JEE Mathematics Functions MCQs Set 19

Answer: (b) λ and μ are integers

Explanation:
1. f(0) = μ must be an integer.
2. f(1) = 1 + λ + μ must be an integer, so λ is an integer.
3. Hence both λ and μ are integers.

Teacher's Note:
1. Substitute the simplest integers 0 and 1 to pin down the constants.
2. Integer-valued at integers forces integer coefficients here because the leading coefficient is 1.

Answer: (c) cos 4

Explanation:
1. [π] = 3, so the first term is cos(3π/2) = 0.
2. πx = π²/2 ≈ 4.93, so [π²/2] = 4 and the second term is cos 4.
3. f(π/2) = cos 4.

Teacher's Note:
1. Evaluate the greatest integer inside each cosine before taking the cosine.
2. π² ≈ 9.87, so π²/2 lies between 4 and 5.

Answer: (c) −1

Explanation:
1. tan⁻¹(−√3) = −π/3.
2. f(−√3) = sin(−π/3) = −√3/2 ≈ −0.866.
3. The greatest integer not exceeding −0.866 is −1.

Teacher's Note:
1. Do not forget that the greatest integer of a negative decimal goes down, not toward zero.
2. [−0.866] = −1.

Answer: (c) \( \frac{(a+1)f(x)+a-1}{(a-1)f(x)+a+1} \)

Explanation:
1. From f = (x − 1)/(x + 1) we get x = (1 + f)/(1 − f).
2. Then ax − 1 = [(a+1)f + a − 1]/(1 − f) and ax + 1 = [(a−1)f + a + 1]/(1 − f).
3. Dividing, f(ax) = [(a+1)f(x) + a − 1] / [(a−1)f(x) + a + 1].

Teacher's Note:
1. Express x in terms of f(x) first, then substitute into f(ax).
2. Common denominators of 1 − f cancel.

Answer: (a) n f(1)

Explanation:
1. The relation says f(x+1) − f(x) = f(x) − f(x−1), so successive differences are equal.
2. Hence f(n) = f(0) + n[f(1) − f(0)] = n f(1).

Teacher's Note:
1. This is the defining property of an arithmetic progression of values.
2. Put x = 1, 2, … to see the pattern.

Answer: (b) nk

Explanation:
1. f(2) = f(1) + f(1) = 2k.
2. By repeated addition f(n) = f(1) + f(1) + … (n times) = nk.

Teacher's Note:
1. Cauchy's additive equation gives f(n) = n f(1) on the natural numbers.
2. Induct on n using f(n + 1) = f(n) + f(1).

Answer: (a) \( \frac{3}{2}(3^n-1) \)

Explanation:
1. f(n) = f(1)ⁿ = 3ⁿ.
2. The sum is 3 + 3² + … + 3ⁿ, a GP with first term 3 and ratio 3.
3. Sum = 3(3ⁿ − 1)/(3 − 1) = (3/2)(3ⁿ − 1).

Teacher's Note:
1. The multiplicative equation gives exponential behaviour, f(n) = f(1)ⁿ.
2. Use the GP sum formula.

Answer: (a) one

Explanation:
1. Put x = y = 0: f(0) = 2f(0) − 1, so f(0) = 1.
2. Put y = 1: f(x + 1) = f(x) + 1 − x − 1 = f(x) − x.
3. So f(n) = 1 − (0 + 1 + … + (n − 1)) = 1 − n(n − 1)/2.
4. f(n) = n gives n² + n − 2 = 0, so n = 1 or −2; only n = 1 is natural.

Teacher's Note:
1. Find f(0) and a recurrence for f(x + 1) first.
2. Reject roots that are not natural numbers.

Answer: (d) 3

Explanation:
1. −2.3 < −1, so f(−2.3) = 1 + 2.3 = 3.3.
2. 3.3 ≥ −1, so f(3.3) = [3.3] = 3.

Teacher's Note:
1. Evaluate the inner function first and then decide which branch applies to the result.
2. Branches are chosen by the input value of each call.

Answer: (d) none of these

Explanation:
1. The first root needs (x + 2)(5 − x) ≥ 0, so −2 ≤ x ≤ 5.
2. The second term needs x² − 4 > 0, so x < −2 or x > 2.
3. The intersection is (2, 5], which is not any of the given sets.

Teacher's Note:
1. Write the conditions for each part of the expression (root, logarithm, inverse function) and intersect them.
2. Check the end points separately.

Answer: (d) (−∞, +∞)

Explanation:
1. The expression needs x¹² − x⁹ + x⁴ − x + 1 > 0.
2. For x ≤ 0 every term is non-negative and the constant 1 makes it positive.
3. For 0 < x < 1: x⁴ − x⁹ > 0 and 1 − x > 0, so the sum is positive; for x ≥ 1: x¹² − x⁹ ≥ 0 and x⁴ − x ≥ 0 so the sum is at least 1.
4. The polynomial is always positive, so the domain is all real numbers.

