JEE Mathematics Functions MCQs Set 16

Answer: (c) \( 0 < f(x) < 1 \)

Explanation:
1. For \( x>5 \) all four factors are positive, so \( f(x)>0 \).
2. \( f(x)-1=\dfrac{(x^2-6x+5)-(x^2-5x+6)}{(x-2)(x-3)}=\dfrac{-(x+1)}{(x-2)(x-3)} < 0 \).
3. So \( 0 < f(x) < 1 \).

Teacher's Note:
1. Compare f(x) with 1 by looking at f(x) - 1.
2. The numerator of f(x) - 1 simplifies to -(x + 1).

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. \( f(1)=2+7-5=4 \), so \( f^{-1}(4)=1 \) provided f has an inverse.
2. \( f'(x)=6x^2+7>0 \), so f is strictly increasing (one-one), and as a cubic it is onto R; so f is invertible.
3. The assertion is true, and it holds because f is one-one and onto, which is the content of the reason.

Teacher's Note:
1. Check that the function is invertible before using the inverse.
2. Then find the x with f(x) = 4.

Answer: (a) 5

Explanation:
1. Put \( t=\dfrac{1-x}{1+x} \); the equation says \( f(t)f\left(\tfrac1t\right)=f(t)+f\left(\tfrac1t\right) \), which holds for \( f(t)=1+t^k \).
2. \( f(3)=28 \) gives \( 3^k=27 \), so \( k=3 \) and \( f(t)=1+t^3 \).
3. \( \sum_{n=1}^{10}(f(n)-1)=\sum n^3=\left(\dfrac{10\cdot11}2\right)^2=3025 \).
4. \( \dfrac{3025}{605}=5 \).

Teacher's Note:
1. The relation is typical for f(t) = 1 + t^k.
2. Then use the formula for the sum of cubes.

Answer: (a) 1

Explanation:
1. The left side is \( (f(x)-1)^3 \).
2. So \( (f(x)-1)^3=x^6 \), giving \( f(x)-1=x^2 \) and \( f(x)=1+x^2 \).
3. \( f(0)=1 \).

Teacher's Note:
1. Recognise the expansion of (f - 1)^3.
2. Take the real cube root of both sides.

Answer: (c) x = 0 or x = 5/3

Explanation:
1. With \( n=[x] \) and \( \{x\}=x-n \): \( 4(x-n)=x+n \), so \( 3x=5n \) and \( x=\dfrac{5n}3 \).
2. We need \( n\le\dfrac{5n}3 < n+1 \): \( n\ge0 \) and \( \dfrac{2n}3 < 1 \Rightarrow n < \dfrac32 \).
3. So \( n=0 \) or \( n=1 \), giving \( x=0 \) or \( x=\dfrac53 \).
4. Check: \( x=\tfrac53 \): \( 4\cdot\tfrac23=\tfrac83=\tfrac53+1 \).

Teacher's Note:
1. Substitute x = n + fractional part and solve for x.
2. Then confirm that x really has integral part n.

Answer: (d) \( \mathbb R-\{4+2\sqrt5,\,4-2\sqrt5\} \) and \( \left(-\infty,-\tfrac14\right]\cup\left[-\tfrac1{20},\infty\right) \)

Explanation:
1. The denominator vanishes when \( x^2-8x-4=0 \), i.e. \( x=4\pm2\sqrt5 \). So the domain is \( \mathbb R-\{4\pm2\sqrt5\} \).
2. Put \( y=f(x) \): \( yx^2-(8y+1)x-(4y+2)=0 \). For \( y=0 \) we get \( x=-2 \), which is allowed. For \( y\ne0 \) the discriminant must be non-negative: \( (8y+1)^2+4y(4y+2)=80y^2+24y+1\ge0 \).
3. The roots of \( 80y^2+24y+1 \) are \( -\tfrac14 \) and \( -\tfrac1{20} \), so \( y\le-\tfrac14 \) or \( y\ge-\tfrac1{20} \) (and \( y=0 \) is included).
4. Range: \( \left(-\infty,-\tfrac14\right]\cup\left[-\tfrac1{20},\infty\right) \).

