JEE Mathematics Functions MCQs Set 24

Answer: (c) \( \frac53 \)

Explanation:
1. f₁(x) = 1/(1 − 1/(1 − x)) = (x − 1)/x and f₂(x) = f₀(f₁(x)) = x, so f₃ = f₀ and the sequence has period 3.
2. f₁₀₀ = f₁ (100 leaves remainder 1 on division by 3), so f₁₀₀(3) = 2/3.
3. f₁(2/3) = −1/2 and f₂(3/2) = 3/2.
4. Sum = 2/3 − 1/2 + 3/2 = 5/3.

Teacher's Note:
1. Compute f₁, f₂, f₃ first; repeated composition often cycles.

Answer: (d) \( g(y)=\frac{y-3}{4} \)

Explanation:
1. Put y = 4x + 3 and solve for x: x = (y − 3)/4.
2. f maps N onto Y one-to-one, so the inverse is g(y) = (y − 3)/4.

Teacher's Note:
1. Inverse: write y = f(x), solve for x, then rename.

Answer: (c) \( \left[-\frac12,\frac12\right] \)

Explanation:
1. For x ≠ 0, |f(x)| = 1/(|x| + 1/|x|) ≤ 1/2 by AM–GM, with equality at |x| = 1.
2. f is odd and takes the value 0 at x = 0 and every value in between, so the range is [−1/2, 1/2].

Teacher's Note:
1. AM–GM on x + 1/x bounds x/(1 + x²).

Answer: (a) injective but not surjective

Explanation:
1. f(x) = 2 + 2/(x − 1) is strictly monotonic on each side of x = 1 and the two branches take values on opposite sides of 2, so f is injective.
2. The value 2 is never attained (2x = 2x − 2 is impossible), so f is not surjective onto R.

Teacher's Note:
1. A Mobius function is one-one; check which value is missed from the range.

Answer: (b) not injective but it is surjective

Explanation:
1. f(1/2) = 1 and f(1) = 0, f(2) = 1/2 = f(2/3), so two different inputs give the same output: f is not injective.
2. For 0 < x < 1, 1/x − 1 takes all values in (0, ∞), and f(1) = 0, so every value of [0, ∞) is attained: f is surjective.

Teacher's Note:
1. Find two inputs with the same output to disprove injectivity.

Answer: (a) \( f_3(x) \)

Explanation:
1. 1 − J(1/x) = 1/(1 − x), so J(1/x) = 1 − 1/(1 − x) = x/(x − 1).
2. Put t = 1/x: J(t) = (1/t)/((1/t) − 1) = 1/(1 − t).
3. Hence J = f₃.

Teacher's Note:
1. Substitute t = 1/x to get J(t) directly.

Answer: (c) onto but not one-one

Explanation:
1. g(n) = n + 1 for odd n and n − 1 for even n.
2. (f∘g)(n) = (n + 1)/2 for odd n and n/2 for even n, so (f∘g)(1) = (f∘g)(2) = 1: not one-one.
3. Every natural number m is obtained (take n = 2m), so f∘g is onto.

Teacher's Note:
1. Tabulate the first few values of f∘g.

Answer: (c) 201

Explanation:
1. f(n) = −100 cannot come from an even n (it would give a positive value).
2. For odd n: −(n − 1)/2 = −100 gives n = 201.

Teacher's Note:
1. A negative output can only come from the odd branch.

Answer: (c) −1

Explanation:
1. g(π) = 0 (π is irrational), and 0 is rational, so f(g(π)) = f(0) = 0.
2. f(e) = 1 (e is irrational), and 1 is rational, so g(f(e)) = g(1) = −1.
3. Sum = 0 + (−1) = −1.

Teacher's Note:
1. Decide rational or irrational at every step of the composition.

Answer: (a) 3

Explanation:
1. log₁₀(log₁₀ 3) = log₁₀(1/log₃ 10) = −log₁₀(log₃ 10). Call u = log₁₀(log₃ 10); the two arguments are u and −u.
2. f(x) − 4 = a sin x + b x^1/3 is odd, so f(u) + f(−u) = 8.
3. f(−u) = 8 − 5 = 3.

Teacher's Note:
1. Split f into an odd part plus a constant.

Answer: (c) (−2, 1)

Explanation:
1. We need 4 − x² ≥ 0 and 1 − x ≠ 0, and the ratio inside ln must be positive.
2. The numerator is positive for −2 < x < 2 (it is 0 at x = ±2, where ln is undefined), and 1 − x > 0 requires x < 1.
3. Domain: (−2, 1).

Teacher's Note:
1. A logarithm needs a strictly positive argument.

Answer: (a) \( f(x)\text{ is odd if }x\ne n\pi \)

Explanation:
1. [(x + π)/π] = 1 + [x/π], so the denominator is [x/π] + 1/2.
2. The numerator x(sin x + tan x) is even.
3. For x ≠ nπ, [−x/π] = −[x/π] − 1, so the denominator at −x is −([x/π] + 1/2): f(−x) = −f(x).
4. So f is odd for x ≠ nπ.

Teacher's Note:
1. For non-integer y, [−y] = −[y] − 1.

Answer: (d) \( (-1,0)\cup(1,2)\cup(2,\infty) \)

Explanation:
1. x³ − x = x(x − 1)(x + 1) > 0 gives x ∈ (−1, 0) ∪ (1, ∞).
2. 4 − x² ≠ 0 removes x = ±2; only x = 2 lies in the set above.
3. Domain: (−1, 0) ∪ (1, 2) ∪ (2, ∞).

Teacher's Note:
1. Intersect the log condition with the denominator condition.

