JEE Mathematics Functions MCQs Set 11

Answer: (c) 0

Explanation:
1. A surjection needs n(A) ≥ n(B).
2. Here n(A) = 4 < n(B) = 6, so no surjection exists.
3. The number is 0.

Teacher's Note:
1. Check the size condition first.
2. If the co-domain is bigger, nothing can be onto.

Answer: (d) 625

Explanation:
1. Total mappings: 5⁴ = 625.
2. A surjection from 4 elements onto 5 elements is impossible, so every mapping is non-surjective.
3. Answer = 625.

Teacher's Note:
1. Count all mappings and subtract the surjections (here zero).
2. Use n(B)^n(A) for all mappings.

Answer: (d) \( (f\circ g)(x) \)

Explanation:
1. (f∘g)(x) = [x − [x]] = [{x}].
2. Since 0 ≤ {x} < 1, the greatest integer is 0.
3. So f∘g is the zero function.
4. The others are (f + g)(x) = x, (fg)(x) = [x]{x} and (f − g)(x) = 2[x] − x, none of which is identically zero.

Teacher's Note:
1. Evaluate each combination at a non-integer such as x = 1.5.
2. Only the composition gives zero.

Answer: (b) −1

Explanation:
1. (f∘f)(x) = K·f(x)/(f(x) + 1) with f(x) = Kx/(x + 1).
2. = K²x/(Kx + x + 1) = K²x/((K + 1)x + 1).
3. For this to equal x we need (K + 1)x + 1 = K².
4. This holds for all x only if K + 1 = 0, then K² = 1, so K = −1.

Teacher's Note:
1. Compare coefficients of x on both sides.
2. K = −1 also makes f its own inverse.

Answer: (b) \( 5^x \)

Explanation:
1. Let y = f⁻¹(x). Then x = f(y) = log₅y, so y = 5ˣ.
2. f⁻¹(x) = 5ˣ.

Teacher's Note:
1. The inverse of the logarithm to base a is the exponential with base a.

Answer: (c) \( \frac{x+7}{10} \)

Explanation:
1. Let y = f⁻¹(x). Then x = 10y − 7, so y = (x + 7)/10.
2. Thus g(x) = (x + 7)/10.

Teacher's Note:
1. Solve the equation y ↦ x for y.
2. Check: f(g(x)) = x.

Answer: (a) 3

Explanation:
1. Let f⁻¹(13) = a. Then f(a) = a² + 4 = 13.
2. a² = 9, and since a ≥ 0, a = 3.

Teacher's Note:
1. The domain of f is [0, ∞), so only the positive root counts.

Answer: (c) \( \phi \)

Explanation:
1. f(x) = 1/2 means x − [x] − 1/2 = 1/2, i.e. {x} = 1.
2. But {x} < 1 always.
3. The set is empty.

Teacher's Note:
1. The fractional part is always less than 1.
2. So the equation has no solution.

Answer: (c) {−1/2}

Explanation:
1. For x ≤ 0, |x| = −x ≥ 0 and f(|x|) = |x| − 1 = −x − 1.
2. −x − 1 = x gives x = −1/2, which satisfies x ≤ 0.
3. The set is {−1/2}.

Teacher's Note:
1. Use the second branch since |x| lies in [0, 2].
2. Check that the solution meets the condition x ≤ 0.

Answer: (b) 1st and 3rd quadrant (including the axes)

Explanation:
1. |a + b| = |a| + |b| holds exactly when a and b have the same sign (or one is zero).
2. So sin x and cos x have the same sign: x in the first or third quadrant (including the axes).

Teacher's Note:
1. Equality in the triangle inequality needs a and b of the same sign.
2. Apply it to sin x and cos x.

Answer: (d) \( \sin^{-1}\left(\cos x^2\right) \)

Explanation:
1. 2ˣ/2^[x] = 2^{x}, with {x} having period 1, so (a) is periodic.
2. sin⁻¹{x} has period 1.
3. sin⁻¹√(cos x) has period 2π.
4. sin⁻¹(cos x²) is not periodic, since cos x² is not periodic.

Teacher's Note:
1. Reduce each function to a combination of periodic functions.
2. x² breaks periodicity.

Answer: (c) 11

Explanation:
1. f(g(x)) is periodic with f non-constant, so g must be linear: g(x) = ax + b.
2. g(2) = 2a + b = 3 and g(4) = 4a + b = 7 give a = 2 and b = −1.
3. g(6) = 12 − 1 = 11.

Teacher's Note:
1. A non-linear polynomial is unbounded and cannot make f(g(x)) periodic.
2. Solve the two conditions for a and b.

