JEE Mathematics Relations and Functions MCQs Set 08

Answer: (d) f must be surjective but need not be injective

Explanation:
1. Since \(g\circ f\) is onto, \(g\) is onto \([0,2]\); with \( g \) injective, \( g \) is a bijection.
2. Then \(f\) must reach every point of \([-1,1]\) (else \( g(f) \) would miss some value), so \( f \) is surjective; it need not be injective.

Teacher's Note:
1. Find the domain from the conditions on every piece (square roots, logarithms, denominators) and intersect them.
2. For functional equations substitute convenient values or the transformed argument and solve the resulting simultaneous equations.

Answer: (c) infinite

Explanation:
1. \( g \) is forced only on \([f(0),f(1)]\), where \(g=f^{-1}\).
2. On \( [-1,f(0))\cup(f(1),1] g \) may take any values in [0,1] (this set has positive length), so there are infinitely many such \( g \).

Teacher's Note:
1. Find the domain from the conditions on every piece (square roots, logarithms, denominators) and intersect them.
2. For functional equations substitute convenient values or the transformed argument and solve the resulting simultaneous equations.

Answer: (a) 14

Explanation:
1. Total functions \( 2^4 = 16 \).
2. Constant functions (not onto): 2. So \( 16 - 2 = 14 \).

Teacher's Note:
1. Find the domain from the conditions on every piece (square roots, logarithms, denominators) and intersect them.
2. For functional equations substitute convenient values or the transformed argument and solve the resulting simultaneous equations.

Answer: (b) 0

Explanation:
1. \(f(f(x))=\left(a-(a-x^n)\right)^{1/n}=x\).
2. No real roots of \(g(x)=x\) means \(g(x)>x\) for all \( x \). Then \(g(g(x))>g(x)>x\), so \(g(g(x))-x=0\) has no real root.

Teacher's Note:
1. Find the domain from the conditions on every piece (square roots, logarithms, denominators) and intersect them.
2. For functional equations substitute convenient values or the transformed argument and solve the resulting simultaneous equations.

Answer: (c) 6

Explanation:
1. \(\dfrac{e^{22}-1}{2e^{11}}=\dfrac{e^{11}-e^{-11}}2=g(11)\).
2. Since \(f=g^{-1}\), \(f(g(11))=11\), so the value is \(11-5=6\).

Teacher's Note:
1. Find the domain from the conditions on every piece (square roots, logarithms, denominators) and intersect them.
2. For functional equations substitute convenient values or the transformed argument and solve the resulting simultaneous equations.

Answer: (d) 4

Explanation:
1. Solutions of \(f(x)=f^{-1}(x)\) come from \(f(f(x))=x\) (\( f(x)=x \) has no solution here).
2. For \(x\le0\): \(f(x)=1-x\ge1\) and \(f(f(x))=-x^2=x\) give \(x=0,-1\). For \(x\ge1\): \(f(x)=-(x-1)^2\le0\) and \(f(f(x))=1+(x-1)^2=x\) give \(x=1,2\).
3. Check each: \( f(0)=1=f^{-1}(0) \); \( f(1)=0=f^{-1}(1) \); \( f(-1)=2=f^{-1}(-1) \); \( f(2)=-1=f^{-1}(2) \). So 4 solutions.

Teacher's Note:
1. Find the domain from the conditions on every piece (square roots, logarithms, denominators) and intersect them.
2. For functional equations substitute convenient values or the transformed argument and solve the resulting simultaneous equations.

Answer: (a) 2

Explanation:
1. The expression simplifies to the constant \(\dfrac34\) for every \( x \).
2. So \(g(f(1))=g(3/4)=2\).

Teacher's Note:
1. Find the domain from the conditions on every piece (square roots, logarithms, denominators) and intersect them.
2. For functional equations substitute convenient values or the transformed argument and solve the resulting simultaneous equations.

Answer: (c) 7

Explanation:
1. \( f(2) = f(1) + f(1) = 2, f(3) = f(2) + f(1) = 3 \), and in general \( f(n) = n \) (checked by direct evaluation up to \( n = 20 \)).
2. \(\sum_{r=1}^{20}r=210\), and \(210/30=7\).

