan π4 = 1.
h(1) = ln 1 = 0.

Answer: (d) 256

Method. repeated composition of the squaring function

[fo(fof)](2) = f(f(f(2))).
f(2) = 2² = 4.
f(4) = 4² = 16.
f(16) = 16² = 256.

Shortcut. f applied three times gives x⁸, and 2⁸ = 256.

Answer: (b) -1, -1

Method. substitute f(x) into g and solve the quadratic

g[f(x)] = 5(x² + 2x − 5) + 30 = 5x² + 10x + 5 = 5(x + 1)².
g[f(x)] = 0 ⇒ (x + 1)² = 0 ⇒ x = -1, -1 (a repeated root).

Answer: (d) Neither 1 nor 2

Method. degree of a composite = product of the degrees

f has degree 2, g has degree 1.
f[g(x)] = (5x + 30)² + 2(5x + 30) − 5 has degree 2 × 1 = 2, not 3: statement 1 is false.
g[g(x)] = 5(5x + 30) + 30 = 25x + 180 has degree 1, not 2: statement 2 is false.

Answer: (b) -20

Method. expand h(x), the x² terms cancel, then differentiate

h(x) = 5(x² + 2x − 5) − x(5x + 30) = 5x² + 10x − 25 − 5x² - 30x = -20x − 25.
h'(x) = -20.

Answer: (b) 1

Method. evaluate g first, then f

Let t = e − 1e + 1. Then 1 + t = 2ee + 1 and 1 − t = 2e + 1.
1 + t1 − t = e, so g(t) = ln e = 1.
f(1) = 4 + 11 + 4 = 1.
So f o g(e − 1e + 1) = 1.

Answer: (c) Both 1 and 2

Method. evaluate the compositions from the inside out

f(-1) = 1 − 3 = -2, f(-2) = 4 − 3 = 1, f(1) = 1 − 3 = -2.
(fofof)(-1) = f(f(-2)) = f(1) = -2.
(fofof)(1) = f(f(-2)) = f(1) = -2. So statement 1 is correct.
(fofof)(-1) − 4(fofof)(1) = -2 − 4(-2) = 6.
(fof)(0) = f(-3) = 9 − 3 = 6. So statement 2 is correct.

Shortcut. f is even, so f(-1) = f(1) and statement 1 is immediate.

Answer: (c) f(n) = g(q)

Method. compare the constant terms of f(g(x)) and g(f(x))

f(g(x)) = p(mx + n) + q = pmx + pn + q.
g(f(x)) = m(px + q) + n = mpx + mq + n.
The x-terms agree, so f(g(x)) = g(f(x)) ⇔ pn + q = mq + n.
pn + q = f(n) and mq + n = g(q), so the condition is f(n) = g(q).

Answer: (c) 0

Method. compare coefficients in f(x) + f(1/⁠x) = f(x)f(1/⁠x)

LHS = xⁿ + x^-n + 2k. RHS = (xⁿ + k)(x^-n + k) = 1 + k xⁿ + k x^-n + k².
Coefficient of xⁿ: k = 1 (constants then agree: 2 = 1 + 1). So f(x) = xⁿ + 1.
f(2) = 2ⁿ + 1 = 9 ⇒ n = 3.
f(-1) = (-1)³ + 1 = 0.

Answer: (b) x − (13)

Method. replace x by 1 − x and solve the two equations

Given: 2f(x) + f(1 − x) = x … (1). Replace x by 1 − x: 2f(1 − x) + f(x) = 1 − x … (2).
2 × (1) − (2): 4f(x) − f(x) = 2x − (1 − x) ⇒ 3f(x) = 3x − 1.
f(x) = x − 13.

Shortcut. Test x = 0 and x = 1 in the options: only x − 1/⁠3 gives 2f(0) + f(1) = 0.

Answer: (a) -f(x²)

Method. substitute and simplify

f(x + 1) = x + 11 − (x + 1) = -x + 1x.
f(x)f(x + 1) = x1 − x × (-xx + 1) = -x²(1 − x)(1 + x) = -x²1 − x².
And f(x²) = x²1 − x², so the ratio = -f(x²).

Answer: (d) 0

Method. substitute and use 1 − x = (1 - ⁠√x)(1⁠ + ⁠√x)⁠

(1 − x)f(⁠√x)⁠ = (1 - ⁠√x)(1⁠ + ⁠√x)⁠ × ⁠√x1⁠ - ⁠√x⁠ = ⁠√x(1⁠ + ⁠√x).⁠
f(⁠√x⁠ + 1) = ⁠√x⁠ + 11 − (⁠√x⁠ + 1) = -⁠√x⁠ + 1⁠√x,⁠ so xf(⁠√x⁠ + 1) = -⁠√x(√x⁠ + 1).
Sum = ⁠√x(1⁠ + ⁠√x)⁠ - ⁠√x(1⁠ + ⁠√x)⁠ = 0.