Teacher's Note:
1. Treat y = f(x) as a quadratic in x and demand a real root.
2. Treat y = 0 separately because the equation becomes linear.

Answer: (d) 1

Explanation:
1. A continuous function that takes only rational values cannot change value (between two rationals it would have to take irrational values), so f is a constant \( c\ne0 \).
2. The equation becomes \( c(x^2-6x+9)=0 \), i.e. \( (x-3)^2=0 \).
3. It has the single root \( x=3 \).

Teacher's Note:
1. Use the intermediate value property to conclude that f is constant.
2. Then divide by the non-zero constant.

Answer: (b) \( 2^{11} \)

Explanation:
1. The sum is odd exactly when an odd number of the six images are odd (7 or 9).
2. Each image is odd in 2 ways and even in 2 ways.
3. Number of functions = \( \sum_{k\ \text{odd}}\binom6k\,2^k2^{6-k}=2^6\left(\binom61+\binom63+\binom65\right)=2^6\cdot2^5=2^{11} \).

Teacher's Note:
1. Count by the parity pattern of the images.
2. The odd-index binomial coefficients add up to half of \( 2^6 \).

Answer: (b) f is one-one and decreasing in \( (0,\infty) \)

Explanation:
1. \( f'(x)=\dfrac{-12x^2}{\left(e^{x^3}-e^{-x^3}\right)^2} < 0 \) for every \( x\ne0 \), so f is decreasing in \( (-\infty,0) \) and in \( (0,\infty) \).
2. For \( x>0 \) we have \( f(x)>1 \) and for \( x < 0 \) we have \( f(x) < -1 \), so the two pieces never share a value and f is one-one.
3. \( f(-x)=-f(x) \), so f is odd, not even.

Teacher's Note:
1. Differentiate and look at the sign.
2. Compare the ranges of the two branches to settle one-one.

Answer: (d) \( -\tfrac12 \)

Explanation:
1. \( \dfrac{x^2-a}{x^2+1}=1-\dfrac{a+1}{x^2+1} \) and \( \dfrac1{x^2+1}\in(0,1] \).
2. If \( a+1>0 \) the expression takes every value in \( [-a,1) \).
3. For f to be onto \( \left[\tfrac\pi6,\tfrac\pi2\right) \) the expression must take every value in \( \left[\tfrac12,1\right) \) and no other, so \( -a=\tfrac12 \).
4. Hence \( a=-\tfrac12 \).

Teacher's Note:
1. Find the range of the inner expression in terms of a.
2. Match it with the interval that sin of the required range gives.

Answer: (c) \( 1+\sqrt{x+2} \)

Explanation:
1. For the branch \( x\ge1 \) we have \( f:[1,\infty)\to[-2,\infty) \), \( f(x)=(x-1)^2-2 \).
2. Put \( y=(x-1)^2-2 \): \( x-1=\pm\sqrt{y+2} \), and \( x\ge1 \) selects the plus sign.
3. So \( f^{-1}(x)=1+\sqrt{x+2} \).

Teacher's Note:
1. To invert a quadratic on a restricted domain, pick the root that lies in that domain.
2. Here the domain is x >= 1, so the positive square root is used.

Answer: (b) 5

Explanation:
1. For \( x\in(\sqrt2,\sqrt3) \): \( \left\{x^2\right\}=x^2-2 \) and \( \left\{\tfrac1x\right\}=\tfrac1x \).
2. So \( \tfrac1x=x^2-2 \), i.e. \( x^3-2x-1=0=(x+1)(x^2-x-1) \). The root in the interval is \( x=\tfrac{1+\sqrt5}{2} \), with \( x^2=x+1 \).
3. \( x^4=(x+1)^2=3x+2 \) and \( \tfrac3x=3(x^2-2)=3x-3 \).
4. Hence \( x^4-\tfrac3x=(3x+2)-(3x-3)=5 \).

