JEE Mathematics Functions MCQs Set 12

Answer: (b) c = 1, d arbitrary

Explanation:
1. f(g(x)) = cx + d + 2 and g(f(x)) = c(x + 2) + d = cx + 2c + d.
2. Equal for all x, so d + 2 = 2c + d, which gives c = 1.
3. d is arbitrary.

Teacher's Note:
1. Compare the constant terms.
2. d cancels out.

Answer: (b) \( 4^n-1,\ 4^n \)

Explanation:
1. g(x) = 3 + 4x, g²(x) = 3 + 4(3 + 4x) = 15 + 16x.
2. g³(x) = 3 + 4(15 + 16x) = 63 + 64x.
3. Pattern: gⁿ(x) = (4ⁿ − 1) + 4ⁿ x.

Teacher's Note:
1. Compute g², g³ and look for the pattern.
2. A = 4ⁿ − 1 and B = 4ⁿ.

Answer: (b) domain, co-domain

Explanation:
1. In f : A → B, A is the domain and B is the co-domain.
2. The range is the subset of B actually attained.

Teacher's Note:
1. Domain = inputs, co-domain = the set the outputs are allowed to lie in.
2. Do not confuse co-domain with range.

Answer: (c) range

Explanation:
1. f(A) is the set of all images f(x), x in A.
2. That set is called the range.

Teacher's Note:
1. The range is always a subset of the co-domain.
2. It equals the co-domain only when f is onto.

Answer: (c) Every element of B has at least one pre-image in A

Explanation:
1. Surjective (onto) means the range is the whole co-domain.
2. Each element of B therefore has at least one pre-image in A.

Teacher's Note:
1. Option (a) describes one-one, not onto.
2. Onto concerns the co-domain.

Answer: (b) n(A) = 1

Explanation:
1. A constant function sends every element of A to the same image.
2. It is one-one only if there are no two distinct elements, i.e. n(A) = 1.

Teacher's Note:
1. Compare images of two different inputs.
2. They are equal for a constant function.

Answer: (a) a singleton set

Explanation:
1. A constant function has exactly one value in its range.
2. If it is onto, the co-domain equals the range, so B has exactly one element.

Teacher's Note:
1. Range = {c}.
2. Onto means B = range.

Answer: (a) f is onto

Explanation:
1. If f∘g is onto, every element of the co-domain of f is f(g(x)) for some x.
2. So every element of the co-domain of f is an image under f, i.e. f is onto.
3. g need not be onto.

Teacher's Note:
1. Follow the outputs backwards from the composite.
2. The outer function must reach the whole co-domain.

Answer: (b) g must be one-one

Explanation:
1. Suppose g(x₁) = g(x₂).
2. Then f(g(x₁)) = f(g(x₂)), and since f∘g is one-one, x₁ = x₂.
3. So g must be one-one; f need not be.

Teacher's Note:
1. Argue by assuming equal images under g.
2. The inner function must be injective.

Answer: (c) one-one and onto

Explanation:
1. A function has an inverse only if each element of the co-domain has exactly one pre-image.
2. That is, f must be one-one (unique) and onto (exists), i.e. bijective.

Teacher's Note:
1. One-one gives uniqueness; onto gives existence.
2. Identity is just one example of a bijection.

Answer: (b) \( I_B \)

Explanation:
1. f(f⁻¹(y)) = y for every y in B.
2. So f∘f⁻¹ is the identity function on B, written I_B.

Teacher's Note:
1. Compare with Question 11: the identity now lives on the co-domain.

Answer: (a) symmetric about the origin

Explanation:
1. For an odd function f(−x) = −f(x), so the point (−x, −y) lies on the graph with (x, y).
2. That is symmetry about the origin.

Teacher's Note:
1. Rotating the graph by 180° about the origin leaves it unchanged.
2. y = x³ is an example.

Answer: (b) symmetric about the y-axis

Explanation:
1. For an even function f(−x) = f(x).
2. The points (x, y) and (−x, y) both lie on the graph, so it is symmetric about the y-axis.

Teacher's Note:
1. Reflect the graph in the y-axis to get the same graph.
2. y = x² is an example.

Answer: (b) symmetric about the y-axis

Explanation:
1. |x| is even: |−x| = |x|.
2. So its graph (a V-shape with vertex at the origin) is symmetric about the y-axis.
3. It is not a straight line.

Teacher's Note:
1. Check the shape: two rays from the origin.
2. Even functions have y-axis symmetry.

Answer: (c) r ≠ p

Explanation:
1. f(x) = |px − q| + r|x| is piecewise linear with kinks at x = 0 and x = q/p.
2. Slope for 0 < x < q/p is r − p.
3. If r = p the slope is zero there, so the minimum is attained on a whole interval.
4. For r ≠ p the minimum is at a single point (at 0 when r > p, at q/p when r < p).

Teacher's Note:
1. Look at the sign of the slope on each piece.
2. A flat piece gives infinitely many minimum points.