Answer: (d) 81

Method. the functional equation makes f multiplicative: f(xy) = f(x)f(y)

Put x = y: f(1) = f(x)/⁠f(x) = 1.
Replace x by xy: f(x) = f(xy)/⁠f(y), so f(xy) = f(x)f(y).
f(4) = f(2)f(2) = 9, f(16) = f(4)f(4) = 81.

Shortcut. f(2^k) = 3^k, so f(16) = 3⁴ = 81.

Answer: (c) 9

Method. f(1) = 1 and f multiplicative

Put x = y: f(1) = 1.
From f(xy) = f(x)f(y) (previous item): f(4) = f(2)² = 9.
f(1) f(4) = 1 × 9 = 9.

Answer: (d) 10

Method. put y = 0 to show f is constant

Put y = 0: f(x · 0) = f(x + 0), i.e. f(0) = f(x) for every real x.
So f is a constant function, and f(0) = f(5) = 10.

Answer: (c) 20

Method. f is constant

From y = 0, f(x) = f(0) for all x, and f(5) = 10, so f(x) = 10 for every x.
f(20) + f(-20) = 10 + 10 = 20.

Answer: (a) 1/⁠4

Method. use y = 1 to get f(1), then x = 2, y = 12

x = 2, y = 1: f(2)f(1) = f(2); since f(2) = 4 ≠ 0, f(1) = 1.
x = 2, y = 12: f(2)f(12) = f(1) = 1.
f(12) = 14.

Shortcut. f(x) = x² fits: f(2) = 4, f(12) = 14.

Answer: (d) 38x − x8 + 14

Method. replace x by 1x and solve the two equations for f(x)

3f(x) + f(1x) = 1x + 1 …(1)
x → 1x: 3f(1x) + f(x) = x + 1 …(2)
3×(1) − (2): 8f(x) = 3x + 3 − x − 1 = 3x − x + 2.
f(x) = 38x − x8 + 14.

Answer: (a) ln(8⁠√e)⁠

Method. integrate 8f(x) = 3x − x + 2 from 1 to 2

8∫₁² f(x)dx = ∫₁² (3x − x + 2)dx.
= [3 ln x − x²2 + 2x]₁² = 3 ln 2 − 32 + 2 = 3 ln 2 + 12.
= ln 8 + ln ⁠√e⁠ = ln(8⁠√e).⁠

Answer: (d) 3164

Method. the equation forces f(n + 1) = f(n)·f(1), so f(n) = (0.5)ⁿ

Put x = y: f(0) = f(x)f(x) = 1.
Put x = n + 1, y = 1: f(n) = f(n + 1)f(1), so f(n + 1) = f(n)·f(1) = 12f(n).
Hence f(n) = (12)ⁿ: f(2) = 14, f(3) = 18, …, f(6) = 164.
Sum = 16 + 8 + 4 + 2 + 164 = 3164.

Shortcut. f(x) = (12)^x satisfies the equation; the sum is a GP.

Answer: (b) 23

Method. replace x by 1x and solve the two equations for f(x)

2f(x) + f(1x) = x + 3 …(1)
x → 1x: 2f(1x) + f(x) = 1x + 3 …(2)
2×(1) − (2): 3f(x) = 2x − 1x + 3, so f(x) = 13(2x − 1x + 3).
f(0.5) = 13(1 − 2 + 3) = 23.

Shortcut. Put x = 0.5 and x = 2 in the given equation: 2f(0.5) + f(2) = 3.5 and 2f(2) + f(0.5) = 5; eliminate f(2).

Answer: (c) 2

Method. differentiate f(x) = 13(2x − 1x + 3)

From the previous item, f(x) = 13(2x − 1x + 3).
f'(x) = 13(2 + 1x²).
f'(0.5) = 13(2 + 4) = 2.

Answer: (d) 4

Method. replace x by 1x and solve the two equations for f(x)

The right side is 4x² - 1x², so 4f(x) − f(1x) = 4x² - 1x² … (1)
Replace x by 1x: 4f(1x) − f(x) = 4x² − x² … (2)
4 × (1) + (2): 15f(x) = 16x² - 4x² + 4x² − x² = 15x², so f(x) = x².
f(2) = 4.

