JEE Mathematics Functions MCQs Set 05

Answer: (b) N, the set of natural numbers

Explanation:
1. f is defined only for non-integer x, where \( \{x\}\in(0,1) \) and \( \sin\{x\}\in(0,\sin1) \).
2. So \( \frac1{\sin\{x\}}\in\left(\frac1{\sin1},\infty\right)\approx(1.19,\infty) \) and every value in it is taken.
3. The greatest integer of such numbers takes every integer value 1, 2, 3, ..., so the range is N.

Teacher's Note:
1. Find the range of {x} first, then of 1/sin{x}.
2. Since 1/sin 1 > 1, the floor starts from 1, not 0.

Answer: (a) 0

Explanation:
1. Replace x by −x: \( x^2f(x)-2f\left(\frac1x\right)=g(-x)=-g(x) \) because f is even. Comparing with the original, g(x) = −g(x), so g ≡ 0.
2. Hence \( x^2f(x)=2f\left(\frac1x\right) \). At x = 5: \( 25f(5)=2f\left(\frac15\right) \); at x = 1/5: \( \frac1{25}f\left(\frac15\right)=2f(5) \).
3. So \( f\left(\frac15\right)=50f(5) \) and \( 25f(5)=100f(5) \), giving f(5) = 0.

Teacher's Note:
1. Use evenness of f and oddness of g to show that g is identically zero.
2. Apply the resulting relation at x and 1/x.

Answer: (a) I and II quadrant

Explanation:
1. f is an absolute value, so f(x) ≥ 0 for every x in its domain.
2. Its graph therefore lies in the upper half-plane. It is also even in x, so it occupies both the first and second quadrants.

Teacher's Note:
1. A modulus is never negative, so the graph lies above or on the x-axis.
2. An even function is symmetric about the y-axis.

Answer: (d) none of these

Explanation:
1. For x ≥ 0, \( y=-\frac{x^2}{1+x^2}\in(-1,0] \), so \( x^2=-\frac{y}{1+y} \) and \( x=\sqrt{\frac{|y|}{1-|y|}} \) for y ≤ 0.
2. For x < 0, y > 0 and \( x=-\sqrt{\frac{y}{1-y}} \).
3. So \( f^{-1}(x)=-\operatorname{sgn}(x)\sqrt{\frac{|x|}{1-|x|}} \), which differs by a sign from (a) and (b); (c) is only valid for x > 0 with the wrong form.

Teacher's Note:
1. Treat x ≥ 0 and x < 0 separately and solve for x in terms of y.
2. The inverse has the sign opposite to its argument.

Answer: (d) none of the above

Explanation:
1. The first term has period \( 2\,n! \) and the second has period \( 2\,(n+1)! \).
2. Since \( 2(n+1)!=(n+1)\cdot2\,n! \), the common period is \( 2(n+1)! \).
3. This is not 2(n!) or (n + 1), so (d) is correct.

Teacher's Note:
1. Periods of the two parts: 2π ÷ (π/n!) and 2π ÷ (π/(n+1)!).
2. The LCM is the larger one here because it is a multiple of the smaller.

Answer: (b) −1

Explanation:
1. \( g(x)=\frac1{1-\frac1{1-x}}=\frac{x-1}{x} \).
2. \( h(x)=\frac1{1-\frac{x-1}{x}}=x \).
3. \( f\,g\,h=\frac1{1-x}\cdot\frac{x-1}x\cdot x=\frac{x-1}{1-x}=-1 \).

Teacher's Note:
1. Compute the nested functions one at a time.
2. f composed three times returns x.

Answer: (b) \( \left[-\frac13,1\right] \)

Explanation:
1. Write \( yx^2-(3y+1)x+(3y+1)=0 \). For real x the discriminant must be non-negative:
2. D = \( (3y+1)^2-4y(3y+1)=(3y+1)(1-y)\ge0 \).
3. This gives \( -\frac13\le y\le1 \).

Teacher's Note:
1. The range of a rational function is found by requiring a real solution x for each y.
2. The denominator x² − 3x + 3 has no real zero, so the function is defined for all x.