Teacher's Note:
1. Split the real line into x ≤ 0, 0 < x < 1 and x ≥ 1 and pair terms.
2. Numerically the minimum value is about 0.51.

Answer: (d) \( \left[\frac{1}{\sqrt2},1\right] \)

Explanation:
1. The inner root needs 1 − x² ≥ 0, so −1 ≤ x ≤ 1.
2. The outer root needs x ≥ √(1 − x²), so x ≥ 0 and x² ≥ 1 − x².
3. That gives x² ≥ 1/2 with x ≥ 0, i.e. x ≥ 1/√2.
4. Combining, the domain is [1/√2, 1].

Teacher's Note:
1. Square the inequality only after checking that both sides are non-negative.
2. x must be non-negative because the right side is non-negative.

Answer: (d) [−1, 1]

Explanation:
1. The innermost root needs 1 − x² ≥ 0, so |x| ≤ 1.
2. Then 0 ≤ √(1 − x²) ≤ 1, so 1 − √(1 − x²) lies in [0, 1] and its root also lies in [0, 1].
3. So 1 − that root is ≥ 0 and the outer root is defined.
4. The domain is [−1, 1].

Teacher's Note:
1. Work from the innermost root outward and note each value stays in [0, 1].
2. Square roots of numbers in [0, 1] stay in [0, 1].

Answer: (b) (0, +∞)

Explanation:
1. The outer log needs log₁₀(1 + x³) > 0.
2. That means 1 + x³ > 1, so x³ > 0 and x > 0.

Teacher's Note:
1. A logarithm is positive when its argument exceeds 1 (base 10).
2. The inner log is then automatically defined.

Answer: (d) none of these

Explanation:
1. We need x² ≥ [x]².
2. For x ≥ 0 we have 0 ≤ [x] ≤ x, so the condition holds.
3. For x < 0 we have [x] ≤ x < 0, so [x]² > x² unless x is an integer.
4. The domain is [0, ∞) together with the negative integers, which is none of the given sets.

Teacher's Note:
1. Test a negative non-integer such as −0.5: x² = 0.25 but [x]² = 1.
2. Negative integers satisfy x² = [x]².

Answer: (d) \( \left(\frac{(4n-1)\pi}{2},\frac{(4n+1)\pi}{2}\right) \)

Explanation:
1. The radicand must be positive, and |cos x| + cos x is positive only when cos x > 0 (it equals 2cos x then, and 0 otherwise).
2. cos x > 0 for x between (4n − 1)π/2 and (4n + 1)π/2.

Teacher's Note:
1. If cos x ≤ 0 the sum |cos x| + cos x is zero.
2. Positive cosine means the interval around 2nπ of half-width π/2.

Answer: (a) (√2, +∞)

Explanation:
1. The base x² − 1 must be positive and not 1, and x > 0, so x > 1 and x ≠ √2.
2. If the base is greater than 1 (x > √2) then log of x ≥ 1 is ≥ 0, which is allowed.
3. If 1 < x < √2 the base lies in (0, 1) and log of x > 1 is negative, which is not allowed.
4. So the domain is (√2, ∞).

Teacher's Note:
1. A logarithm with base less than 1 is negative for arguments greater than 1.
2. Treat the base greater than 1 and base less than 1 cases separately.

Answer: (c) {2, 3}

Explanation:
1. x must be an integer with 0 ≤ 2x − 1 ≤ 16 − x, giving 1 ≤ x ≤ 17/3.
2. Also 0 ≤ 4x − 5 ≤ 20 − 3x, giving 5/4 ≤ x ≤ 25/7.
3. Integers satisfying both: x = 2, 3.

Teacher's Note:
1. For nCr and nPr we need integers with 0 ≤ r ≤ n.
2. Combine all four inequalities.

Answer: (c) {1, −1}

Explanation:
1. The root needs |x| ≤ 1.
2. For the inverse sine, |(1 + x²)/(2x)| ≤ 1; but |1 + x²| ≥ 2|x| always, with equality only when |x| = 1.
3. So x = 1 or x = −1.

Teacher's Note:
1. Use the AM–GM inequality: 1 + x² ≥ 2|x|.
2. Equality holds only at x = ±1.

Answer: (a) (−∞, −3] ∪ [3, +∞)

Explanation:
1. sec⁻¹u needs |u| ≥ 1, so |1 − |x||/2 ≥ 1, i.e. |1 − |x|| ≥ 2.
2. Then 1 − |x| ≥ 2 (impossible) or 1 − |x| ≤ −2, so |x| ≥ 3.
3. For these x, u ≤ −1 and sec⁻¹u lies in (π/2, π], which is positive, so the root is real.

Teacher's Note:
1. Remember the range of sec⁻¹ is [0, π] without π/2.
2. Both signs of |1 − |x|| must be checked.