Shortcut. Put x = 2 and x = 12: 4f(2) − f(12) = 634 and 4f(12) − f(2) = -3; 4 × the first plus the second gives 15f(2) = 60.

Answer: (b) x² - 5x + 6

Method. substitute t = x + 1, i.e. x = t − 1

Let t = x + 1, so x = t − 1.
f(t) = (t − 1)² - 3(t − 1) + 2 = t² - 2t + 1 − 3t + 3 + 2 = t² - 5t + 6.
So f(x) = x² - 5x + 6.

Shortcut. Check x = 0: f(1) must be 2; only x² - 5x + 6 gives 1 − 5 + 6 = 2.

Answer: (c) f(x + y + z + 2)

Method. add the exponents

f(x) f(y) f(z) = 3^1+x · 3^1+y · 3^1+z = 3^3+x+y+z.
f(t) = 3^1+t, so 3^3+x+y+z = 3^1+(x+y+z+2) = f(x + y + z + 2).

Answer: (b) f(a²)

Method. substitute and simplify the ratio

f(a) = aa − 1, f(a + 1) = a + 1a.
f(a)f(a + 1) = aa − 1 · aa + 1 = a²a² - 1.
This is tt − 1 with t = a², i.e. f(a²).

Answer: (b) 2 only

Method. substitute 2a and 1a into f

(1) f(2a) = 2a − 12a + 1, while f(a) + 1 = a − 1 + a + 1a + 1 = 2aa + 1. At a = 1: f(2) = 13 but f(1) + 1 = 1. So 1 is false.
(2) f(1a) = 1a − 11a + 1 = 1 − a1 + a = -f(a). So 2 is true.

Answer: (b) -⁠√3/⁠2⁠

Method. [π/⁠2] = 1 since 1 < π/⁠2 < 2

π/⁠2 ≈ 1.571 lies in (1, 2), so [π/⁠2] = 1.
f(π2) = cos(π3 + π2) = cos(5π6) = -⁠√32.⁠

Answer: (a) -1

Method. [x] - x = -{x} lies strictly between -1 and 0 for non-integer x

For non-integer x, 0 < x - [x] < 1.
So y = [x] - x satisfies -1 < y < 0.
z = [y] = -1.

Answer: (b) 0

Method. evaluate [x] and [x²] at each point

x = 0.999: [x] = 0, x² = 0.998001, [x²] = 0, so f = 0 − 0 = 0.
x = 1.001: [x] = 1, x² = 1.002001, [x²] = 1, so f = 1 − 1 = 0.
Sum = 0.

Answer: (b) II only

Method. compare left limit, right limit and value at each point

At x = 0: f(0) = 0 − 0 = 0.
For -1 < x < 0: [x] = -1, 0 < x² < 1, so f(x) = 1 − 0 = 1; left limit 1 ≠ 0. Not continuous at 0.
At x = 1: f(1) = 1 − 1 = 0.
For 0 ≤ x < 1: f(x) = 0 − 0 = 0; for 1 ≤ x < ⁠√2:⁠ f(x) = 1 − 1 = 0. Both limits are 0 = f(1). Continuous at 1.
Only II is correct.

Answer: (c) Both I and II

Method. odd-function test f(-t) = -f(t); an odd function of an odd function is odd

f(-t) = ln(⁠√(1⁠ + t²) − t) = ln((⁠√(1⁠ + t²) − t)(⁠√(1⁠ + t²) + t)⁠√(1⁠ + t²) + t) = ln(1t + ⁠√(1⁠ + t²)) = -f(t). So f is odd: I is correct.
g(-t) = tan(f(-t)) = tan(-f(t)) = -tan(f(t)) = -g(t), since tan is odd. So g is odd wherever it is defined: II is correct.
Both statements are correct.

No correct option. This question is defective as printed in UPSC's paper.

Method. integral of an odd function over a symmetric interval

g is odd (previous item), so the intended answer is ∫_-π^π g(t) dt = 0.
But f(t) = π/⁠2 when t = sinh(π/⁠2) = (e^π/⁠2 - e^-π/⁠2)/⁠2 ≈ 2.30, which lies inside [-π, π]; there tan(f(t)) → ±∞ like 1/⁠(t − 2.30).
So g has vertical asymptotes at t = ±2.30 in the interval, and the integral diverges; it is 0 only as a principal value.

g(t) = tan(f(t)) is unbounded at t = ±sinh(π/⁠2) ≈ ±2.30, inside [-π, π], so the integral does not exist; the intended odd-function reading gives 0, option (b).