Answer: (c) \( [0,\pi] \)

Explanation:
1. cos⁻¹(t) needs |t| ≤ 1.
2. Here 2 + sin x > 0, so 2/(2 + sin x) ≤ 1 gives 2 ≤ 2 + sin x, i.e. sin x ≥ 0.
3. In [0, 2π] this means x ∈ [0, π].

Teacher's Note:
1. The argument is positive, so only the upper bound 1 is a restriction.
2. Solve sin x ≥ 0.

Answer: (c) {3, 4, 5}

Explanation:
1. ⁿP_r needs n ≥ r ≥ 0 with n and r integers.
2. x − 3 ≥ 0 gives x ≥ 3; 7 − x ≥ x − 3 gives x ≤ 5; x must be an integer.
3. Domain = {3, 4, 5}.

Teacher's Note:
1. Both indices must be non-negative integers and n ≥ r.
2. List the integers satisfying the inequalities.

Answer: (b) {1, 4, 7}

Explanation:
1. f(n) > 2 can happen only for n = 3k + 1, where f = 10 − n.
2. 10 − n > 2 gives n < 8, so n = 1, 4, 7.
3. For n = 3k, f = 2 (not > 2); for n = 3k + 2, f = 0.

Teacher's Note:
1. Check each of the three branches.
2. Only the middle branch can exceed 2.

Answer: (c) \( 2\tan\alpha \)

Explanation:
1. Let t = √(x² + x) > 0. The expression is t + tan²α/t.
2. By AM-GM: t + tan²α/t ≥ 2√(tan²α) = 2 tan α, as tan α > 0 for α in (0, π/2).

Teacher's Note:
1. Apply AM ≥ GM to the two positive terms.
2. Equality holds when t = tan α.

Answer: (c) {−3, −2}

Explanation:
1. f(x) = |x| lies in [0, 8/5) for −8/5 < x < 8/5.
2. g(f(x)) = [|x| − 3], where |x| − 3 ∈ [−3, −7/5).
3. The integer parts are −3 (for [−3, −2)) and −2 (for [−2, −7/5)).
4. The set is {−3, −2}.

Teacher's Note:
1. Find the range of |x| first.
2. Take greatest integers of values between −3 and −1.4.

Answer: (c) \( [0,\infty) \)

Explanation:
1. The expression equals ½[(x − 2y)² + (2y − 3z)² + (3z − x)²].
2. It is ≥ 0 and equals 0 when x = 2y = 3z.
3. It can be made arbitrarily large, so the range is [0, ∞).

Teacher's Note:
1. Write the quadratic form as a sum of squares.
2. Expand to verify.

Answer: (b) \( D_f=Z,\ R_f=\{1,-1\} \)

Explanation:
1. (−1)ˣ is real only for integer x (for general real x it is complex).
2. So the maximum domain is Z and the range is {1, −1}.

Teacher's Note:
1. Non-integer powers of −1 are not real.
2. The sign alternates between even and odd x.

Answer: (d) neither one-one nor onto

Explanation:
1. f(x) = e^{2ix} = cos 2x + i sin 2x.
2. f(x + π) = f(x), so f is many-one.
3. |f(x)| = 1, so f is not onto C.
4. Neither one-one nor onto.

Teacher's Note:
1. Euler's formula shows the periodicity and the unit modulus.
2. Not every complex number has modulus 1.

Answer: (c) one-one onto

Explanation:
1. n = 1, 2, 3, 4, 5, 6 give 0, −1, 1, −2, 2, −3, ...
2. Every integer appears exactly once (non-negative integers from odd n, negative integers from even n).
3. So f is one-one and onto.

Teacher's Note:
1. List the first few values to see the pattern.
2. Odd n give 0, 1, 2, ...; even n give −1, −2, ...

Answer: (b) onto but not one-one

Explanation:
1. det A = k has solutions for every real k (for example diag(k, 1)), so f is onto.
2. det diag(1, 1) = det diag(−1, −1) = 1, so f is many-one.

Teacher's Note:
1. Show an example matrix for the onto part.
2. Use two different matrices with equal determinants for many-one.

Answer: (a) \( x^3+3x \)

Explanation:
1. (x − 1/x)³ = x³ − 1/x³ − 3(x − 1/x).
2. So x³ − 1/x³ = (x − 1/x)³ + 3(x − 1/x).
3. With t = g(x) = x − 1/x this gives f(t) = t³ + 3t.

Teacher's Note:
1. Cube x − 1/x and rearrange.
2. Replace x − 1/x by t.

Answer: (b) \( \sqrt x \)

Explanation:
1. g(sin²x) = |sin x| = √(sin²x).
2. So g(t) = √t.

Teacher's Note:
1. Write |sin x| as the square root of sin²x.
2. Then replace sin²x by t.