JEE Mathematics Functions MCQs Set 10

Answer: (c) \( \log_e\left(x+\sqrt{x^2-1}\right) \)

Explanation:
1. Let y = (eˣ + e⁻ˣ)/2 with x ≥ 0. Then e^(2x) − 2y eˣ + 1 = 0 so eˣ = y ± √(y² − 1).
2. For x ≥ 0, eˣ ≥ 1, so eˣ = y + √(y² − 1).
3. x = ln(y + √(y² − 1)), so f⁻¹(x) = ln(x + √(x² − 1)).

Teacher's Note:
1. Solve the quadratic in eˣ and choose the root giving x ≥ 0.
2. This is cosh⁻¹.

Answer: (b) \( 6+\sqrt[4]{3-\sin^3x} \)

Explanation:
1. Let y = f(x): sin y = (3 − (x − 6)⁴)^(1/3).
2. Cube both sides: sin³y = 3 − (x − 6)⁴, so (x − 6)⁴ = 3 − sin³y.
3. With x ≥ 6, x = 6 + (3 − sin³y)^(1/4).

Teacher's Note:
1. Invert step by step.
2. The restriction x ≥ 6 selects the positive fourth root.

Answer: (d) not defined

Explanation:
1. f(2) is defined by both x² (giving 4) and 3x (giving 6), so f assigns two different values to x = 2.
2. A function must have exactly one image, so f(2) is not defined.

Teacher's Note:
1. The two pieces overlap at x = 2 with different values.
2. Branches should meet only at one end with a single rule.

Answer: (b) a = 2, b = −3

Explanation:
1. f(−1) = −a + b = −5 and f(3) = 3a + b = 3.
2. Subtract: 4a = 8, a = 2.
3. Then b = −5 + a = −3.

Teacher's Note:
1. Two conditions fix the two constants.
2. Both are integers, as required.

Answer: (b) {0, 3, 8}

Explanation:
1. f² means the square of each value: 3², 2², 1² = 9, 4, 1.
2. f² − 1 gives 8, 3, 0.
3. The range is {0, 3, 8}.

Teacher's Note:
1. The range of a finite function is the set of outputs.
2. Do not confuse f² with f∘f.

Answer: (d) \( \frac12 \)

Explanation:
1. f is a constant function with value 1/2.
2. So f(−2/5) = 1/2.

Teacher's Note:
1. Constant functions ignore the input.

Answer: (c) 0

Explanation:
1. √(π/3) ≈ 1.02, so [√(π/3)] = 1.
2. f = sin(π/3 − (π/3)) = sin 0 = 0.

Teacher's Note:
1. Evaluate the greatest integer first.
2. (√(π/3))² = π/3.

Answer: (d) \( \{(a,5),(b,4),(c,10)\} \)

Explanation:
1. f² + g² adds the squares of the second components.
2. a: 1² + (−2)² = 5. b: (−2)² + 0² = 4. c: 3² + 1² = 10.
3. So f² + g² = {(a, 5), (b, 4), (c, 10)}.

Teacher's Note:
1. Square the outputs first, then add.
2. The first components stay the same.

Answer: (c) h, f

Explanation:
1. f: each of 2, 3, 4 has exactly one image, so f is a function.
2. g: 2 has two images (2 and 4), so g is not a function.
3. h: each of 1 to 5 has one image, so h is a function.

Teacher's Note:
1. A relation is a function when no input has two outputs.
2. Check every repeated first component.

Answer: (d) d = 0

Explanation:
1. For an odd function f(−x) = −f(x).
2. ax⁵, bx³ and cx are odd for any a, b, c.
3. The constant d is even, so f is odd only if d = 0.

Teacher's Note:
1. Sum of odd and even parts: the even part must vanish.
2. The constant term is the only even term.

Answer: (a) R

Explanation:
1. [x] is defined for all real x, and x is defined for all real x.
2. So [x] + x is defined on R.

Teacher's Note:
1. Sums of functions with domain R have domain R.
2. The greatest integer function has no restriction.

Answer: (b) R

Explanation:
1. |x − 1| is defined for every real x.

Teacher's Note:
1. The absolute value of any real number exists.
2. Its range is [0, ∞).