Answer: (a) 15

Explanation:
1. Put x = 1: \( f(y)=\frac{f(1)}y \), so \( f(x)=\frac kx \).
2. f(30) = 20 gives k = 600, so \( f(40)=\frac{600}{40}=15 \).

Teacher's Note:
1. Put x = 1 to discover the form of f.
2. Then use the given value to find the constant.

Answer: (a) \( \frac{26}5 \)

Explanation:
1. If x ≥ 0 the first equation gives y = 10 and the second gives x + 20 = 12, x = −8: contradiction. So x < 0 and the first equation becomes \( -2x+y=10 \).
2. If y < 0 the second equation gives x = 12, impossible, so y ≥ 0 and it becomes \( x+2y=12 \).
3. Solve: \( x=-\frac85,\ y=\frac{34}5 \), so \( x+y=\frac{26}5 \).

Teacher's Note:
1. Handle each modulus by cases on the sign of x and y.
2. Check that the solution satisfies the assumed signs.

Answer: (d) \( \left(-\frac13,0\right)\cup\left(\frac13,1\right)\cup(1,\infty) \)

Explanation:
1. With n = [x] and t = {x}, the base is \( b=n+3t \). Since \( n^2-5n+7>0 \) always and \( n^2-5n+6=(n-2)(n-3)\ge0 \) for every integer n, the argument is always ≥ 1.
2. So the logarithm is non-negative when b > 1. (If 0 < b < 1 we would need n = 2 or 3, but then b ≥ 2, impossible.)
3. b > 1: n ≥ 2 always; n = 1 needs t > 0; n = 0 needs t > 1/3; n = −1 needs t > 2/3; n ≤ −2 impossible. This gives \( \left(-\frac13,0\right)\cup\left(\frac13,1\right)\cup(1,\infty) \).

Teacher's Note:
1. The argument of the logarithm is always at least 1, so the base must exceed 1.
2. Take each integer part n separately.

Answer: (a) 2

Explanation:
1. By AM-GM, \( x^n+y^n+(n-2)\cdot1\ge n\sqrt[n]{x^ny^n}=n|xy|\ge nxy \).
2. So g(x, y) ≥ 0, and g = 0 only when \( |x|=|y|=1 \) and xy > 0.
3. Hence the solutions are (1, 1) and (−1, −1): exactly 2.

Teacher's Note:
1. Apply AM-GM to x^n, y^n and (n − 2) ones.
2. Equality requires every term to be equal to 1 and xy to be positive.

Answer: (d) \( \left[-\frac13,\frac13\right) \)

Explanation:
1. Let [x] = [3x] = k. For k = 0: 0 ≤ x < 1 and 0 ≤ 3x < 1, so \( 0\le x<\frac13 \).
2. For k = −1: −1 ≤ x < 0 and −1 ≤ 3x < 0, so \( -\frac13\le x<0 \). For |k| ≥ 1 otherwise there is no solution since [3x] ≥ 3[x] ≠ [x].
3. The union is \( \left[-\frac13,\frac13\right) \).

Teacher's Note:
1. Set both greatest integers equal to k and solve each inequality.
2. Check k = 0 and k = −1; other values of k fail.

Answer: (d) \( \{1,2\} \)

Explanation:
1. On (0, π/3): sec x ∈ (1, 2), so [sec x] = 1. tan x ∈ (0, √3), so \( [\tan x+1]\in\{1,2\} \).
2. cos x ∈ (1/2, 1), so \( [\cos x+1]=1 \) and \( [\cos x+2]=2 \).
3. sin x ∈ (0, √3/2): \( [\sin x+1]=1 \) and \( [\sin x+2]=2 \). Both cases occur, so the range is {1, 2}.

Teacher's Note:
1. Work from the innermost greatest integer outwards.
2. tan x + 1 crosses 2 at x = π/4, so two cases arise.

Answer: (c) A is true, but R is false.

Explanation:
1. \( f(x)=x^2+\tan^{-1}x \) is increasing for x > 0 and unbounded, hence non-periodic: A is true.
2. R is false: for example, x and −x + sin x are non-periodic but their sum sin x is periodic.

Teacher's Note:
1. A periodic continuous function must be bounded; x² + tan⁻¹ x is not.
2. One counter-example disproves 'always'.