JEE Mathematics Functions MCQs Set 20

Answer: (b) \( x\in\left[-2,-\frac12\right]\cup\left[\frac12,2\right] \)

Explanation:
1. The exponential is always positive, so only cos⁻¹(log₄ x²) needs a domain.
2. −1 ≤ log₄ x² ≤ 1 gives 1/4 ≤ x² ≤ 4.
3. So 1/2 ≤ |x| ≤ 2.

Teacher's Note:
1. Square roots of an exponential never cause trouble.
2. x² ≥ 1/4 and x² ≤ 4 gives two intervals symmetric about 0.

Answer: (d) none of these

Explanation:
1. logₑ x needs x > 0.
2. The outer log needs |logₑ x| > 0, so logₑ x ≠ 0, i.e. x ≠ 1.
3. The domain is (0, 1) ∪ (1, ∞), which is none of the given sets.

Teacher's Note:
1. The modulus removes the sign but still excludes zero.
2. ln x = 0 only at x = 1.

Answer: (b) R − Z

Explanation:
1. x − [x] is the fractional part of x.
2. It is positive exactly when x is not an integer.
3. So the domain is R − Z.

Teacher's Note:
1. The fractional part lies in [0, 1) and is zero at integers.
2. A log needs a strictly positive argument.

Answer: (a) [0, 1)

Explanation:
1. We need −1 ≤ x + [x] ≤ 1.
2. On [0, 1): [x] = 0, so x + [x] = x lies in [0, 1), which is allowed.
3. On [−1, 0): [x] = −1 and x − 1 < −1 except at x = 0, which is not in the interval, so no x works there.
4. For x ≥ 1 the sum is at least 2. So the domain is [0, 1).

Teacher's Note:
1. Take each unit interval [n, n + 1) separately.
2. x + [x] jumps by 2 at each integer.

Answer: (b) (0, 1) ∪ (1, +∞)

Explanation:
1. g(x) = logₓ 5 is defined for x > 0, x ≠ 1.
2. f(x) = ln 25 / ln x² = 2 ln 5 / (2 ln|x|) = log_{|x|} 5.
3. For x > 0 this equals logₓ 5, so the two functions agree on (0, 1) ∪ (1, ∞).

Teacher's Note:
1. Equality of functions needs equal values on the common domain.
2. The common domain is where both logs are defined.

Answer: (a) [1, +∞)

Explanation:
1. Let t = x² + 1 ≥ 1. Then f = t + 1/t − 1.
2. By AM–GM t + 1/t ≥ 2 with equality at t = 1 (x = 0), so f ≥ 1.
3. f grows without bound as x → ∞, so the range is [1, ∞).

Teacher's Note:
1. Shift the variable so the AM–GM inequality can be used.
2. Check that the equality case t = 1 is attainable.

Answer: (d) none of these

Explanation:
1. u = x²/(1 + x²) takes every value in [0, 1) (u = 0 at x = 0 and u → 1 as x → ∞ without reaching it).
2. cos⁻¹ is decreasing, so f takes values in (0, π/2].
3. The range is (0, π/2], which excludes 0, so none of the given intervals is correct.

Teacher's Note:
1. Find the range of the inner fraction first.
2. The value 1 is never reached, so cos⁻¹ never equals 0.

Answer: (a) [0, 3]

Explanation:
1. The domain is −3 ≤ x ≤ 3, where 9 − x² lies between 0 and 9.
2. The principal square root then takes values from 0 to 3.

Teacher's Note:
1. Use the domain to find the range of the radicand and then take the root.
2. A principal root is never negative.

Answer: (b) [1, 5]

Explanation:
1. For 1 ≤ x ≤ 2, f = 1 (the minimum).
2. At x = −1: f = 2 + 3 = 5; at x = 3: f = 2 + 1 = 3.
3. f is piecewise linear, so the maximum is 5 and the range is [1, 5].

Teacher's Note:
1. Piecewise linear functions attain extreme values at the corners and end points.
2. Evaluate f at −1, 1, 2 and 3.

Answer: (b) (−∞, 2]

Explanation:
1. 5 + 4x − x² = 9 − (x − 2)², which takes every value in (0, 9] on its domain.
2. log₃ of (0, 9] covers (−∞, 2].
3. So y ≤ 2.

Teacher's Note:
1. Complete the square to find the range of the argument.
2. log₃ 9 = 2 and log₃ tends to −∞ as the argument tends to 0.

