JEE Mathematics Functions MCQs Set 07

Answer: (b) 2

Explanation:
1. \( f(g(x))=\frac{rx}{1-x+rx} \) and \( g(f(x))=\frac{r\frac{x}{1+x}}{1-\frac{x}{1+x}}=rx \).
2. Equation: rx = rx/(1 − x + rx), so rx = 0 or 1 − x + rx = 1.
3. The second case gives x(r − 1) = 0: it holds for every x only if r = 1. The first gives x = 0 only unless r = 0, when it holds for all x.
4. So r ∈ {0, 1}: two elements.

Teacher's Note:
1. For infinitely many solutions, the equation must reduce to an identity for the parameter.
2. Check r = 0 and r = 1 separately.

Answer: (c) 2

Explanation:
1. On [0, 1): [x] = 0, so x² − 3x = 0 gives x = 0 (valid).
2. On [1, 2): x² − 3x + 1 = 0 gives x = (3 ± √5)/2 = 0.38 or 2.62, neither in [1, 2).
3. On [2, 3): x² − 3x + 2 = 0 gives x = 1 or 2; only x = 2 is valid. At x = 3: 9 − 9 + 3 ≠ 0.
4. Solutions: x = 0 and x = 2.

Teacher's Note:
1. Split the interval by the value of [x] and solve the quadratic in each piece.
2. Check each root against the interval it came from.

Answer: (c) a finite set

Explanation:
1. Replace x by 1/x: \( f\left(\frac1x\right)+f(x)=\frac{1+x^3}{x^2} \).
2. Both left sides are the same, so \( \frac{x^3+1}{x}=\frac{x^3+1}{x^2}\Rightarrow (x^3+1)(x-1)=0 \).
3. So the domain can contain only x = 1 and x = −1: a finite set.

Teacher's Note:
1. The left side is symmetric under x → 1/x, which forces the right side to be symmetric too.
2. That condition holds only at a few points, restricting the domain.

Answer: (d) \( \frac{35}{3} \)

Explanation:
1. \( (\tan x)\operatorname{sgn}x \) is even, so the odd part must vanish: both brackets must be zero.
2. \( [a]^2-5[a]+4=0\Rightarrow [a]=1\text{ or }4 \); \( 6\{a\}^2-5\{a\}+1=0\Rightarrow \{a\}=\frac12\text{ or }\frac13 \).
3. a ∈ {1.5, 1 + 1/3, 4.5, 4 + 1/3}. Sum = 6 + 17/3 = 35/3.

Teacher's Note:
1. Even function plus odd terms means the odd terms must have zero coefficients.
2. Combine every allowed integer part with every allowed fractional part.

Answer: (c) \( \pi \)

Explanation:
1. The numerator has period π/2, but \( |\sin x-\cos x|=\sqrt2\,|\sin(x-\tfrac{\pi}{4})| \) has period π.
2. Under x → x + π/2, sin x → cos x and cos x → −sin x, so the denominator becomes |sin x + cos x|, which is different. Hence π/2 is not a period.
3. The period of the quotient is π.

Teacher's Note:
1. The period of a quotient is the LCM of the periods, if it works.
2. Test the smaller candidate π/2 explicitly before accepting π.

Answer: (d) \( (-\infty,1] \)

Explanation:
1. Domain: 1 − 2x ≥ 0, so D1 = (−∞, 1/2].
2. Let t = √(1 − 2x) ≥ 0, so x = (1 − t²)/2 and f = t + (1 − t²)/2.
3. This is a downward parabola in t with maximum 1 at t = 1, and it tends to −∞ as t grows. So the range is (−∞, 1].

Teacher's Note:
1. Substitute t = √(1 − 2x) to turn the function into a quadratic in t.
2. For an onto function the codomain D2 equals the range.

