JEE Mathematics Functions MCQs Set 15

Answer: (d) one-one and onto

Explanation:
1. \( (f-g)(x)=x \) for rational x and \( -x \) for irrational x.
2. One-one: rational x map to rational values and irrational x map to irrational values, and within each class the map is injective.
3. Onto: a rational r is the image of r, and an irrational s is the image of -s.
4. So f - g is one-one and onto.

Teacher's Note:
1. Check what happens to rational and irrational inputs separately.
2. Rational values come from rational inputs and irrational values from irrational inputs.

Answer: (b) \( f\left(\tfrac\pi2\right)=-1 \)

Explanation:
1. \( \pi^2\approx9.87 \), so \( [\pi^2]=9 \) and \( [-\pi^2]=-10 \).
2. \( f(x)=\cos9x+\cos10x \).
3. \( f\left(\tfrac\pi2\right)=\cos\tfrac{9\pi}2+\cos5\pi=0-1=-1 \).
4. \( f(\pm\pi)=\cos9\pi+\cos10\pi=-1+1=0 \), and \( f\left(\tfrac\pi4\right)=\cos\tfrac{9\pi}4+\cos\tfrac{5\pi}2=\tfrac1{\sqrt2} \). Only the first statement is true.

Teacher's Note:
1. Work out the two greatest integers first: 9 and -10.
2. Then substitute each value of x.

Answer: (a) (3, 4)

Explanation:
1. The quadratic \( x^2-7x+13 \) has negative discriminant, so it is always positive.
2. For the root in the denominator, \( \log_{1/2}\left(x^2-7x+13\right)>0 \), and as the base is less than 1 this means \( x^2-7x+13 < 1 \).
3. \( x^2-7x+12 < 0 \) gives \( (x-3)(x-4) < 0 \), so \( 3 < x < 4 \).

Teacher's Note:
1. A logarithm to a base below 1 is positive for arguments below 1.
2. The denominator cannot be zero, so the inequality is strict.

Answer: (a) \( \left(0,10^{-1/2}\right)-\{10^{-2}\} \)

Explanation:
1. \( \log_{10}x \) needs \( x>0 \), and then \( \dfrac{2\log_{10}x+1}{-x}>0 \) needs \( 2\log_{10}x+1 < 0 \), i.e. \( x < 10^{-1/2} \).
2. The base \( 100x \) must be positive and not 1, so \( x\ne\tfrac1{100}=10^{-2} \).
3. Domain: \( \left(0,10^{-1/2}\right)-\{10^{-2}\} \).

Teacher's Note:
1. List the conditions: positive argument, positive base, base not equal to 1.
2. Do not forget to remove the point where the base equals 1.

Answer: (a) Domain is R and range is \( \left[\log_{10}\tfrac{11}3,\infty\right) \)

Explanation:
1. The discriminant of \( 3x^2-4x+5 \) is \( 16-60 < 0 \), so the quadratic is always positive and the domain is R.
2. Its minimum is at \( x=\tfrac23 \): \( 3\cdot\tfrac49-\tfrac83+5=\tfrac{11}3 \).
3. The quadratic takes every value from \( \tfrac{11}3 \) to infinity, so the range of f is \( \left[\log_{10}\tfrac{11}3,\infty\right) \).

Teacher's Note:
1. Use the discriminant to see that the argument is always positive.
2. The range follows from the minimum of the quadratic.

Answer: (d) \( \left(1,\tfrac73\right] \)

Explanation:
1. \( f(x)=1+\dfrac1{x^2+x+1} \).
2. \( x^2+x+1=\left(x+\tfrac12\right)^2+\tfrac34\in\left[\tfrac34,\infty\right) \).
3. So \( \dfrac1{x^2+x+1}\in\left(0,\tfrac43\right] \) and \( f\in\left(1,\tfrac73\right] \).

Teacher's Note:
1. Split the fraction as 1 + a simpler fraction.
2. Find the minimum of the denominator by completing the square.

Answer: (b) Domain \( [0,1] \), range \( \left[0,\tfrac\pi2\right] \)

Explanation:
1. With \( x=\tan\theta \): \( \dfrac{2x}{1+x^2}=\sin2\theta \) and \( \dfrac{1-x^2}{1+x^2}=\cos2\theta \).
2. \( g(x)=\cos^{-1}(\cos2\theta)=2|\theta| \) for \( |\theta| < \tfrac\pi2 \).
3. \( f(x)=\sin^{-1}(\sin2\theta)=2\theta \) only for \( |\theta|\le\tfrac\pi4 \), i.e. \( |x|\le1 \).
4. f = g needs \( \theta\ge0 \), so \( x\in[0,1] \), and then the common value \( 2\tan^{-1}x \) ranges over \( \left[0,\tfrac\pi2\right] \).

