JEE Mathematics Functions MCQs Set 08

Answer: (d) 8

Explanation:
1. The equation means |g(x)| = 3/2 or |g(x)| = 1/2, that is g(x) = ±3/2 or g(x) = ±1/2.
2. For |x| > 2 the graph is a line of slope 2 reaching every value: g = 1/2 and 3/2 give one point each for x > 2 (x = 2.25, 2.75); g = −1/2 and −3/2 give one point each for x < −2. That is 4 solutions.
3. Between −2 and 2 the wave has maximum 1 and minimum −1, so g = ±3/2 gives nothing, while g = 1/2 occurs twice on (−2, 0) and g = −1/2 occurs twice on (0, 2): 4 more solutions.
4. Total = 8.

Teacher's Note:
1. Split the equation into the four cases g = ±1/2 and g = ±3/2.
2. Count the intersections of each horizontal line with the graph.

Answer: (c) \( (x_0,\ g(f(x_0))) \)

Explanation:
1. A = (x₀, f(x₀)). The horizontal line y = f(x₀) meets y = x at B = (f(x₀), f(x₀)).
2. The vertical line through B is x = f(x₀), which meets y = g(x) at C = (f(x₀), g(f(x₀))).
3. The horizontal line through C has y = g(f(x₀)) and meets x = x₀ at D = (x₀, g(f(x₀))).

Teacher's Note:
1. Follow the construction one step at a time.
2. The line y = x converts a height into an abscissa.

Answer: (a) (i) and (ii) only

Explanation:
1. Try a quadratic f(x) = px² + qx + r and compare coefficients of x², xy, y², x, y and the constant: x²: 5p = 9p + 2, so p = −1/2 (the xy and y² terms agree); the terms in x and y give q = 0; the constants give 2r = r + 1, so r = 1.
2. So f(x) = 1 − x²/2, which is the (unique) solution of the relation.
3. It is even, so f(1) = f(−1): many-one (i) true. It tends to −∞, so it has no minimum (ii) true. It is even, so (iii) is false, and it is unbounded below, so (iv) is false.

Teacher's Note:
1. Look for a polynomial solution by comparing coefficients.
2. Then test each property on the function found.

Answer: (a) (B) and (C)

Explanation:
1. A function is homogeneous if f(tx, ty) = tⁿ f(x, y) for some n.
2. (A) scales by t (degree 1). (D) depends only on y/x (degree 0).
3. (B): numerator scales by t², denominator is x + y² which does not scale by a single power: not homogeneous. (C): cos x and sin x do not scale: not homogeneous.

Teacher's Note:
1. Replace x, y by tx, ty and see whether a common power of t factors out.
2. Trigonometric functions of x alone destroy homogeneity.

Answer: (a) (B) and (C) only

Explanation:
1. The condition gives f(x) = 1 ± xⁿ. With f(2) = 9 we need 1 + 2³ = 9, so f(x) = x³ + 1.
2. (A) 2f(4) = 130 and 3f(6) = 651: false. (B) 14f(1) = 28 and f(3) = 28: true.
3. (C) 9f(3) = 252 and 2f(5) = 252: true. (D) f(10) = 1001, f(11) = 1332: false.

Teacher's Note:
1. The relation is the known polynomial identity f(x) = 1 ± xⁿ.
2. Use f(2) = 9 to fix the sign and the power.

Answer: (a) 12 and 210

Explanation:
1. f(g(50)) = a(50b + a) + b and g(f(50)) = b(50a + b) + a, so the difference is \( a^2-b^2-a+b=(a-b)(a+b-1)=28 \).
2. The factors a − b and a + b − 1 have opposite parity, so the pairs are (1, 28) and (4, 7) with a > b.
3. (1, 28): a + b = 29, a − b = 1 gives a = 15, b = 14, ab = 210. (4, 7): a + b = 8, a − b = 4 gives a = 6, b = 2, ab = 12.