Answer: (c) \( R-\{3n\pi:n\in Z\} \)

Explanation:
1. cot θ is undefined when θ = nπ.
2. Here θ = x/3, so x/3 ≠ nπ, i.e. x ≠ 3nπ.
3. The domain is R − {3nπ : n ∈ Z}.

Teacher's Note:
1. Substitute the argument in the condition for cot.
2. Do not forget the factor 3.

Answer: (b) \( \left[\frac{71}{12},\infty\right) \)

Explanation:
1. 3x² + 7x + 10 has minimum at x = −7/6.
2. Minimum value = 10 − 49/12 = 71/12.
3. The parabola opens upward, so the range is [71/12, ∞).

Teacher's Note:
1. Minimum of ax² + bx + c is c − b²/4a.
2. Here 10 − 49/12 = 71/12.

Answer: (a) \( (0,\infty) \)

Explanation:
1. e^x takes all positive values and never reaches 0.
2. So its range is (0, ∞).

Teacher's Note:
1. The exponential function is always positive.
2. It is one-one and onto (0, ∞).

Answer: (b) [−1, 3]

Explanation:
1. sin x − √3 cos x = 2 sin(x − π/3), which takes values in [−2, 2].
2. Adding 1 gives [−1, 3].
3. For f to be onto, S must be [−1, 3].

Teacher's Note:
1. Use the R-formula a sin x + b cos x = R sin(x + α).
2. Here R = 2.

Answer: (c) bijective

Explanation:
1. f(x) = ax + b with a ≠ 0 is strictly monotonic, hence one-one.
2. For any y ∈ R, x = (y − b)/a is a real pre-image, so f is onto.
3. So f is bijective.

Teacher's Note:
1. A non-constant linear function is always a bijection on R.
2. Solve y = ax + b for x.

Answer: (d) not a bijection

Explanation:
1. f(1) = f(−1) = 1, so f is not one-one.
2. The values of x² are non-negative perfect squares, so 2 is not attained and f is not onto.
3. So f is not a bijection.

Teacher's Note:
1. One counter-example to injectivity is enough.
2. Z → Z squares skip many integers.

Answer: (d) identity function

Explanation:
1. For integer x, [x] = x.
2. So f(x) = x, which is the identity function on Z.

Teacher's Note:
1. The greatest integer of an integer is the integer itself.

Answer: (b) onto but not one-one

Explanation:
1. f(x) = x² on (0, ∞) is one-one because x > 0.
2. The range is (0, ∞), but the co-domain is [0, ∞) which also contains 0.
3. 0 has no pre-image in (0, ∞), so f is not onto.

Teacher's Note:
1. Check whether 0 is attained.
2. The domain is open at 0.

Answer: (a) one-one but not onto

Explanation:
1. f(x) = 2x/(1 + x) = 2 − 2/(1 + x), which is strictly increasing for x ≥ 0.
2. So f is one-one.
3. The range is [0, 2), so the value 2 is not attained and f is not onto [0, 2].

Teacher's Note:
1. Rewrite as 2 − 2/(1 + x) to see the monotonic behaviour.
2. The supremum 2 is never reached.

Answer: (c) 4

Explanation:
1. A function from A to A has n(A)^n(A) choices.
2. nⁿ = 256 = 4⁴.
3. So n(A) = 4.

Teacher's Note:
1. The total number of functions from A to A is nⁿ.
2. 256 = 4⁴.

Answer: (c) \( {}^5P_3 \)

Explanation:
1. One-one functions from a 3-element set into a 5-element set: 5 choices for the first, 4 for the second, 3 for the third.
2. Number = 5 × 4 × 3 = ⁵P₃.

Teacher's Note:
1. Injective maps from A into B number ⁿ(ᴮ)P_n(ᴬ).
2. The count is not 5³ or 3⁵.

Answer: (c) 5

Explanation:
1. Total functions from A to B: 5² = 25.
2. One-one functions: ⁵P₂ = 20.
3. Many-one functions = 25 − 20 = 5.

Teacher's Note:
1. Every function is either one-one or many-one.
2. Subtract the injective maps from all maps.