JEE Mathematics Relations and Functions MCQs Set 07

Answer: (c) 0

Explanation:
1. Let \(\log x=u,\ \log y=v\). Then \(f(x/y)=\cos(u-v)\) and \(f(xy)=\cos(u+v)\).
2. \(\tfrac12[\cos(u-v)+\cos(u+v)]=\cos u\cos v=f(x)f(y)\), so the expression is 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: (d) \([-5,-\sqrt{21}]\cup[\sqrt{21},5]\cup\{0\}\)

Explanation:
1. Inner root needs \(x^4-21x^2\ge0\), i.e. \(x^2\ge21\) or \(x=0\).
2. Outer root needs \(\sqrt{x^4-21x^2}\le10\), i.e. \(x^4-21x^2-100\le0\), which gives \((x^2-25)(x^2+4)\le0\) and \(|x|\le5\).
3. Combine the two conditions.

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) 4

Explanation:
1. Put \(x=7\): \(\tfrac{7+59}{7-1}=11\), so \(3f(7)+2f(11)=100\).
2. Put \(x=11\): \(\tfrac{70}{10}=7\), so \(3f(11)+2f(7)=140\).
3. Eliminate \( f(11) \): \( 9f(7)+6f(11)=300 \) and \( 4f(7)+6f(11)=280 \), so \( 5f(7)=20 \) and \( f(7)=4 \).

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) \(\dfrac{2a+b}{2(a^2-b^2)}\)

Explanation:
1. Put \(x=1\): \(af(2)+bf(1/2)=1\).
2. Put \(x=-\tfrac12\): \(af(1/2)+bf(2)=-\tfrac12\).
3. Eliminate \( f(1/2) \): \(f(2)(a^2-b^2)=a+\tfrac b2\), so \(f(2)=\dfrac{2a+b}{2(a^2-b^2)}\).

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) \(\left[-\dfrac12,1\right)\)

Explanation:
1. \(3+4x-4x^2\ge0\) gives \((2x-3)(2x+1)\le0\), so \(-\tfrac12\le x\le\tfrac32\).
2. \( f \) is given for \(-1<x<1\). The intersection is \([-\tfrac12,1)\).

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: (a) \((8,10)\)

Explanation:
1. Innermost: \(18x-x^2-77>0\). Next: \(\log_3(\cdot)>0\) requires \(18x-x^2-77>1\).
2. Outermost: \(\log_5\{\log_3(\cdot)\}>0\) requires \(\log_3(\cdot)>1\), i.e. \(18x-x^2-77>3\).
3. \(x^2-18x+80<0\) gives \((x-8)(x-10)<0\), so \(8<x<10\).

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) \(2g(x)g(y)\)

Explanation:
1. \(g(x)=\dfrac{a^x+a^{-x}}2\).
2. \(g(x+y)+g(x-y)=\dfrac{a^{x+y}+a^{-x-y}+a^{x-y}+a^{y-x}}2=\dfrac{(a^x+a^{-x})(a^y+a^{-y})}{2}=2g(x)g(y)\).

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) odd and periodic

Explanation:
1. The first condition gives \(f(2\lambda-t)=f(t)\), so \(f(2\lambda+x)=f(-x)\).
2. The second gives \(f(2\lambda+x)=-f(2\lambda-x)=-f(x)\). Hence \(f(-x)=-f(x)\): \( f \) is odd.
3. Also \(f(x+2\lambda)=-f(x)\), so \(f(x+4\lambda)=f(x)\): \( f \) is periodic.

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) \(\{\pm1,\pm5\}\)

Explanation:
1. \(\cos nx\) has period \(2\pi/|n|\) and \(\sin(5x/n)\) has period \(2\pi|n|/5\); the product must repeat after \(2\pi\).
2. For \(n=\pm1\): periods \(2\pi\) and \(2\pi/5\) fit; for \(n=\pm5\): \(2\pi/5\) and \(2\pi\) fit. For other \( n \) (for example 2, 10) the product contains a term of period \( 4 \pi \), and a numerical check of \( n \) up to 10 confirms this.
3. So the set is {+1,-1,+5,-5}.

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) \((-e,-1)\cup(1,e)\)

Explanation:
1. Need \(0<\ln|x|<1\).
2. This gives \(1<|x|<e\), i.e. \(x\in(-e,-1)\cup(1,e)\).

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) 4

Explanation:
1. For \(x\le1\) the range is \((-\infty,5]\).
2. For \(x>1\) the range is \((a^2+1,\infty)\) if \(a\ne0\) (and the single value 1 if \( a = 0 \), which fails).
3. To cover all of \( R \) we need \(a^2+1\le5\), so \(0<|a|\le2\): \( a = -2, -1, 1, 2 \), i.e. 4 values.