Teacher's Note:
1. Factorise the difference as (a − b)(a + b − 1).
2. List factor pairs of 28 that give positive integers a, b.

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

Explanation:
1. (f∘g)(x) = sgn(x(1 − x²)) is odd because g is odd, so (A) is false. It jumps at x = 0, ±1 (sign changes), so it is discontinuous there: (C) is true.
2. (g∘f)(x) = sgn x · (1 − sgn² x), which is 0 for x ≠ 0 and g(0) = 0 at x = 0. So g∘f ≡ 0, which is both even and odd, and continuous and differentiable everywhere: (B) and (D) are true.

Teacher's Note:
1. Compute both compositions explicitly.
2. The zero function is both even and odd.

Answer: (a) (ii) and (iv)

Explanation:
1. Replace x by 1 − x: the arguments x² + x + 3 and x² − 3x + 5 swap places, giving f(x²−3x+5) + 2f(x²+x+3) = 6x² − 2x + 13.
2. Solving the two equations: f(x² + x + 3) = 2x² + 2x + 3 = 2(x² + x + 3) − 3, so f(x) = 2x − 3.
3. f is increasing (not decreasing), has the root x = 3/2 ∈ (0, 2), is not odd (f(0) = −3), and is invertible (one-one and onto). So (ii) and (iv).

Teacher's Note:
1. The substitution x → 1 − x swaps the two arguments.
2. Solve the two linear equations for f at the two arguments.

Answer: (a) (i) and (iii)

Explanation:
1. The base [2 tan πx] must be positive and not 1, so 2 tan πx ≥ 2, i.e. tan πx ≥ 1 and x ∈ [n + 1/4, n + 1/2).
2. The argument must be positive: \( \frac{(x+3)(x-1)}{(2x-3)(2x+1)}\gt0\Rightarrow x\in(-\infty,-3)\cup\left(-\frac12,1\right)\cup\left(\frac32,\infty\right) \).
3. n = 0: [1/4, 1/2) ⊂ (−1/2, 1) ✓. n = −5: [−19/4, −9/2) ⊂ (−∞, −3) ✓. n = −1: [−3/4, −1/2) lies where the argument is negative ✗. n = 1: [5/4, 3/2) lies in (1, 3/2) where it is negative ✗.

Teacher's Note:
1. Find where the base is greater than 1 and where the logarithm argument is positive.
2. Intersect the intervals with each given n.

Answer: (c) 3

Explanation:
1. Expanding: the determinant equals \( x^3-x^2\sin x-x\cos x+\sin x\cos x=(x-\sin x)(x^2-\cos x) \).
2. x − sin x = 0 only at x = 0, since x − sin x is strictly increasing.
3. x² = cos x has exactly two solutions, ±x₀ with x₀ ≈ 0.82 (even function; at x = 0 the left side is below, at x = 1 above).
4. Total number of solutions = 3.

Teacher's Note:
1. Factorise the determinant after expanding.
2. For x² = cos x, use the symmetry about the y-axis.

Answer: (a) (i) and (ii)

Explanation:
1. f(0) = max(1, 1, 0) = 1 and g(1) = max(1, 0) = 1: (i) true.
2. f(1) = 1 + sin 1 ≈ 1.84, so g(f(1)) = max(1, 0.84) = 1: (ii) true.
3. g(1) = 1 and g(0) = 1, so f(g(1)) = f(g(0)) = f(1) = 1 + sin 1, which is neither 1 nor sin 1: (iii) and (iv) are false.

Teacher's Note:
1. Compute inner functions first.
2. Use f(1) = 1 + sin 1 ≈ 1.84.

Answer: (a) (i) and (iii)

Explanation:
1. The parabola y = 4x²/9 meets the circle x² + y² = 25 where 16x⁴/81 + x² = 25, i.e. 16t² + 81t − 2025 = 0 with t = x²: t = 9, so x = ±3 (and y = 4).
2. The domain of the common region is [−3, 3] and the largest y is 5 (at x = 0), the smallest 0: range [0, 5].
3. For x = 0, y takes every value from 0 to 5, so R ∩ R′ is not a function. So (i) and (iii).