Teacher's Note:
1. Replace the fractional parts using the interval, then reduce powers with x^2 = x + 1.
2. The answer does not depend on solving for x numerically.

Answer: (c) \( [-5,-4]\cup[1,2] \)

Explanation:
1. We need \( 1\le\log_2(x^2+3x-2)\le3 \), i.e. \( 2\le x^2+3x-2\le8 \).
2. \( x^2+3x-4\ge0 \) gives \( x\le-4 \) or \( x\ge1 \).
3. \( x^2+3x-10\le0 \) gives \( -5\le x\le2 \).
4. Intersection: \( [-5,-4]\cup[1,2] \).

Teacher's Note:
1. The inner expression must lie in the domain of the outer function.
2. Solve the two quadratic inequalities and intersect.

Answer: (a) 1

Explanation:
1. Put \( x=y=1 \): \( f(1)^2+f(3)^2=2f(1) \), so \( (f(1)-1)^2=0 \) and \( f(1)=1 \).
2. Put \( y=1 \): \( f(x)+f\left(\tfrac3x\right)=2f(x) \), so \( f\left(\tfrac3x\right)=f(x) \).
3. The equation now reads \( 2f(x)f(y)=2f(xy) \), so f is multiplicative and \( f(x)=f(\sqrt x)^2\ge0 \).
4. Also \( f(x)f\left(\tfrac3x\right)=f(3)=1 \) gives \( f(x)^2=1 \), so \( f(x)=1 \) for all \( x>0 \).
5. Hence \( f(97)=1 \).

Teacher's Note:
1. Use the special values x = y = 1 and y = 1 first.
2. Positivity of a square (f(x) = f(sqrt x)^2) removes the -1 possibility.

Answer: (d) \( (-\infty,0]\cup[5,\infty) \)

Explanation:
1. \( f^2-f-6\ge0 \Rightarrow (f-3)(f+2)\ge0 \Rightarrow f\le-2 \) or \( f\ge3 \).
2. \( f\ge3 \) is the complement of \( f < 3 \), so \( x\in(-\infty,0] \).
3. \( f\le-2 \) is the complement of \( f>-2 \), so \( x\in[5,\infty) \).
4. Answer: \( (-\infty,0]\cup[5,\infty) \).

Teacher's Note:
1. Factorise the quadratic in f(x).
2. Use complements of the given solution sets.

Answer: (a) -17

Explanation:
1. \( f(x)+5=ax^7+bx^3+cx \) is an odd function.
2. So \( f(-x)+5=-(f(x)+5) \), i.e. \( f(-x)+f(x)=-10 \).
3. With \( x=7 \): \( 7+f(7)=-10 \), so \( f(7)=-17 \).

Teacher's Note:
1. Remove the constant to expose an odd function.
2. Odd functions add up to zero at x and -x.

Answer: (a) 1

Explanation:
1. For integer coefficients, \( f(m)-f(n)=a_1(m-n)+a_2(m^2-n^2)+\cdots+a_k(m^k-n^k) \).
2. Every term is divisible by \( m-n \), so \( (m-n) \) divides 1.
3. Therefore \( m-n=\pm1 \) and \( |m-n|=1 \).

Teacher's Note:
1. Use the fact that m - n divides m^r - n^r.
2. An integer dividing 1 can only be 1 or -1.

Answer: (a) 1

Explanation:
1. \( \cos(xy)=x \) with integer x forces \( x\in\{-1,0,1\} \).
2. \( x=0 \): \( \cos0=1\ne0 \), rejected.
3. \( x=-1 \): \( \cos(-y)=-1 \) needs \( y=(2k+1)\pi \), not an integer.
4. \( x=1 \): \( \cos y=1 \) and \( \tan y=y \) give \( \sin y=y\cos y=y \), which for a real y holds only at \( y=0 \); then \( \cos0=1 \) works.
5. The only pair is \( (1,0) \).

