JEE Mathematics Functions MCQs Set 09

Answer: (b) (2, −1)

Explanation:
1. f(1) = a + b = 1 and f(0) = b = −1.
2. So a = 2 and b = −1.
3. Check: f(2) = 3 and f(−1) = −3, which match the given pairs.

Teacher's Note:
1. Substitute two known pairs to get two linear equations.
2. Then verify with the remaining pairs.

Answer: (d) \( 2f(x) \)

Explanation:
1. f(3x) = 6x + 3|x| and f(−x) = −2x + |x|.
2. f(3x) − f(−x) − 4x = 6x + 3|x| + 2x − |x| − 4x = 4x + 2|x|.
3. This equals 2(2x + |x|) = 2f(x).

Teacher's Note:
1. Expand each term separately and collect like terms.
2. |3x| = 3|x| and |−x| = |x|.

Answer: (c) \( 2^n-1 \)

Explanation:
1. f(1) = 1, f(2) = 3, f(3) = 7, f(4) = 15 suggests f(n) = 2ⁿ − 1.
2. Proof: f(n + 1) + 1 = 2(f(n) + 1), so f(n) + 1 is a geometric sequence with first term 2 and ratio 2.
3. Hence f(n) = 2ⁿ − 1.

Teacher's Note:
1. Generate the first few terms to guess the pattern.
2. Adding 1 to both sides makes the recurrence multiplicative.

Answer: (b) 7

Explanation:
1. 7^(ln x) = x^(ln 7) because both equal e^(ln7 · ln x).
2. So f(x) = 7 · 7^(ln x)/x^(ln 7) = 7.
3. f is constant, so f(2015) = 7.

Teacher's Note:
1. Use a^(ln b) = b^(ln a).
2. The function collapses to a constant.

Answer: (b) \( \frac{25}{12} \)

Explanation:
1. At x = 0: f = 6 and g = 3.
2. 2f/(3g) = 12/9 = 4/3 and 3g/(2f) = 9/12 = 3/4.
3. Sum = 4/3 + 3/4 = 25/12.

Teacher's Note:
1. Read the values from the ordered pairs at x = 0.
2. Add the two fractions with a common denominator.

Answer: (d) 0

Explanation:
1. Let a = log x and b = log y. Then f(xy) = sin(a + b) and f(x/y) = sin(a − b).
2. Sum = 2 sin a cos b.
3. Subtracting 2f(x)cos(log y) = 2 sin a cos b gives 0.

Teacher's Note:
1. Use sin(a + b) + sin(a − b) = 2 sin a cos b.
2. log(xy) = log x + log y and log(x/y) = log x − log y.

Answer: (b) an even function

Explanation:
1. x/(eˣ − 1) + x/2 = (x/2)·(eˣ + 1)/(eˣ − 1) = (x/2) coth(x/2).
2. This is an even function, since x and coth(x/2) are both odd.
3. Adding the constant 1 keeps it even.

Teacher's Note:
1. Combine the first two terms over a common denominator.
2. A product of two odd functions is even.

Answer: (a) 1

Explanation:
1. sin π[x] = 0 for all x, because [x] is an integer.
2. e^(cos{x}) depends only on {x}, which has period 1.
3. So f has period 1.

Teacher's Note:
1. The term sin π[x] vanishes identically.
2. The period comes from the fractional part.

Answer: (b) 2k

Explanation:
1. f(x + k) = −f(x).
2. Then f(x + 2k) = −f(x + k) = f(x).
3. So 2k is a period (the smallest one if f is not constant).

Teacher's Note:
1. Apply the condition twice.
2. The sign changes cancel after two shifts.

Answer: (a) \( 0 < x\le1 \)

Explanation:
1. sin⁻¹x > 0 requires 0 < x ≤ 1.
2. For the logarithm to be defined its argument must be positive, so sin⁻¹x > 0.
3. Therefore the domain is 0 < x ≤ 1 (x = 0 is excluded).

Teacher's Note:
1. The argument of a logarithm must be strictly positive.
2. sin⁻¹0 = 0 is not allowed.

Answer: (b) \( \left\{\frac12,-2\right\} \)

Explanation:
1. Set 2x² − 1 = 1 − 3x.
2. 2x² + 3x − 2 = 0, so (2x − 1)(x + 2) = 0.
3. x = 1/2 or x = −2.

Teacher's Note:
1. Two functions are equal only at points where their values agree.
2. Solve the equation for those points.

Answer: (c) Domain = R − {4}, Range = {−1}

Explanation:
1. For x ≠ 4, f(x) = (4 − x)/(x − 4) = −1.
2. The domain excludes 4, so the domain is R − {4} and the range is {−1}.