Answer: (b) N, the set of natural numbers

Explanation:
1. {x} ∈ (0, 1) is needed (sin 0 = 0 makes the expression undefined), so sin{x} ∈ (0, sin 1) and 1/sin{x} takes every value in (1/sin 1, ∞) = (1.19..., ∞).
2. The greatest integer of these values is every integer ≥ 1.
3. Range = N.

Teacher's Note:
1. Fractional part lies in [0, 1) and the sine is increasing there.

Answer: (a) {0, 1, 2, 3, 4, 5, 6}

Explanation:
1. Put t = 3^x ∈ (0, 3). Then 9^x − 3^x + 1 = t² − t + 1.
2. On (0, 3) this has minimum 3/4 (at t = 1/2) and supremum 7 (not attained), so it takes all values in [3/4, 7).
3. The greatest integer takes the values 0, 1, 2, 3, 4, 5, 6.

Teacher's Note:
1. The value 3/4 gives integer part 0, so 0 is in the range.

Answer: (c) 3

Explanation:
1. The numerator has period 2π/n and tan(x/n) has period nπ.
2. The period of f is the LCM of 2π/n and nπ. For n = 3 these are 2π/3 and 3π, whose LCM is 6π.
3. n = 1 and n = 2 give 2π, so n = 3.

Teacher's Note:
1. Test n = 1, 2, 3 by computing the LCM of the two periods.

Answer: (a) I and II quadrants

Explanation:
1. f is an absolute value, so f(x) ≥ 0 for every x in its domain (x ≠ 0).
2. f depends on |x|, so it is even and has points for both positive and negative x, all with y ≥ 0.
3. Hence the graph lies in the I and II quadrants.

Teacher's Note:
1. An outer modulus forces y ≥ 0; evenness gives both signs of x.

Answer: (a) \( \left[\sqrt2,\frac{1+\sqrt5}{2}\right] \)

Explanation:
1. −1 ≤ x − x² ≤ 1: x² − x − 1 ≤ 0 gives (1 − √5)/2 ≤ x ≤ (1 + √5)/2; the other side always holds.
2. 1 − 1/|x| ≥ 0 gives |x| ≥ 1.
3. [x² − 1] ≠ 0 means x² − 1 ∉ [0, 1), so with |x| ≥ 1 we need x² ≥ 2, i.e. |x| ≥ √2.
4. Combining: √2 ≤ x ≤ (1 + √5)/2 (the negative part falls outside the first condition).

Teacher's Note:
1. Intersect all three conditions on a number line.

Answer: (c) f(f(1)) = 1

Explanation:
1. f(0) = max{1, 1, 0} = 1 and g(1) = 1: (a) is correct.
2. f(1) = 1 + sin 1 ≈ 1.84 and g(1.84) = max{1, 0.84} = 1: (b) is correct.
3. f(f(1)) = f(1.84) = 1 + sin(1.84) ≈ 1.96 ≠ 1, so (c) is incorrect.
4. g(0) = 1, so f(g(0)) = f(1) = 1 + sin 1: (d) is correct.

Teacher's Note:
1. Evaluate each composition numerically.

Answer: (d) 1

Explanation:
1. Rewrite as (x − π)² + 1 = sin(x/2).
2. The left side is at least 1 and sin(x/2) is at most 1, so both must equal 1.
3. That needs x = π, and then sin(π/2) = 1 holds. Exactly one solution.

Teacher's Note:
1. Compare the minimum of the left side with the maximum of the right side.

Answer: (b) 6

Explanation:
1. k odd gives f(k) = k + 3 (even), and f(k + 3) = (k + 3)/2.
2. If (k + 3)/2 were odd, the third step gives (k + 3)/2 + 3 = 27, so (k + 3)/2 = 24, which is even: contradiction.
3. So (k + 3)/2 is even and the third step gives (k + 3)/4 = 27, so k = 105.
4. Digit sum = 1 + 0 + 5 = 6.

Teacher's Note:
1. Check the parity at every stage of the iteration.

Answer: (b) even

Explanation:
1. For −1 < x < 1, 0 ≤ x⁴ < 1, so [x⁴] = 0.
2. f(x) = 1/√(1 − x²), and f(−x) = f(x): f is even.

Teacher's Note:
1. Evaluate the bracket on the given domain first.

Answer: (a) one-one but not onto

Explanation:
1. f'(x) = 2 + cos 2x > 0 between integers, and at each integer [x] jumps up by 1, so f is strictly increasing: one-one.
2. At each integer f jumps upwards, so values in the gaps are never attained and f is not onto R.

Teacher's Note:
1. A strictly increasing function with jump discontinuities misses values.

Answer: (d) −1

Explanation:
1. Let s = |x| + 1 ≥ 1. Then |x| − 2 − x² = −(s² − 3s + 4), so f = −(s + 4/s − 3).
2. By AM–GM, s + 4/s ≥ 4 with equality at s = 2, so f ≤ −(4 − 3) = −1.
3. The maximum value is −1 (at x = ±1).

Teacher's Note:
1. Substitute s = |x| + 1 to reach the form s + 4/s.

Answer: (c) 8001

Explanation:
1. The infinite series is 2/(1 − 1/3) = 3, so f(x) = 3^{log₃(1 + f(x)/f(1/x))} = 1 + f(x)/f(1/x).
2. Hence f(x)f(1/x) = f(x) + f(1/x), whose polynomial solutions are f(x) = 1 ± xⁿ; f(10) = 1001 gives f(x) = 1 + x³.
3. f(20) = 1 + 8000 = 8001.

Teacher's Note:
1. The relation f(x)f(1/x) = f(x) + f(1/x) characterises 1 + xⁿ.