Answer: (c) f(x) can be expressed as a sum or a difference of an odd and an even function

Explanation:
1. \( f(x)=\frac{f(x)+f(-x)}{2}+\frac{f(x)-f(-x)}{2} \).
2. The first part is even and the second is odd (the domain [−a, a] is symmetric).
3. So every such function is the sum of an even and an odd function.

Teacher's Note:
1. The even and odd parts of a function are (f(x) ± f(−x))/2.
2. This works because −x lies in the domain whenever x does.

Answer: (c) \( f_1(x) \)

Explanation:
1. n = 1 (odd): \( f_3(x)=f_2(f_1(x))=\frac{1}{1-\frac{x}{x-1}}=1-x \).
2. n = 2 (even): \( f_4(x)=f_3(f_2(x))=1-\frac{1}{1-x}=\frac{-x}{1-x}=\frac{x}{x-1} \).
3. This equals f1(x).

Teacher's Note:
1. Use the recursion exactly as stated: odd n composes with f1, even n composes with f2.
2. Simplify each composition into a single fraction.

Answer: (d) [3, ∞)

Explanation:
1. Need x ≥ 3. Put t = √(x − 3) ≥ 0, so x − 2 = t² + 1.
2. Then x − 2 − 2√(x − 3) = (t − 1)² ≥ 0 and x − 2 + 2√(x − 3) = (t + 1)² ≥ 0, so both roots are always defined.
3. Domain: [3, ∞).

Teacher's Note:
1. Rewrite each radicand as a perfect square using t = √(x − 3).
2. Perfect squares are never negative, so only x ≥ 3 is needed.

Answer: (d) 1004

Explanation:
1. \( f(x)+f(1-x)=\frac{e^{2x-1}}{1+e^{2x-1}}+\frac{e^{1-2x}}{1+e^{1-2x}}=\frac{e^{2x-1}}{1+e^{2x-1}}+\frac{1}{e^{2x-1}+1}=1 \).
2. The 2008 terms form 1004 pairs (k and 2009 − k), each adding to 1.
3. Sum = 1004.

Teacher's Note:
1. Look for the pairing f(x) + f(1 − x) = 1.
2. There are 2008 terms, which is an even number, so all terms pair up.

Answer: (a) 2

Explanation:
1. If x < 1: f(x) = 0 and every further application keeps 0, so f⁴(x) = 0 = x only for x = 0.
2. If x = 1: f(1) = 0, then 0, so f⁴(1) = 0 ≠ 1.
3. If x > 1: f(x) − 2 = 2(x − 2) as long as the iterates stay ≥ 1, so f⁴(x) − 2 = 16(x − 2), which equals x − 2 only for x = 2. (Iterates that fall below 1 collapse to 0 and cannot return to x.)
4. Solutions: x = 0 and x = 2.

Teacher's Note:
1. Treat the two pieces of the function separately and track where the iterates land.
2. Fixed points of f are also fixed points of f⁴: x = 0 and x = 2.

Answer: (c) −2

Explanation:
1. g(x) = x² − 2x − 1 + 3x − 1 = x² + x − 2.
2. y = g(|x|) = |x|² + |x| − 2 = x² + |x| − 2.
3. Both x² and |x| are at least 0, with minimum at x = 0, so the least value is −2.

Teacher's Note:
1. Replacing x by |x| makes the function even and its minimum occurs where |x| is smallest.
2. The vertex of x² + x − 2 is at negative x, which is not reachable once |x| is used.

Answer: (a) \( 1+\sqrt{x+2} \)

Explanation:
1. Here a = 1 and b = −2 (the vertex is at (1, −2)). On [1, ∞), y = (x − 1)² − 2, so (x − 1)² = y + 2.
2. Since x ≥ 1, x − 1 = +√(y + 2), i.e. x = 1 + √(y + 2).
3. So f⁻¹(x) = 1 + √(x + 2).

Teacher's Note:
1. Solve for x in terms of y and choose the sign of the root according to the domain.
2. On [1, ∞) the value x − 1 is non-negative, so the positive root is used.