Teacher's Note:
1. Substitute x = tan(theta) to simplify both functions.
2. Compare the two expressions on each sign of x.

Answer: (d) -1

Explanation:
1. \( f(f(x))=\dfrac{\alpha\cdot\frac{\alpha x}{x+1}}{\frac{\alpha x}{x+1}+1}=\dfrac{\alpha^2x}{(\alpha+1)x+1} \).
2. Setting this equal to x gives \( \alpha^2x=(\alpha+1)x^2+x \) for all x.
3. So \( \alpha+1=0 \) and \( \alpha^2=1 \), hence \( \alpha=-1 \).

Teacher's Note:
1. Compute the composite and compare coefficients.
2. Both conditions must hold for the identity to be valid for all x.

Answer: (c) \( f(f^{-1}(B))=B \) only if \( B\subseteq f(X) \)

Explanation:
1. \( f(f^{-1}(B)) \) consists of the images of those x with \( f(x)\in B \), so it equals \( B\cap f(X) \).
2. It equals B exactly when every element of B is an image, i.e. \( B\subseteq f(X) \).
3. Statements (a), (b) and (d) fail in general (for example, a constant function breaks them).

Teacher's Note:
1. Always compute f(f^-1(B)) as B intersected with the range of f.
2. One-one and onto properties decide when the other identities hold.

Answer: (b) \( \tfrac{11}3 \)

Explanation:
1. \( \{x+1\}=\{x\} \) and \( [x+1]=[x]+1 \), so the equation becomes \( x-[x]+2x=4[x]-2 \), i.e. \( x=\dfrac{5[x]-2}3 \).
2. With \( n=[x] \) we need \( n\le\dfrac{5n-2}3 < n+1 \), which gives \( n\ge1 \) and \( n < 2.5 \), so \( n=1,2 \).
3. \( n=1 \): \( x=1 \); \( n=2 \): \( x=\tfrac83 \).
4. Sum \( =1+\tfrac83=\tfrac{11}3 \).

Teacher's Note:
1. Replace the fractional part by x minus the integer part.
2. Then solve for x in terms of n = [x] and check consistency.

Answer: (a) an odd function

Explanation:
1. For \( 0 < |x| < 1 \): \( f(-x)=x^2\sin\left(-\tfrac\pi x\right)=-f(x) \).
2. For \( |x|\ge1 \): \( f(-x)=(-x)|-x|=-x|x|=-f(x) \).
3. Also \( f(0)=0 \). So \( f(-x)=-f(x) \) throughout and f is odd.

Teacher's Note:
1. Check each piece of the definition separately.
2. An odd function has f(0) = 0 if it is defined at 0.

Answer: (c) \( \left[-\tfrac\pi2,0\right] \)

Explanation:
1. \( \dfrac{1+\cos2x}2=\cos^2x \), so \( f(x)=|\cos x| \).
2. On \( \left[-\tfrac\pi2,0\right] \), \( \cos x\ge0 \) and increases from 0 to 1, so f is one-one.
3. On \( [0,\pi] \), \( [-\pi,\pi] \) and \( [-\pi,0] \), f takes some values twice (for example \( |\cos(\pi/3)|=|\cos(2\pi/3)| \)).

Teacher's Note:
1. Simplify the root to |cos x|.
2. A function is one-one on an interval where it is strictly monotonic.

Answer: (d) {-1, 0}

Explanation:
1. For an integer x, \( [x]+[-x]=x-x=0 \).
2. For a non-integer x, \( [-x]=-[x]-1 \), so the sum is \( -1 \).
3. The range is \( \{-1,0\} \), so \( B=\{-1,0\} \).

Teacher's Note:
1. Treat integers and non-integers separately.
2. The range of a step function is a finite set of values.

Answer: (c) \( \sqrt{2+\sqrt{4+\log_2x}} \)

Explanation:
1. Put \( y=2^{x^4-4x^2} \): \( \log_2y=x^4-4x^2=(x^2-2)^2-4 \).
2. So \( (x^2-2)^2=4+\log_2y \) and, since \( x\ge2 \) gives \( x^2-2\ge2>0 \), \( x^2=2+\sqrt{4+\log_2y} \).
3. Taking the positive root (x >= 2): \( x=\sqrt{2+\sqrt{4+\log_2y}} \).

Teacher's Note:
1. Take logarithms to base 2 and complete the square.
2. Choose the signs that keep x in the given domain.