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) \([0,2]\)

Explanation:
1. \(\sqrt2f(x)=2\sin(x-\pi/4)\in[-2,2]\), so \(\sqrt2f+3\in[1,5]\).
2. \(\log_{\sqrt5}1=0\) and \(\log_{\sqrt5}5=2\); the function is increasing, so the range is \([0,2]\).

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: (b) 5.5

Explanation:
1. \(\phi(x)+\phi(-x)=\dfrac1{1+e^{-x}}+\dfrac{e^x}{e^x+1}=1\).
2. The terms pair up into 5 pairs summing to 1 each, plus \(\phi(0)=\tfrac12\): \( S = 5.5 \).

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) 9

Explanation:
1. Put \(x=-1\): \(f(1)=f(-1)+f(2)=-3+f(2)\), so \(f(2)=6\).
2. Put \(x=1\): \(f(3)=f(1)+f(2)=3+6=9\).

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) S is an equivalence relation but R is not

Explanation:
1. \( R \): taking \(w=0\) gives \((0,y)\in R\) for every \( y \), but \((y,0)\notin R\) if \(y\ne0\). So \( R \) is not symmetric.
2. \( S \): \(qm=pn\) means \(\tfrac mn=\tfrac pq\); equality of fractions is reflexive, symmetric and transitive.

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) reflexive only

Explanation:
1. Reflexive: \(x-x+\sqrt5=\sqrt5\) is irrational.
2. Not symmetric: \(x-y=\sqrt5\) gives \(2\sqrt5\) (irrational) but \(y-x+\sqrt5=0\) (rational).
3. Not transitive: \(x-y=\sqrt2\) and \(y-z=-\sqrt2-\sqrt5\) give irrational values, but \(x-z+\sqrt5=0\).

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: (a) \([-1,3]\)

Explanation:
1. \(\sin x-\sqrt3\cos x=2\sin(x-\pi/3)\in[-2,2]\).
2. Adding 1 gives the range [-1, 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) 45

Explanation:
1. All functions: \( 3^4 = 81 \).
2. Onto functions: \( 3^4 - 3\cdot 2^4 + 3\cdot 1^4 = 81 - 48 + 3 = 36 \).
3. Not onto: \( 81 - 36 = 45 \).

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) many-one and into

Explanation:
1. \(x-x^4+x^7-x^8-1<0\) for every \(x>0\) (its maximum on x>0 is about \( -0.51 \), checked numerically).
2. So \( f(x) = -1 \) for all \( x \) > 0: a constant function, which is many-one and not onto.

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. \(f(x)=x+|x|\) equals 0 for every \(x\le0\), so it is not injective.
2. Its values are never negative, 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) \(\left[-\dfrac\pi4,\dfrac\pi4\right]\)

Explanation:
1. \(g(f(x))=(\sin x+\cos x)^2-1=\sin2x\).
2. \(\sin2x\) is one-one and increasing for \(2x\in[-\pi/2,\pi/2]\), i.e. \(x\in[-\pi/4,\pi/4]\).

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

Explanation:
1. \(g(x)+1=4(x+1)\), so \(g^n(x)+1=4^n(x+1)\).
2. Inverse: \(y=4^n(x+1)-1\Rightarrow x=(y+1)4^{-n}-1\).

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: (c) \(\{(4m+1)\tfrac\pi2:m\in I\}\cup\bigcup_{n\in I}\left[2n\pi+\tfrac{7\pi}6,2n\pi+\tfrac{11\pi}6\right]\)

Explanation:
1. Domain of \( f \): \(3|t|-t-2\ge0\): for \(t\ge0\) we need \(t\ge1\); for \(t<0\) we need \(t\le-\tfrac12\).
2. With \(t=\sin x\in[-1,1]\): \(\sin x=1\) or \(-1\le\sin x\le-\tfrac12\).
3. \(\sin x=1\) at \(x=2n\pi+\tfrac\pi2\); \(\sin x\le-\tfrac12\) on \([2n\pi+\tfrac{7\pi}6,2n\pi+\tfrac{11\pi}6]\).

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) \(\pm\sqrt{n\pi},\ n\in\{0,1,2,\ldots\}\)

Explanation:
1. Let \(s=\sin(\sin x^2)\). LHS = \(s^2\), RHS = \(s\), so \(s=0\) or \(s=1\).
2. \(s=1\) is impossible since \(\sin(\sin x^2)\le\sin1<1\). So \(\sin(\sin x^2)=0\), i.e. \(\sin x^2=0\), giving \(x^2=n\pi\).

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.