Teacher's Note:
1. Find the intersection points of the parabola and the circle.
2. A relation is a function only if each x has exactly one y.

Answer: (a) 0

Explanation:
1. For |x| ≥ 1: x² − 1 = x(x − 2) gives x = 1/2, which contradicts |x| ≥ 1.
2. For |x| < 1: 1 − x² = x(x − 2) gives 2x² − 2x − 1 = 0, so x = (1 ± √3)/2; only (1 − √3)/2 ≈ −0.37 lies in (−1, 1), and it is not an integer.
3. So there is no integral solution: the number is 0.

Teacher's Note:
1. Split the modulus at |x| = 1.
2. Check each root against its case and for integrality.

Answer: (a) 0

Explanation:
1. Each factor is ≥ 0, so f ≥ 0.
2. At x = −π/2, sin x = −1 and the first factor is 0, so f = 0 is attained.
3. The minimum value is 0.

Teacher's Note:
1. A product of non-negative factors is non-negative.
2. Look for a point where one factor vanishes.

Answer: (b) 2

Explanation:
1. \( f(x)=\frac12\sin\pi x \), which has fundamental period 2π/π = 2.

Teacher's Note:
1. Use the double-angle formula.
2. A function sin(kx) has period 2π/k.

Answer: (a) [−1, 1]

Explanation:
1. We need −2 ≤ |x| + 1 ≤ 2.
2. Since |x| + 1 ≥ 1, the condition is |x| + 1 ≤ 2, i.e. |x| ≤ 1: the domain is [−1, 1].

Teacher's Note:
1. Impose the domain of f on its argument.
2. The absolute value is never negative.

Answer: (c) [−4, 3]

Explanation:
1. Let u = |sin x| and v = |cos x| with u² + v² = 1, u, v ≥ 0 (the first quadrant arc).
2. f = 3u − 4v is increasing in the angle on this arc: at |cos x| = 1 (u = 0) f = −4; at |sin x| = 1 (v = 0) f = 3.
3. The range is [−4, 3].

Teacher's Note:
1. Restrict to the first-quadrant arc of the unit circle.
2. The function is monotonic there, so check the end points.

Answer: (c) R − {0}

Explanation:
1. f needs x ≠ 0, and the outer f needs f(x) ≠ 0.
2. x + 1/x = 0 has no real solution, so only x = 0 is excluded: the domain is R − {0}.

Teacher's Note:
1. The outer function excludes the value 0.
2. Check whether the inner function can take the value 0.

Answer: (a) R − {−2, −5/3}

Explanation:
1. The inner g needs x ≠ −2, and the outer g needs g(x) ≠ −2.
2. (x + 1)/(x + 2) = −2 gives x + 1 = −2x − 4, i.e. x = −5/3. The domain is R − {−2, −5/3}.

Teacher's Note:
1. Exclude where g is undefined and where it equals the excluded input −2.
2. Solve g(x) = −2.

Answer: (a) P, R and S only

Explanation:
1. The range is (0, 1], containing exactly one natural number, 1: (P) holds. The domain is R: (R). h(−x) = h(x): (S).
2. h is not periodic because it decays to 0: (Q) fails.

Teacher's Note:
1. Find the range from |x| ≥ 0.
2. A decaying function cannot be periodic.

Answer: (a) all four values

Explanation:
1. f is many-one on an interval exactly when its vertex x = −3a/2 lies strictly inside the interval.
2. The condition −3 < −3a/2 < 5 gives −10/3 < a < 2, which holds for each of −2, −1, 0 and 1.

Teacher's Note:
1. Find where the vertex lies for each value of a.
2. A parabola is many-one on any interval containing its vertex in the interior.

Answer: (d) 1

Explanation:
1. The minimum of x² + 6x + b is b − 9, at x = −3.
2. For the range to be exactly [−8, ∞) we need b − 9 = −8, so b = 1.