Answer: (a) \( f\left(\dfrac xy\right) \)

Explanation:
1. The relation is \( f(xy)+f\left(\tfrac xy\right)=2f(x) \).
2. Putting \( x=1 \): \( f(y)+f\left(\tfrac1y\right)=0 \), so \( f\left(\tfrac yx\right)=-f\left(\tfrac xy\right) \).
3. Interchange x and y in the relation: \( f(xy)+f\left(\tfrac yx\right)=2f(y) \).
4. Subtracting the two relations: \( f\left(\tfrac xy\right)-f\left(\tfrac yx\right)=2f(x)-2f(y) \), i.e. \( 2f\left(\tfrac xy\right)=2(f(x)-f(y)) \).
5. Hence \( f(x)-f(y)=f\left(\tfrac xy\right) \)

Teacher's Note:
1. Swap the roles of x and y and subtract.
2. The relation f(1/y) = -f(y) comes from putting x = 1.

Answer: (b) \( \tfrac23 \)

Explanation:
1. From \( f(x)-f(y)=f\left(\tfrac xy\right) \) and the differentiability at 1, f is a logarithmic function; with \( f(1)=0 \) and \( f'(1)=2 \) we get \( f(x)=2\ln x \).
2. \( f'(x)=\dfrac2x \), so \( f'(3)=\tfrac23 \).

Teacher's Note:
1. Identify f first.
2. Then differentiate.

Answer: (d) 22

Explanation:
1. The graph is symmetric about \( x=2 \) and about \( x=10 \), so f has period \( 2(10-2)=16 \).
2. \( f(0)=5 \) gives \( f(4)=5 \) (reflection about x = 2) and then \( f(16k)=f(16k+4)=5 \).
3. In \( [0,170] \): \( x=0,16,\dots,160 \) (11 values) and \( x=4,20,\dots,164 \) (11 values).
4. Minimum number: \( 11+11=22 \).

Teacher's Note:
1. Two vertical symmetry lines imply a period equal to twice their distance.
2. Count the values coming from each of the two starting points, 0 and 4.

Answer: (a) 1

Explanation:
1. From the two symmetries, f has period 16, and therefore also 32 and 48.
2. If 1 were a period, then \( f(2)=f(2+4\cdot1)=f(6) \), contradicting \( f(2)\ne f(6) \).
3. So 1 cannot be a period.

Teacher's Note:
1. Multiples of a period are periods too.
2. Use the given inequality to rule out a very small period.

Answer: (a) \( x-2y-4=0 \)

Explanation:
1. \( \dfrac{x^2}4-ax=\dfrac14(x-2a)^2-a^2 \), so the vertex is at \( x=2a \) with \( y=-a^2+a^2+a-2=a-2 \).
2. Eliminate a: \( a=\dfrac x2 \), so \( y=\dfrac x2-2 \), i.e. \( x-2y-4=0 \).

Teacher's Note:
1. Complete the square to read off the vertex coordinates.
2. Express the two coordinates in terms of the parameter and eliminate it.

Answer: (c) 997

Explanation:
1. Pair r with \( 1996-r \): each pair adds up to 1.
2. There are 1995 terms: 997 pairs (r = 1 to 997 with r = 1995 to 999) and the middle term \( r=998 \), \( f\left(\tfrac12\right)=\dfrac{3}{3+3}=\tfrac12 \).
3. Sum \( =997+\tfrac12=997.5 \), and its integer part is 997.

Teacher's Note:
1. Use the pairing property of g_a.
2. Do not forget the unpaired middle term.

Answer: (b) an even number

Explanation:
1. Pair r with \( 1997-r \): each pair has sum 1.
2. There are 1996 terms, i.e. 998 pairs, and no middle term.
3. Sum \( =998 \), an even number.

Teacher's Note:
1. An even number of terms pairs up completely.
2. Each pair contributes exactly 1.

Answer: (b) 2n - 1

Explanation:
1. Pair r with \( 2n-r \): \( n-1 \) pairs each with sum 1, plus the middle term \( r=n \) with \( f\left(\tfrac12\right)=\tfrac12 \).
2. \( \sum f=n-1+\tfrac12=n-\tfrac12 \).
3. Multiplying by 2 gives \( 2n-1 \).

Teacher's Note:
1. Use the pairing property and count the unpaired middle term.
2. Then double the sum.

Answer: (c) 987

Explanation:
1. \( f(0)=\dfrac1{1+\sqrt a} \).
2. The remaining terms \( r=1,\dots,2n \) pair as \( r \) and \( 2n+1-r \): n pairs, each adding to 1, so they sum to n.
3. So the whole sum is \( \dfrac1{1+\sqrt a}+n \), and \( n=987 \).

Teacher's Note:
1. Separate the term r = 0.
2. The other 2n terms form n pairs.

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

Explanation:
1. For \( 3 < x < 5 \): \( x-1>0 \), \( x-5 < 0 \), so the numerator is negative.
2. \( x-2>0 \) and \( x-3>0 \), so the denominator is positive.
3. Hence \( f(x) < 0 \).

Teacher's Note:
1. Study the sign of each linear factor on the interval.
2. The quotient is negative when numerator and denominator have opposite signs.