Answer: (a) (−2, −1)

Explanation:
1. f(|x|) = x² − 2|x| − 1. For positive x, x² − 2x − 1 = k has two positive roots when both roots are positive and distinct.
2. The roots are 1 ± √(k + 2). They are distinct if k > −2 and both positive if 1 − √(k + 2) > 0, i.e. k + 2 < 1, k < −1.
3. Each positive root gives two real roots ±x, so we get 4 roots for −2 < k < −1.

Teacher's Note:
1. f(|x|) is the right half of the graph reflected in the y-axis.
2. Four roots need two distinct positive roots of the quadratic.

Answer: (c) 2φ(2n + 1)

Explanation:
1. 8n + 4 = 4(2n + 1), and 4 and 2n + 1 are relatively prime (2n + 1 is odd).
2. φ(8n + 4) = φ(4)·φ(2n + 1), and φ(4) = φ(2²) = 2¹(2 − 1) = 2.
3. So φ(8n + 4) = 2φ(2n + 1).

Teacher's Note:
1. Factor out the power of 2 and use multiplicativity for coprime factors.
2. φ(2²) = 2 is the key value.

Answer: (b) 2

Explanation:
1. For an odd prime p, φ(pⁿ) = p^{n−1}(p − 1) is even. For p = 2, φ(2ⁿ) = 2^{n−1}, which is odd only for n = 1.
2. So φ(n) is odd only if n has no odd prime factor and the power of 2 is at most 1: n = 1 or n = 2.
3. Both n = 1 (φ = 1) and n = 2 (φ = 1) work: 2 numbers.

Teacher's Note:
1. A product of integers is odd only when every factor is odd.
2. Check the contribution of each prime power to the product.

Answer: (b) 4

Explanation:
1. \( \varphi(7^n)=7^{n-1}\cdot 6=2058 \).
2. \( 7^{n-1}=343=7^3 \), so n − 1 = 3.
3. n = 4.

Teacher's Note:
1. Apply the formula for a prime power directly.
2. Divide 2058 by 6 to get 343, which is 7 cubed.

Answer: (a) (A), (B) and (D)

Explanation:
1. [x + 1] = [x] + 1 keeps increasing, so it is aperiodic.
2. sin x² has zeros at √(nπ), which are not equally spaced, so it is aperiodic.
3. sin² x has period π, so it is periodic.
4. sin⁻¹ x is defined only on [−1, 1], so it is aperiodic.

Teacher's Note:
1. A periodic function must be defined on the whole real line and repeat exactly.
2. Look at the spacing of the zeros for a function like sin x².

Answer: (a) (A) and (D)

Explanation:
1. x − [x] = {x} has period 1.
2. sin(1/x) oscillates faster near 0 and is not periodic; x cos x has growing amplitude; neither repeats.
3. sin⁻¹(sin x) is a triangular wave of period 2π.
4. So (A) and (D) are periodic.

Teacher's Note:
1. Check whether the function repeats its values after a fixed shift.
2. Growing amplitude or non-uniform spacing rules out periodicity.

Answer: (a) (A), (C) and (D)

Explanation:
1. (A) is strictly increasing, so one-one.
2. (B) x + 1/x decreases on (0, 1) and increases on (1, ∞), so g(1/2) = g(2): not one-one.
3. (C) The vertex is at x = −2, so on x > 0 it is strictly increasing: one-one.
4. (D) e⁻ˣ is strictly decreasing: one-one.

Teacher's Note:
1. Monotonic functions are one-one.
2. x + 1/x takes the same value at x and 1/x.

Answer: (a) (A), (C) and (D)

Explanation:
1. |sin x| and cos² x have period π.
2. cos(sin(x + π)) = cos(−sin x) = cos(sin x), so cos(sin x) has period π.
3. [x + π] is not periodic (it increases by 1 whenever x increases by 1), so (B) is not.

Teacher's Note:
1. Test f(x + π) = f(x) for each function.
2. Greatest integer functions of a linear expression are aperiodic.