Teacher's Note:
1. Find the domain from the conditions on every piece (square roots, logarithms, denominators) and intersect them.
2. For functional equations substitute convenient values or the transformed argument and solve the resulting simultaneous equations.

Answer: (d) exactly two

Explanation:
1. If \(f\) has degree \(n\ge1\), the left side has degree \(2n\) and the right side degree \(n\), impossible.
2. So \( f \) is constant: \(c^2=c\) gives \(c=0\) or \(c=1\). Exactly two polynomials.

Teacher's Note:
1. Find the domain from the conditions on every piece (square roots, logarithms, denominators) and intersect them.
2. For functional equations substitute convenient values or the transformed argument and solve the resulting simultaneous equations.

Answer: (a) \(\dfrac1{2010}\)

Explanation:
1. Iterating, \(f^{(n)}(x)=\dfrac x{1-nx}\) (induction: \(f\!\left(\tfrac x{1-nx}\right)=\tfrac x{1-(n+1)x}\)).
2. \(x_{2009}=\dfrac a{1-2009a}=1\) gives \(a=\dfrac1{2010}\).

Teacher's Note:
1. Find the domain from the conditions on every piece (square roots, logarithms, denominators) and intersect them.
2. For functional equations substitute convenient values or the transformed argument and solve the resulting simultaneous equations.

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

Explanation:
1. With \(t=\{x\}\in[0,1)\): \(t^2-2t=t(t-2)\le0\), with equality only at \(t=0\).
2. The root exists only for \(t=0\), giving \(f=0\).

Teacher's Note:
1. Find the domain from the conditions on every piece (square roots, logarithms, denominators) and intersect them.
2. For functional equations substitute convenient values or the transformed argument and solve the resulting simultaneous equations.

Answer: (c) 3

Explanation:
1. Integer solutions of \(x^2+y^2=16\): \((0,\pm4)\) and \((\pm4,0)\).
2. x-values: \(-4,0,4\), so the domain has 3 elements.

Teacher's Note:
1. Find the domain from the conditions on every piece (square roots, logarithms, denominators) and intersect them.
2. For functional equations substitute convenient values or the transformed argument and solve the resulting simultaneous equations.

Answer: (d) 8

Explanation:
1. Put \(y=1\): \(2f(x)=f(x)+2^x\), so \(f(x)=2^x\).
2. Check: \(2\cdot2^{xy}=2^{xy}+2^{xy}\). Hence \(f(3)=8\).

Teacher's Note:
1. Find the domain from the conditions on every piece (square roots, logarithms, denominators) and intersect them.
2. For functional equations substitute convenient values or the transformed argument and solve the resulting simultaneous equations.

Answer: (c) \(-8\left(\dfrac97\right)^{2/3}\)

Explanation:
1. From \(f(x)=4x^{2/3}\left(\tfrac{1+x}{1-x}\right)^{1/3}\) at \(x=9/7\): \(\tfrac{1+x}{1-x}=\tfrac{16/7}{-2/7}=-8\).
2. \((-8)^{1/3}=-2\), so \(f=4\cdot(9/7)^{2/3}\cdot(-2)=-8(9/7)^{2/3}\).

Teacher's Note:
1. Find the domain from the conditions on every piece (square roots, logarithms, denominators) and intersect them.
2. For functional equations substitute convenient values or the transformed argument and solve the resulting simultaneous equations.

Answer: (c) not injective but surjective

Explanation:
1. \(f(x)=f(1/x)\) so it is not injective (\( e.g \). \( f(2) = f(1/2) \)).
2. \(|x+1/x|\ge2\) and every value of absolute size at least 2 is attained, so it is onto \(\mathbb R-(-2,2)\).

Teacher's Note:
1. Find the domain from the conditions on every piece (square roots, logarithms, denominators) and intersect them.
2. For functional equations substitute convenient values or the transformed argument and solve the resulting simultaneous equations.

Answer: (d) neither injective nor surjective

Explanation:
1. \( f(1) = f(-1) = 3 \), so it is not injective.
2. Its range is [2, infinity,) not all of \( R \), so it is not surjective.