Answer: (c) {(x, b), (y, a), (z, c)}

Explanation:
1. Test each option against the three conditions.
2. Option 1: all three conditions are true. Option 2: (i) and (iii) are true. Option 4: (i) is true and (ii) is true.
3. Option 3: f(x) = b makes (i) false, f(y) = a makes (ii) false, f(z) = c makes (iii) true, so exactly one is true.

Teacher's Note:
1. With only six one-one maps, checking each is quick.
2. Count the true conditions for every candidate.

Answer: (d) f is many-one and onto

Explanation:
1. f(x) = (x − 1)(x − 2)(x − 3), so f(1) = f(2) = f(3) = 0 and f is many-one.
2. A cubic polynomial takes every real value, so f is onto.

Teacher's Note:
1. Find the roots to test injectivity.
2. Every odd-degree polynomial is onto R.

Answer: (b) f is one-one and onto

Explanation:
1. f′(x) = 3x² + 2x + 3 + cos x ≥ 3x² + 2x + 2.
2. The quadratic 3x² + 2x + 2 has discriminant 4 − 24 < 0, so it is always positive and f′(x) > 0.
3. So f is strictly increasing, hence one-one; it tends to ±∞, so it is onto.

Teacher's Note:
1. Show injectivity through a positive derivative.
2. A continuous function with limits −∞ and +∞ is onto R.

Answer: (d) many-one and into

Explanation:
1. For x ≥ 0, f = 2·6ˣ; for x < 0, f = 6ˣ + 6⁻ˣ.
2. f(−1) = 1/6 + 6 and f(x) for x > 0 can equal 6 + 1/6 as well (2·6ˣ = 6 + 1/6 has a solution), so f is many-one.
3. f(x) ≥ 2 for all x, so the range is not R and f is into.

Teacher's Note:
1. Compare values of the two branches at a common level.
2. The minimum value is 2, attained at x = 0.

Answer: (d) none of these

Explanation:
1. f′(x) = p + cos x. For f to be monotonic and so bijective we need p + cos x ≥ 0 for all x or ≤ 0 for all x.
2. That means p ≥ 1 or p ≤ −1 (at p = ±1 the derivative vanishes only at isolated points).
3. So the set is (−∞, −1] ∪ [1, ∞), which is none of the given sets.

Teacher's Note:
1. A bijective continuous function on R must be strictly monotonic.
2. The extreme values of cos x are ±1.

Answer: (b) one-one and onto

Explanation:
1. Where cos x ≥ 0: f′ = 2 − sin x ≥ 1; where cos x < 0: f′ = 2 + sin x ≥ 1.
2. So f is strictly increasing, hence one-one.
3. f → ±∞ at the extremes and is continuous, so it is onto R.

Teacher's Note:
1. The two branches both have derivative at least 1.
2. Continuity at the points where cos x = 0 is automatic.

Answer: (c) many-one and into

Explanation:
1. f(−x) = f(x), so f is even and many-one.
2. f = 1 − 5/(x² + 1), where x² + 1 ≥ 1 gives f in [−4, 1).
3. The range is smaller than R, so f is into.

Teacher's Note:
1. Even functions fail to be one-one.
2. Rewrite as 1 minus a fraction to find the range.

Answer: (c) 1/4

Explanation:
1. For f to be onto [0, π/2) we need x² + x + k to take all values in [0, ∞).
2. The minimum of x² + x + k is k − 1/4, at x = −1/2.
3. It must equal 0, so k = 1/4.

Teacher's Note:
1. tan⁻¹ of a non-negative number lies in [0, π/2).
2. Equate the minimum of the quadratic to 0.

Answer: (d) none of these

Explanation:
1. √(x²) = |x|, so f(x) = x + |x| equals 2x for x ≥ 0 and 0 for x < 0.
2. All negative numbers map to 0, so f is not injective.
3. Its range is [0, ∞), so f is not surjective either.

Teacher's Note:
1. Write the function without the root.
2. Constant on a whole interval means not one-one.

Answer: (d) f(x) + f(−x)

Explanation:
1. For g(x) = f(x) + f(−x): g(−x) = f(−x) + f(x) = g(x), so g is even.
2. f(x) − f(−x) is odd.
3. [x] − x = −{x} is not even, and e^(3−2x) tan²x is not symmetric because of the exponential factor.

Teacher's Note:
1. Test each candidate by replacing x with −x.
2. f(x) + f(−x) is the standard even part construction.