JEE Mathematics Functions MCQs Set 22

Answer: (d) a periodic function of period 2

Explanation:
1. nx + n − [nx + n] = {nx}, the fractional part of nx, which has period 1/n.
2. tan(πx/2) has period 2.
3. 2 is a whole multiple (2n times) of 1/n, so the sum repeats with period 2, and the tangent alone forbids anything smaller.

Teacher's Note:
1. Find the period of each term and take the least common multiple.
2. A fractional part {nx} repeats every 1/n.

Answer: (c) a periodic function of period 2

Explanation:
1. f(x + 2) = f(x) for every x, so 2 is a period.
2. f(0) = 0 and f(1) = 1 differ, so 1 is not a period; f(0) = 0 and f(1/2) = 3/4 differ, so 1/2 is not a period.
3. The least period is 2.

Teacher's Note:
1. Test the candidate periods with values at 0, 1/2 and 1.
2. The defining relation itself gives period 2.

Answer: (b) a constant function

Explanation:
1. Expand with sin(x + π/3) = ½ sin x + (√3/2) cos x and cos(x + π/3) = ½ cos x − (√3/2) sin x.
2. The sin x cos x terms cancel, leaving f = sin²x + ¼ sin²x + ¾ cos²x + ½ cos²x = (5/4)(sin²x + cos²x) = 5/4.
3. Hence (gof)(x) = g(5/4) = 1, a constant.

Teacher's Note:
1. Reduce f(x) with double-angle formulas; the x-dependent terms cancel.
2. A constant inner function makes the composite constant.

Answer: (d) log |x|

Explanation:
1. We need g(xⁿ) = n g(x) for all x in the domain.
2. For g(x) = log|x|: log|xⁿ| = n log|x| ✓.
3. The other options fail, e.g. g = eˣ gives e^(xⁿ) ≠ n eˣ.

Teacher's Note:
1. The function must turn powers into multiples, which is the logarithm.
2. Test a simple case such as x = 2, n = 2.

Answer: (a) \( f(x)=\sin^2x,\ g(x)=\sqrt{x} \)

Explanation:
1. Option 1: g(f(x)) = √(sin²x) = |sin x| ✓ and f(g(x)) = sin²(√x) ✓.
2. Option 2: f(g(x)) = sin|x|, not sin²√x.
3. Option 3: g(f(x)) = sin|x|, not |sin x|.

Teacher's Note:
1. Compose each pair both ways and compare.
2. √(sin²x) = |sin x|, not sin x.

Answer: (c) a straight line

Explanation:
1. f(f(x)) = 1/(1 − 1/(1 − x)) = (x − 1)/x.
2. f(f(f(x))) = 1/(1 − (x − 1)/x) = x.
3. So y = x, a straight line (the part with x > 1).

Teacher's Note:
1. Compose step by step and simplify.
2. f applied three times returns x.

Answer: (a) (1, 8)

Explanation:
1. Let f = ax² + bx + 6 (since f(0) = 6).
2. f(−3) = 9a − 3b + 6 = 6 gives b = 3a; f(2) = 4a + 2b + 6 = 11 gives 10a = 5.
3. a = 1/2, b = 3/2, so f(1) = 1/2 + 3/2 + 6 = 8.

Teacher's Note:
1. Use f(0) to fix the constant term first.
2. Two conditions then give a and b.

Answer: (c) [−6, 1]

Explanation:
1. We need g(x) in the domain of f: −5 ≤ |2x + 5| ≤ 7.
2. The left inequality always holds; |2x + 5| ≤ 7 gives −7 ≤ 2x + 5 ≤ 7.
3. So −6 ≤ x ≤ 1.

Teacher's Note:
1. The inner function's output must lie in the outer function's domain.
2. |u| ≤ 7 means −7 ≤ u ≤ 7.

Answer: (b) \( 1-\sqrt{1-x} \)

Explanation:
1. y = x(2 − x) = 1 − (x − 1)², so (x − 1)² = 1 − y.
2. Since x ≤ 1, x − 1 = −√(1 − y), i.e. x = 1 − √(1 − y).
3. Interchanging, f⁻¹(x) = 1 − √(1 − x).

Teacher's Note:
1. Complete the square, then choose the sign using the domain.
2. The domain x ≤ 1 makes x − 1 non-positive.

Answer: (b) \( \text{is given by }\frac{x+5}{3} \)

Explanation:
1. f is a bijection on R, so the inverse exists.
2. y = 3x − 5 gives x = (y + 5)/3, so f⁻¹(x) = (x + 5)/3.

Teacher's Note:
1. Do not confuse the inverse with the reciprocal.
2. Solve y = f(x) for x.