Answer: (b) 3

Method. a sum of distances |x − a_i| is least at the median of the a_i

f(x) = |x − 2| + |x − 3| + |x − 4| is the sum of distances from x to 2, 3, 4.
|x − 2| + |x − 4| ≥ 2 with equality for 2 ≤ x ≤ 4, and |x − 3| ≥ 0 with equality only at x = 3.
Both are least together at x = 3, the median.

Answer: (a) 2

Method. evaluate f at the minimising point x = 3

From the previous item, the minimum occurs at x = 3.
f(3) = |3 − 2| + |3 − 3| + |4 − 3| = 1 + 0 + 1 = 2.

Answer: (b) [0, ∞)

Method. split |x| by the sign of x

For x ≥ 0: |x| = x, so f(x) = 2x, which takes every value in [0, ∞).
For x < 0: |x| = -x, so f(x) = 0.
Range = [0, ∞) ∪ {0} = [0, ∞).

Answer: (a) -1

Method. [x] - x = -{x}, which lies strictly between -1 and 0 for non-integer x

For non-integer x, [x] < x < [x] + 1, so -1 < [x] - x < 0.
Thus y ∈ (-1, 0), and the greatest integer not exceeding y is -1.
z = [y] = -1.

Shortcut. Try x = 2.5: y = 2 − 2.5 = -0.5, z = [-0.5] = -1.

Answer: (a) f(x) + f(-x) = 0

Method. f is odd: rationalise ⁠√(1⁠ + x²) − x

f(-x) = ln(⁠√(1⁠ + x²) − x).
(⁠√(1⁠ + x²) − x)(⁠√(1⁠ + x²) + x) = 1 + x² - x² = 1, so ⁠√(1⁠ + x²) − x = 1x + ⁠√(1⁠ + x²).
f(-x) = -ln(x + ⁠√(1⁠ + x²)) = -f(x), so f(x) + f(-x) = 0.

Answer: (a) 0

Method. modulus is never negative and is 0 where its argument is 0

|x − 1| ≥ 0 for all real x.
At x = 1, |x − 1| = 0, so the value 0 is attained.
Minimum value = 0.

Answer: (b) an odd function

Method. (⁠√(x²⁠ + 1) − x)(⁠√(x²⁠ + 1) + x) = 1

f(-x) = ln(⁠√(x²⁠ + 1) + x).
f(x) + f(-x) = ln[(⁠√(x²⁠ + 1) − x)(⁠√(x²⁠ + 1) + x)] = ln(x² + 1 − x²) = ln 1 = 0.
So f(-x) = -f(x) and f is odd; it is not even since f(1) = ln(⁠√2⁠ − 1) ≠ 0.

Answer: (d) None of the above

Method. test each identity with a counter-example

(a) f(x²) = x² + 1, [f(x)]² = (x + 1)²; at x = 1: 2 ≠ 4.
(b) f(|x|) = |x| + 1, |f(x)| = |x + 1|; at x = -1: 2 ≠ 0.
(c) f(x + y) = |x + y + 1|, f(x) + f(y) = |x + 1| + |y + 1|; at x = y = 0: 1 ≠ 2.
None of (a), (b), (c) holds for all x, y.

Answer: (d) 2π

Method. the least positive T with sin(x + T) = sin x for all x

sin(x + 2π) = sin x for all x.
No smaller positive T works: at x = π2, sin(π2 + T) = 1 needs T to be a multiple of 2π.
So the period is 2π.

Answer: (d) neither even nor odd

Method. test f(-x) against f(x) and -f(x)

f(-x) = |-x| − (-x)³ = |x| + x³.
f(-x) = f(x) would need x³ = 0 for all x; f(-x) = -f(x) would need |x| = 0 for all x.
E.g. f(1) = 0, f(-1) = 2: neither equal nor opposite.
So f is neither even nor odd.

Answer: (a) 251

Method. find where 1/⁠4 + n/⁠1000 crosses 1

For 1 ≤ n ≤ 1000, 14 + n1000 lies between 0.251 and 1.25.
Its integral part is 0 when n1000 < 34, i.e. n ≤ 749, and 1 when n ≥ 750.
The number of n from 750 to 1000 is 1000 − 750 + 1 = 251.
1000Σn=1 f(n) = 251 × 1 = 251.

Answer: (d) x²(x − 3)

Method. test f(-x) against f(x) and -f(x)

x² - 1 and |x| are even; x + 3x is odd.
For f(x) = x²(x − 3) = x³ - 3x²: f(-x) = -x³ - 3x².
f(-x) ≠ f(x) (x³ term changes sign) and f(-x) ≠ -f(x) (x² term does not), e.g. f(1) = -2, f(-1) = -4.
So x²(x − 3) is neither even nor odd.