Teacher's Note:
1. Cancel the common factor only after noting the excluded point.
2. The function is a constant on its domain.

Answer: (a) \( [0,1) \)

Explanation:
1. x²/(x² + 1) = 1 − 1/(x² + 1).
2. x² + 1 ≥ 1, so 0 < 1/(x² + 1) ≤ 1.
3. Hence f takes all values in [0, 1).

Teacher's Note:
1. Write the fraction as 1 minus a fraction to see its bounds.
2. The value 1 is never reached.

Answer: (b) \( 2\log|x| \)

Explanation:
1. log x² is defined for all x ≠ 0, but log x is defined only for x > 0.
2. log x² = 2 log|x| for every x ≠ 0, so 2 log|x| is the equivalent function (same domain and same values).

Teacher's Note:
1. Equal functions must have the same domain.
2. The absolute value keeps the domain symmetric.

Answer: (b) Two

Explanation:
1. A linear function f(x) = ax + b that maps [−1, 1] onto [0, 2] is either increasing or decreasing.
2. Increasing: f(−1) = 0, f(1) = 2 gives f(x) = x + 1.
3. Decreasing: f(−1) = 2, f(1) = 0 gives f(x) = 1 − x.
4. So there are two.

Teacher's Note:
1. The endpoints of the domain go to the endpoints of the range.
2. Each choice of orientation gives one function.

Answer: (c) \( (1,4) \)

Explanation:
1. log₃x is increasing.
2. For 3 < x < 81, log₃x ranges over the open interval (1, 4).
3. For a surjection the co-domain must equal the range, so B = (1, 4).

Teacher's Note:
1. A strictly increasing function maps an open interval onto an open interval.
2. End values are not attained.

Answer: (c) many-one, onto

Explanation:
1. f(1) = 0 and f(3) = 0, so f is many-one.
2. The range is {0, 1}, which equals the co-domain, so f is onto.

Teacher's Note:
1. Compare images of two different inputs.
2. Compare the range with the co-domain.

Answer: (b) onto but not one-one

Explanation:
1. f(1) = f(−1) = 1/2, so f is not one-one.
2. For x ∈ R, 1/(x² + 1) takes every value in (0, 1], which is the co-domain, so f is onto.

Teacher's Note:
1. Even functions are not one-one.
2. 1/(x² + 1) has maximum 1 at x = 0 and tends to 0.

Answer: (d) 10

Explanation:
1. A one-one function from a set with 4 elements to B uses ⁿP₄ ways, where n = n(B).
2. ⁿP₄ = 5040 = 10 × 9 × 8 × 7.
3. So n = 10.

Teacher's Note:
1. Count injective maps as n(n − 1)(n − 2)(n − 3).
2. Factorise 5040.

Answer: (b) 8

Explanation:
1. Each of the 3 elements of A can be sent to either of the 2 elements of B.
2. Number of functions = 2³ = 8.

Teacher's Note:
1. Number of functions from A to B is n(B)^n(A).
2. Not n(A)^n(B).

Answer: (a) 21

Explanation:
1. Total functions from A to A = 3³ = 27.
2. Bijections = 3! = 6.
3. Non-bijective mappings = 27 − 6 = 21.

Teacher's Note:
1. For finite sets of the same size, one-one is the same as onto.
2. Subtract bijections from all functions.

Answer: (c) ℓ is 27

Explanation:
1. Functions from A to B: 3³ = 27, so ℓ = 27.
2. One-one functions: 3! = 6, so m = 6.
3. Only 'ℓ is 27' is correct.

Teacher's Note:
1. Compute both counts before reading the options.
2. One-one functions between two 3-element sets are bijections.

Answer: (a) −1

Explanation:
1. π is irrational, so g(π) = 0 and f(0) = 0, giving (f∘g)(π) = 0.
2. e is irrational, so f(e) = 1, and g(1) = −1 because 1 is rational, giving (g∘f)(e) = −1.
3. Sum = 0 + (−1) = −1.

Teacher's Note:
1. Decide rational or irrational at each step of the composition.
2. Work from the inside out.

Answer: (a) x

Explanation:
1. f(f(x)) = (a − (f(x))ⁿ)^(1/n).
2. (f(x))ⁿ = a − xⁿ, so a − (f(x))ⁿ = xⁿ.
3. Hence f∘f(x) = (xⁿ)^(1/n) = x.

Teacher's Note:
1. Substitute f(x) into itself and simplify the inner power.
2. The expression returns x.