Teacher's Note:
1. Surjective onto [−8, ∞) means the minimum value equals −8.
2. Complete the square to find the minimum.

Answer: (a) 0 only

Explanation:
1. x² + 4x + 5 = (x + 2)² + 1 ≥ 1, so [·] is a positive integer n ≥ 1 and log₂ n ≥ 0.
2. For the square root we need log_{1/2}(log₂ n) ≥ 0, i.e. 0 < log₂ n ≤ 1, so n = 2. Then log₂ 2 = 1, log_{1/2} 1 = 0 and f(x) = 0.
3. The range is {0}, so only 0 occurs.

Teacher's Note:
1. Find which integer values of [x² + 4x + 5] keep the radicand non-negative.
2. Only one such integer exists.

Answer: (a) none of them

Explanation:
1. Let a = tan²θ > 0 and b = tan²φ > 0. The expression is \( \frac{(1+a)^2}{b}+\frac{1+b}{a} \).
2. Its minimum over a, b > 0 is about 4.73 (e.g. a = b = t gives t + 3 + 2/t ≥ 3 + 2√2 ≈ 5.83 and unequal a, b lower it only to about 4.73).
3. So the expression always exceeds 4 and cannot equal 0, 1, 2 or 4.

Teacher's Note:
1. Rewrite in terms of a = tan²θ and b = tan²φ.
2. Estimate the minimum with AM-GM or a numerical search.

Answer: (a) all four values

Explanation:
1. 3 sin x + 4 cos x = 5 sin(x + φ), so f′(x) = a + 5 cos(x + φ).
2. For |a| < 5, f′ changes sign, so f rises and falls: it is many-one. All four values satisfy |a| < 5.

Teacher's Note:
1. Differentiate and compare a with the amplitude 5.
2. A derivative that changes sign gives a many-one function.

Answer: (b) 4

Explanation:
1. Since [t] ≤ t, the left side is at most x/2 + 2x/3 + y/4 + 4y/5 = 7x/6 + 21y/20, with equality only if x/2, 2x/3, y/4 and 4y/5 are all integers.
2. x/2 and 2x/3 integers means x is a multiple of 6: x = 6, 12, 18, 24 (4 values). y/4 and 4y/5 integers means y is a multiple of 20: y = 20 (1 value).
3. Number of pairs = 4 × 1 = 4.

Teacher's Note:
1. Compare each greatest integer term with its argument.
2. Equality needs every argument to be an integer.

Answer: (b) 15

Explanation:
1. f′(x) = 3(x − 2)(x − 4). Three distinct real roots need f(2) = 20 + c > 0 and f(4) = 16 + c < 0, i.e. −20 < c < −16.
2. The roots lie in (1, 2) (since f(1) = 16 + c < 0 < f(2)), in (2, 4), and in (4, 5) (since f(5) = 20 + c > 0): so [α] = 1 and [γ] = 4.
3. The middle root β lies in (2, 3) when f(3) = 18 + c < 0 (c < −18), giving [β] = 2, and in [3, 4) when c ≥ −18, giving [β] = 3.
4. Possible sums: 1 + 2 + 4 = 7 and 1 + 3 + 4 = 8. Their sum is 15.

Teacher's Note:
1. Use the turning points of the cubic to locate the roots.
2. The middle root can fall on either side of 3.

Answer: (b) 18

Explanation:
1. If k ≤ x < k + 1 then [x] = k and x[x] = kx takes every value in [k², k² + k).
2. This interval contains k natural numbers (k², …, k² + k − 1), and the intervals for different k do not overlap. The largest k with k² + k − 1 < 2009 is k = 44 (44² + 44 = 1980; for k = 45 the numbers start at 2025).
3. N = 1 + 2 + … + 44 = 990, and the sum of its digits is 9 + 9 + 0 = 18.

Teacher's Note:
1. Split x into integer blocks [k, k + 1).
2. Add the counts for k = 1 to 44.