Answer: (b) \( \frac12\left(1+\sqrt{1+4\log_2x}\right) \)

Explanation:
1. y = 2^(x² − x) gives x² − x = log₂ y.
2. x = [1 ± √(1 + 4 log₂ y)]/2; since x ≥ 1 we take the plus sign.
3. f⁻¹(x) = ½(1 + √(1 + 4 log₂ x)).

Teacher's Note:
1. Solve the quadratic in x and use x ≥ 1 to choose the root.
2. The minus root is at most 1/2 when y ≥ 1.

Answer: (c) not defined

Explanation:
1. f(x) = {x} is periodic, so f(0) = f(1) = 0 and f is many-one.
2. A many-one function has no inverse.

Teacher's Note:
1. Periodic functions are never one-one.
2. Check two inputs with equal outputs.

Answer: (a) \( \frac12\log\frac{1+x}{1-x} \)

Explanation:
1. y = (e²ˣ − 1)/(e²ˣ + 1) gives e²ˣ = (1 + y)/(1 − y).
2. x = ½ log[(1 + y)/(1 − y)].
3. Interchange x and y.

Teacher's Note:
1. Multiply numerator and denominator by eˣ.
2. This is the inverse hyperbolic tangent.

Answer: (c) 0

Explanation:
1. Let u = x + y, v = x − y; then xy = (u² − v²)/4, so f(u, v) = (u² − v²)/4.
2. f(x, y) = (x² − y²)/4 and f(y, x) = (y² − x²)/4.
3. Their sum is 0, so the mean is 0.

Teacher's Note:
1. Rewrite the argument pair as new variables u and v.
2. Swapping the arguments changes the sign.

Answer: (b) f(2 + x) = f(2 − x)

Explanation:
1. Points at equal distance x on either side of x = 2 have equal heights.
2. That is f(2 + x) = f(2 − x).

Teacher's Note:
1. Reflection about x = a sends x to 2a − x.
2. Here 2a − x = 4 − x, equivalent to f(2 + t) = f(2 − t).

Answer: (a) All of 1, 2 and 3

Explanation:
1. f(3/2) = 9/4, f(9/4) = 3/2, f(3/2) = 9/4, so statement 1 holds.
2. f(5/2) = 2, f(2) = 1, f(1) = 1, so 1 + 1 = 2 = f(5/2): statement 2 holds.
3. f(1) = 1 and f(f(1)) = 1: statement 3 holds.

Teacher's Note:
1. Evaluate each composition one branch at a time.
2. Pick the branch from the size of the input at every step.

Answer: (b) 1, 2 and 3 only

Explanation:
1. Expanding with cos(x + π/3) = ½ cos x − (√3/2) sin x, the sin x cos x terms cancel and f = cos²x(1 + ¼ − ½) + ¾ sin²x = ¾.
2. So f is constant, hence even, f(π/8) = f(π/4) and f has every real number as a period, so it is periodic (statement 4 is false).

Teacher's Note:
1. Check numerically at two or three values of x.
2. A constant function is even and has every real number as a period.

Answer: (c) f(x + y) = f(x) + f{y + g(x)}

Explanation:
1. x = [x] + g(x), so [x + y] = [x] + [y + g(x)] because [x] is an integer.
2. Option 1 fails for x = y = 0.5 (1 ≠ 0); option 2 fails for the same values (0 ≠ 1).

Teacher's Note:
1. An integer can be pulled out of a greatest integer function.
2. Test x = y = 0.5.

Answer: (c) 3

Explanation:
1. Let x = n², n² = 5q + r, and n = q + r, so n² = 5n − 4r.
2. r = 0: n = 5 (x = 25); r = 1: n² − 5n + 4 = 0, n = 1 or 4 (x = 1, 16); r = 2, 3, 4 give no real n.
3. The values are x = 1, 16, 25, so there are 3.

Teacher's Note:
1. Substitute q = n − r in the division relation.
2. Test r = 0, 1, 2, 3, 4 separately.

Answer: (a) 1 and 3 only

Explanation:
1. Let t = 3x + 1/x. Then t³ = 27x³ + 1/x³ + 9t, so f = t³ − 9t.
2. For both roots t = 2, so f = 8 − 18 = −10.
3. Hence f(α) = f(β) = −10: statements 1 and 3 are true, statement 2 is false.

Teacher's Note:
1. Cube the expression 3x + 1/x to bring in the cubes.
2. Both roots give the same value of t.

Answer: (b) f(x) = π − x for π/2 ≤ x ≤ π

Explanation:
1. sin⁻¹(sin x) = x only for x in [−π/2, π/2].
2. For x in [π/2, π], sin x = sin(π − x) with π − x in [0, π/2], so f(x) = π − x.

Teacher's Note:
1. Reduce x to the principal range [−π/2, π/2].
2. Use sin(π − x) = sin x.