Teacher's Note:
1. Use the integer condition to limit x to -1, 0, 1.
2. Then test each value.

Answer: (c) 8

Explanation:
1. The graph of f is an arch with vertex \( (1,2) \), mapping \( [0,2] \) onto \( [0,2] \).
2. \( f(f(f(x))) \) consists of \( 2^{3-1}=4 \) equal arches, each rising from 0 to 2 and falling back to 0.
3. The line \( y=\tfrac x2 \) stays between 0 and 1 on \( [0,2] \), so each arch crosses it twice.
4. The first arch starts on the line at the origin, so the origin is one of its two crossings.
5. Total: \( 4\times2=8 \) solutions.

Teacher's Note:
1. Think of the iterates as repeated arches.
2. Count intersections with the line, or confirm numerically.

Answer: (b) 13

Explanation:
1. \( \{kx\}=kx-[kx] \), and \( \sum_{k=1}^{12}kx=78x \).
2. So the equation reduces to \( [x]+[2x]+\cdots+[12x]=0 \).
3. All terms are non-negative, so each is 0; the largest, \( [12x] \), is 0 only for \( 12x < 1 \), i.e. \( x < \tfrac1{12} \).
4. With \( x=\tfrac1n \): \( n>12 \) and \( \tfrac1n\ge\tfrac1{25} \) gives \( n\le25 \). So \( n=13,14,\dots,25 \): 13 values.

Teacher's Note:
1. Cancel the linear parts to get a sum of floors equal to zero.
2. Remember x = 1/12 itself fails because [12x] = 1.

Answer: (a) \( f(a+x)=f(a-x) \)

Explanation:
1. Points at equal distances x on both sides of the line \( x=a \) are \( a+x \) and \( a-x \).
2. Symmetry about the line means the function takes the same value at both points: \( f(a+x)=f(a-x) \).

Teacher's Note:
1. Reflect the point a + x in the vertical line x = a to get a - x.
2. The even function f(x) = f(-x) is the special case a = 0.

Answer: (c) \( x=a \)

Explanation:
1. Put \( x=a+t \). Then \( f(a+t)=f(2a-a-t)=f(a-t) \).
2. So the values at \( a+t \) and \( a-t \) are equal for every t, which means the graph is symmetric about the line \( x=a \).

Teacher's Note:
1. Replace x by a + t to see the pairing of points.
2. The line of symmetry is the midpoint of x and 2a - x.

Answer: (b) one-one and into

Explanation:
1. Let \( g(x)=x^3-3x+2 \). \( g'(x)=3(x^2-1)\ge0 \) for \( x\le-1 \), so g is increasing and f is one-one.
2. As \( x\to-\infty \), \( g\to-\infty \) and \( f\to0 \). At \( x=-1 \), \( g=-1+3+2=4 \), so \( f(-1)=e^4 \).
3. The range is \( (0,e^4] \), which is smaller than the co-domain \( (0,e^5] \), so f is into.

Teacher's Note:
1. Monotonic functions are one-one.
2. Compare the actual range with the stated co-domain.

Answer: (a) 3

Explanation:
1. For \( x < -1 \): \( f(x)=-3x \). For \( -1\le x\le1 \): \( f(x)=x+4 \). For \( x>1 \): \( f(x)=5x \).
2. f decreases from \( +\infty \) down to \( f(-1)=3 \) and then increases (it is \( x+4 \) on \( [-1,1] \) and \( 5x \) beyond).
3. The minimum value 3 is reached only at \( x=-1 \), so \( f(x)=3 \) has exactly one solution. For \( k>3 \) there are two solutions.

Teacher's Note:
1. Remove the modulus signs piece by piece.
2. The line y = k meets a V-shaped graph once only at the lowest point.