JEE Mathematics Relation and Functions MCQs Set 15

Answer: (c) \( [1,\infty) \)

Explanation:
1. \( (f\circ g\circ h)(x)=\dfrac{2}{\cos\sqrt{x+3}+1} \), defined for \( x\ge-3 \) with \( \cos\sqrt{x+3}\ne-1 \).
2. \( \sqrt{x+3} \) takes every value in \( [0,\infty) \), so \( \cos\sqrt{x+3} \) takes every value in \( [-1,1] \) (except where the denominator vanishes).
3. The denominator lies in \( (0,2] \), so the function takes all values in \( [1,\infty) \).

Teacher's Note:
1. Work from the inside out: find the values of the inner function first.
2. Exclude points where the denominator is zero before finding the range.

Answer: (a) reflexive and symmetric

Explanation:
1. Reflexive: \( |a-a|=0\le1 \).
2. Symmetric: \( |a-b|=|b-a| \).
3. Not transitive: \( a=2,b=2.5,c=3.4 \) gives \( |a-b|=0.5\le1 \), \( |b-c|=0.9\le1 \), but \( |a-c|=1.4>1 \).

Teacher's Note:
1. A single counterexample disproves transitivity.
2. Closeness relations are usually reflexive and symmetric but not transitive.

Answer: (b) \( (x+1)4^{-n}-1 \)

Explanation:
1. \( g^2(x)=3+4(3+4x)=15+16x=(4^2-1)+4^2x \) and \( g^3(x)=63+64x=(4^3-1)+4^3x \).
2. In general \( g^n(x)=(4^n-1)+4^nx \).
3. Solving \( y=(4^n-1)+4^nx \) for \( x \): \( x=(y+1-4^n)4^{-n}=(y+1)4^{-n}-1 \).

Teacher's Note:
1. Find the pattern for the first two or three compositions and then generalise.
2. Check with \( n=1 \): the inverse of \( 3+4x \) is \( (x-3)/4=(x+1)/4-1 \).

Answer: (c) transitive and reflexive

Explanation:
1. \( xRy \) means \( x(x-y)=0 \), i.e. \( x=0 \) or \( x=y \).
2. Reflexive: \( x=x \). Transitive: if \( xRy \) and \( yRz \), then either \( x=0 \) (so \( xRz \)), or \( x=y\ne0 \) and then \( yRz \) forces \( y=z \), so \( x=z \).
3. Not symmetric: \( 0R1 \) holds but \( 1R0 \) does not. So \( R \) is reflexive and transitive but not symmetric.

Teacher's Note:
1. Translate the equation into a plain condition before testing the three properties.
2. A relation needs all three properties to be an equivalence relation.

Answer: (d) \( f \) is neither one-one nor onto

Explanation:
1. For \( x\ge0 \): \( f(x)=\dfrac{e^x-e^{-x}}{e^x+e^{-x}}=\tanh x\in[0,1) \). For \( x<0 \): \( f(x)=\dfrac{e^{-x}-e^{-x}}{e^x+e^{-x}}=0 \).
2. So \( f(0)=f(-1)=0 \): not one-one.
3. The range is \( [0,1) \), not all of \( \mathbb R \): not onto.

Teacher's Note:
1. Split the function into the cases \( x\ge0 \) and \( x<0 \) when an absolute value is present.
2. Two equal outputs for different inputs rules out one-one.

Answer: (c) ii, iii and iv only

Explanation:
1. If \( f(a_1)=f(a_2) \), then \( g(f(a_1))=g(f(a_2)) \); so one-one \( g\circ f \) forces \( a_1=a_2 \): \( f \) is one-one (ii) is true, but \( g \) need not be one-one (it may merge values outside the range of \( f \)), so (i) is false.
2. If \( g\circ f \) is onto \( C \), every element of \( C \) is reached through \( g \), so \( g \) is onto; with one-one \( f \) from (ii), (iii) is true.
3. A composition of one-one functions is one-one, so (iv) is true.

Teacher's Note:
1. A counterexample for (i): let \( g \) map two elements outside the range of \( f \) to the same value.
2. Remember: the inner function inherits injectivity, the outer function inherits surjectivity.