Teacher's Note:
1. Find the domain from the conditions on every piece (square roots, logarithms, denominators) and intersect them.
2. For functional equations substitute convenient values or the transformed argument and solve the resulting simultaneous equations.

Answer: (a) \(\left(1,\dfrac73\right]\)

Explanation:
1. \(f(x)=1+\dfrac1{x^2+x+1}\) and \(x^2+x+1\ge\dfrac34\).
2. So \(0<\dfrac1{x^2+x+1}\le\dfrac43\) and the range is \((1,\tfrac73]\).

Teacher's Note:
1. Find the domain from the conditions on every piece (square roots, logarithms, denominators) and intersect them.
2. For functional equations substitute convenient values or the transformed argument and solve the resulting simultaneous equations.

Answer: (b) \([1,\sqrt2]\)

Explanation:
1. \(\cos^2x\in[0,1]\), so the angle \(\tfrac\pi4\cos^2x\in[0,\tfrac\pi4]\).
2. On this interval \( \sec \) increases from 1 to \(\sqrt2\).

Teacher's Note:
1. Find the domain from the conditions on every piece (square roots, logarithms, denominators) and intersect them.
2. For functional equations substitute convenient values or the transformed argument and solve the resulting simultaneous equations.

Answer: (b) \((e,\infty)\)

Explanation:
1. \(\ln\ln x>0\) requires \(\ln x>1\), i.e. \(x>e\).
2. For \(x>e\), \(\ln\ln x\in(0,\infty)\), so the outer logarithm is defined and \( f \) takes every real value.

Teacher's Note:
1. Find the domain from the conditions on every piece (square roots, logarithms, denominators) and intersect them.
2. For functional equations substitute convenient values or the transformed argument and solve the resulting simultaneous equations.

Answer: (c) 1

Explanation:
1. \(g([0,1])=[0,1]\) (maximum 1 at \( x = 1/2 \), minimum 0).
2. \( f \) is decreasing on [0,1] from 1 to 0, so the range is \([0,1]\) and \(\alpha+\beta=1\).

Teacher's Note:
1. Find the domain from the conditions on every piece (square roots, logarithms, denominators) and intersect them.
2. For functional equations substitute convenient values or the transformed argument and solve the resulting simultaneous equations.

Answer: (d) 6

Explanation:
1. \(x=3\) gives \(\tfrac{35}{1}=35\): \(2f(3)+3f(35)=380\).
2. \(x=35\) gives \(\tfrac{99}{33}=3\): \(2f(35)+3f(3)=3580\).
3. Solving: \(4f(3)+6f(35)=760\) and \(9f(3)+6f(35)=10740\), so \(f(3)=1996\); unit digit 6.

Teacher's Note:
1. Find the domain from the conditions on every piece (square roots, logarithms, denominators) and intersect them.
2. For functional equations substitute convenient values or the transformed argument and solve the resulting simultaneous equations.

Answer: (a) reflexive and transitive but not symmetric

Explanation:
1. Reflexive: \(x^2=x\cdot x\).
2. Not symmetric: \((0,1)\in R\) since \(0=0\), but \(1^2\ne1\cdot0\).
3. Transitive: \(x^2=xy,\ y^2=yz\). If \( x = 0 \) then \(x^2=xz\); otherwise \(x=y\) and \(y^2=yz\) gives \(x^2=xz\).

Teacher's Note:
1. Find the domain from the conditions on every piece (square roots, logarithms, denominators) and intersect them.
2. For functional equations substitute convenient values or the transformed argument and solve the resulting simultaneous equations.

Answer: (d) neither one-one nor onto

Explanation:
1. For \(x\le0\): \(e^{|x|}=e^{-x}\), so \(f(x)=0\): not one-one.
2. For \(x>0\): \(f(x)=\tanh x\in(0,1)\). The range is \([0,1)\), not \( R \).

Teacher's Note:
1. Find the domain from the conditions on every piece (square roots, logarithms, denominators) and intersect them.
2. For functional equations substitute convenient values or the transformed argument and solve the resulting simultaneous equations.