Answer: (d) \( |c|>|b|\sqrt2 \)

Explanation:
1. min f = 2c² − b² (at x = −b) and max g = b² + c² (at x = −c).
2. min f > max g gives 2c² − b² > b² + c², i.e. c² > 2b².
3. So |c| > |b|√2.

Teacher's Note:
1. Complete the square for each quadratic.
2. Compare the extreme values.

Answer: (b) 4p

Explanation:
1. Let f(x) = u. Then f(x + p) = (u − 5)/(u − 3).
2. f(x + 2p) = ((u−5)/(u−3) − 5)/((u−5)/(u−3) − 3) = (−4u + 10)/(−2u + 4) = (2u − 5)/(u − 2).
3. f(x + 4p) = f(x + 2p + 2p) simplifies to u.

Teacher's Note:
1. Apply the recurrence repeatedly.
2. f(x + 2p) ≠ f(x) because (u − 2)² = −1 has no real solution.

Answer: (b) 7

Explanation:
1. f(7752) = 2·10³ + f(775); f(775) = 5·10² + f(77); f(77) = 7·10 + f(7); f(7) = 7.
2. So f(7752) = 2000 + 500 + 70 + 7 = 2577, giving a = 2, b = 5, c = 7, d = 7.
3. (c + d) − (a + b) = 14 − 7 = 7.

Teacher's Note:
1. Unroll the recursion: it reverses the digits of x.
2. Read off the digits carefully.

Answer: (c) one-one and into

Explanation:
1. For x in (1/2, 1), x/(1 − x) increases from 1 to ∞.
2. So ln(x/(1 − x)) increases from 0 to ∞ and f is strictly increasing, hence one-one.
3. The range is (0, ∞), which does not include 0 of the co-domain, so f is into.

Teacher's Note:
1. Study monotonicity and the range separately.
2. The range is an open interval.

Answer: (c) 6

Explanation:
1. Need 0 < log₆ y ≤ 1 for log₀.₅ to be non-negative, with y = (x² + x)/(x + 4).
2. That is 1 < y ≤ 6.
3. y > 1: x² > 4 (x > −4), so x > 2 (or −4 < x < −2).
4. y ≤ 6: (x − 8)(x + 3) ≤ 0, so −3 ≤ x ≤ 8.
5. Positive integers: 3, 4, 5, 6, 7, 8, which is 6 integers.

Teacher's Note:
1. The square root needs log₀.₅(·) ≥ 0, which means the argument lies in (0, 1].
2. Solve the two inequalities and intersect.

Answer: (b) 9

Explanation:
1. The relation is satisfied by f(x) = 1 ± xⁿ.
2. f(4) = 65 gives 4ⁿ = 64 (with the plus sign), so n = 3 and f(x) = x³ + 1.
3. f(2) = 8 + 1 = 9.

Teacher's Note:
1. Try f(x) = 1 + xⁿ and check that it satisfies the equation.
2. The sign minus would give a negative value at 4.

Answer: (b) \( \sqrt3 \)

Explanation:
1. f′(x) = −(1 + 2 cos x)/sin²x.
2. f′ = 0 at cos x = −1/2, i.e. x = 2π/3.
3. f(2π/3) = (2 − 1/2)/(√3/2) = √3, which is the minimum, since f′ changes from negative to positive.

Teacher's Note:
1. Differentiate and solve cos x = −1/2 in (0, π).
2. Check the sign change of f′.

Answer: (a) even

Explanation:
1. Interchange x and y: f(y) + f(x) = f(y + x) f(y − x).
2. Subtract from the original: f(x + y)[f(x − y) − f(y − x)] = 0.
3. Putting y = 2x gives f(x) = f(−x) for every non-trivial solution.
4. So f is even.

Teacher's Note:
1. Swap the roles of x and y and compare.
2. The zero function is also covered (it is even).

Answer: (d) 3

Explanation:
1. For x < 0: RHS = x + |x| = 0, so |cos x| = 0 and x = −π/2 or −3π/2 within |x| ≤ 2π (2 roots).
2. For x ≥ 0: |cos x| = 2x. As 2x ≤ 1 we need x ≤ 1/2, where cos x = 2x has exactly one root.
3. Total = 3 roots.

Teacher's Note:
1. Handle negative and positive x separately.
2. On [0, 1/2] cos x decreases from 1 and 2x increases from 0, so they cross once.

Answer: (a) \( \frac{2x^2+5}{2x-3} \)

Explanation:
1. f∘f(x) = x, so f is its own inverse.
2. f₂₀₀₈ = x (even number of applications) and f₂₀₀₉ = f(x).
3. Sum = x + (3x + 5)/(2x − 3) = (2x² + 5)/(2x − 3).

Teacher's Note:
1. Check that f(f(x)) = x using the form (ax + b)/(cx − a).
2. Even compositions give the identity.