Answer: (d) ii, iii and iv only

Explanation:
1. \( \ln\ln\ln x \) must be positive: \( \ln\ln x>1 \), \( \ln x>e \), \( x>e^e \). So the domain is \( (e^e,\infty) \).
2. Each logarithm is increasing, so \( f \) is increasing and hence one-one.
3. As \( x\to e^e \) from the right \( f\to-\infty \), and \( f\to\infty \) as \( x\to\infty \), with \( f \) continuous, so the range is \( \mathbb R \): onto, not into.

Teacher's Note:
1. A composition of increasing functions is increasing.
2. Find the domain by peeling off one logarithm at a time.

Answer: (c) 5

Explanation:
1. Put \( y=x \): \( f(x+f(x))=f(x)+x \), which suggests \( f(t)=t \).
2. Check \( f(x)=x \): \( f(x+y)=x+y \), so the equation holds. (A constant shift \( f(x)=x+c \) gives \( c=0 \).)
3. Then \( f(1000)=1000=200k \), so \( k=5 \).

Teacher's Note:
1. A linear guess \( f(x)=ax+b \) quickly gives \( a=1,b=0 \).
2. Always verify the guessed function in the original equation.

Answer: (b) 1

Explanation:
1. \( g(x)=4x(1-x)=1-(2x-1)^2 \) takes every value in \( [0,1] \) for \( x\in[0,1] \).
2. \( f(t)=\dfrac{1-t}{1+t} \) is decreasing on \( [0,1] \), from \( f(0)=1 \) to \( f(1)=0 \).
3. So the range of \( f\circ g \) is \( [0,1] \): \( \alpha+\beta=1 \).

Teacher's Note:
1. Find the range of the inner function first, then apply the outer function to that interval.
2. A monotonic outer function maps the end-points to the end-points.

Answer: (c) 6

Explanation:
1. Let \( T(x)=\dfrac{2x+29}{x-2} \). Then \( T(T(x))=\dfrac{2(2x+29)+29(x-2)}{(2x+29)-2(x-2)}=\dfrac{33x}{33}=x \), so \( T \) is its own inverse.
2. Replace \( x \) by \( T \): \( 2f(T)+3f(x)=100T+80 \). Eliminating \( f(T) \): \( 5f(x)=300T+240-(200x+160) \), so \( f(x)=60T+16-40x \).
3. At \( x=3 \): \( T=35 \), so \( f(3)=2100+16-120=1996 \), unit digit 6.

Teacher's Note:
1. Whenever a functional equation involves a self-inverse substitution, apply it twice to get a second equation.
2. Solve the two equations as simultaneous linear equations in \( f(x) \) and \( f(T) \).

Answer: (d) \( \frac{1}{2010} \)

Explanation:
1. \( \dfrac1{x_{n+1}}=\dfrac{1-x_n}{x_n}=\dfrac1{x_n}-1 \), so the reciprocals form an arithmetic progression with common difference \( -1 \).
2. \( \dfrac1{x_{2009}}=\dfrac1a-2009=1 \), so \( \dfrac1a=2010 \) and \( a=\dfrac1{2010} \).

Teacher's Note:
1. Turn the iteration into an arithmetic progression by taking reciprocals.
2. Count carefully: \( x_{2009} \) is 2009 steps after \( x_0 \).

Answer: (d) \( \{0\} \)

Explanation:
1. Let \( t=\{x\}\in[0,1) \). The quantity under the root is \( t(t-2) \).
2. For \( 0<t<1 \), \( t>0 \) and \( t-2<0 \), so \( t(t-2)<0 \) and \( f \) is not real.
3. Only \( t=0 \) works, giving \( f(x)=0 \). The range is \( \{0\} \).

Teacher's Note:
1. Check where the expression under the square root is non-negative.
2. The function is defined only at integers.

Answer: (c) i, iii and iv only

Explanation:
1. Need \( |x|^2-5|x|+6\ge0 \): \( (|x|-2)(|x|-3)\ge0 \), so \( |x|\le2 \) or \( |x|\ge3 \).
2. Need \( |x|^2-2|x|-8\le0 \): \( (|x|-4)(|x|+2)\le0 \), so \( |x|\le4 \).
3. Together \( |x|\in[0,2]\cup[3,4] \), i.e. \( x\in[-4,-3]\cup[-2,2]\cup[3,4] \). The interval \( [-3,-2] \) is excluded.

Teacher's Note:
1. Substitute \( t=|x| \) and solve two quadratic inequalities.
2. Intersect the two conditions; the gap \( 2<|x|<3 \) is not allowed.