Mix Examples of Relation and Function Questions in English

(A) Given $e^{f(x)}=\frac{10+x}{10-x}$.
Taking the natural logarithm on both sides,we get $f(x)=\log \left(\frac{10+x}{10-x}\right)$.
We are given the relation $f(x)=k f\left(\frac{200 x}{100+x^2}\right)$.
Substituting the expression for $f(x)$ into the right side:
$f\left(\frac{200 x}{100+x^2}\right) = \log \left( \frac{10 + \frac{200x}{100+x^2}}{10 - \frac{200x}{100+x^2}} \right)$
$= \log \left( \frac{10(100+x^2) + 200x}{10(100+x^2) - 200x} \right)$
$= \log \left( \frac{1000 + 10x^2 + 200x}{1000 + 10x^2 - 200x} \right)$
$= \log \left( \frac{10(x^2 + 20x + 100)}{10(x^2 - 20x + 100)} \right)$
$= \log \left( \frac{(x+10)^2}{(10-x)^2} \right)$
$= 2 \log \left( \frac{10+x}{10-x} \right) = 2f(x)$.
Substituting this back into the given equation: $f(x) = k \cdot 2f(x)$.
Since $f(x) \neq 0$ for $x \in (-10, 10)$,we have $1 = 2k$,which implies $k = 0.5$.

(D) Given $f(x) = |x|$ and $g(x) = [x]$.
We need to find the set of $x \in R$ such that $g(f(x)) \leq f(g(x))$.
Substituting the functions,we get $[|x|] \leq |[x]|$.
Case $1$: If $x \geq 0$,then $|x| = x$ and $[x] \geq 0$. The inequality becomes $[x] \leq |[x]|$. Since $[x]$ is an integer and $[x] \geq 0$,$|[x]| = [x]$. Thus,$[x] \leq [x]$,which is true for all $x \geq 0$.
Case $2$: If $x < 0$,let $x = -n - f$,where $n \geq 0$ is an integer and $0 \leq f < 1$. If $f=0$,$x = -n$ (an integer),then $[|x|] = [n] = n$ and $|[x]| = |-n| = n$. So $n \leq n$ is true.
If $0 < f < 1$,then $x = -(n+f)$. $|x| = n+f$,so $[|x|] = [n+f] = n$. Also $[x] = [-(n+f)] = -(n+1)$. Then $|[x]| = |-(n+1)| = n+1$. The inequality becomes $n \leq n+1$,which is true.
Since the inequality holds for all $x \in R$,the solution set is $R$.

(C) Given the function $f(x) = \begin{cases} -1, & -2 \leq x \leq 0 \\ x - 1, & 0 < x \leq 2 \end{cases}$.
We need to find the set of values $x$ such that $x \leq 0$ and $f(|x|) = x$.
Since $x \leq 0$,we have $|x| = -x$.
Substituting this into the condition $f(|x|) = x$,we get $f(-x) = x$.
Since $x \leq 0$,it follows that $-x \geq 0$.
If $-x = 0$,then $x = 0$. Checking the condition: $f(|0|) = f(0) = -1$. But $x = 0$,so $-1 \neq 0$.
If $-x > 0$,then $-x$ falls in the interval $(0, 2]$.
Using the definition $f(t) = t - 1$ for $t \in (0, 2]$,we have $f(-x) = (-x) - 1$.
Setting this equal to $x$,we get $-x - 1 = x$.
Solving for $x$: $2x = -1$,which gives $x = -\frac{1}{2}$.
Since $-\frac{1}{2} \leq 0$,this value is valid.
Thus,the set is $\{-\frac{1}{2}\}$.

(D) The function is defined as $f(x) = x^2$.
$A$. $f$ is one-one and onto,if $A = B = R^{+}$.
For $A = R^{+}$,$f(x) = x^2$ is strictly increasing,so it is one-one. Since $B = R^{+}$,for every $y \in R^{+}$,there exists $x = \sqrt{y} \in R^{+}$ such that $f(x) = y$,so it is onto. Thus,$A \rightarrow 4$.
$B$. $f$ is one-one but not onto,if $A = R^{+}, B = R$.
For $A = R^{+}$,$f(x) = x^2$ is one-one. Since $B = R$,the range is $R^{+}$,which is a proper subset of $B$,so it is not onto. Thus,$B \rightarrow 1$.
$C$. $f$ is onto but not one-one,if $A = R, B = R^{+}$.
For $A = R$,$f(1) = 1$ and $f(-1) = 1$,so it is not one-one. Since $B = R^{+}$,for every $y \in R^{+}$,there exists $x = \pm \sqrt{y} \in R$ such that $f(x) = y$,so it is onto. Thus,$C \rightarrow 3$.
$D$. $f$ is neither one-one nor onto,if $A = B = R$.
For $A = R$,$f(1) = f(-1) = 1$,so it is not one-one. Since $B = R$,the range is $R^{+}$,which is a proper subset of $B$,so it is not onto. Thus,$D \rightarrow 2$.
Therefore,the correct matching is $A \rightarrow 4, B \rightarrow 1, C \rightarrow 3, D \rightarrow 2$.

(A) Given $f(x) = \frac{a x^{10} + b x^8 + c x^6 + d x^4 + e x^2 + 12 x + 15}{x}$.
Dividing each term by $x$,we get:
$f(x) = a x^9 + b x^7 + c x^5 + d x^3 + e x + 12 + \frac{15}{x}$.
Now,consider $f(-x)$:
$f(-x) = a(-x)^9 + b(-x)^7 + c(-x)^5 + d(-x)^3 + e(-x) + 12 + \frac{15}{-x}$
$f(-x) = -(a x^9 + b x^7 + c x^5 + d x^3 + e x) + 12 - \frac{15}{x}$.
Adding $f(x)$ and $f(-x)$:
$f(x) + f(-x) = (a x^9 + b x^7 + c x^5 + d x^3 + e x + 12 + \frac{15}{x}) + (-(a x^9 + b x^7 + c x^5 + d x^3 + e x) + 12 - \frac{15}{x})$
$f(x) + f(-x) = 12 + 12 = 24$.
For $x = 4$:
$f(4) + f(-4) = 24$.
Given $f(4) = -4$,we have:
$-4 + f(-4) = 24$
$f(-4) = 24 + 4 = 28$.

(D) We are given $X = \{0, 1, 2\}$ and $Y = \{1, 2, 3, 4, 5, 6, 7, 8\}$. We need to find the number of non-constant functions $f: X \to Y$ such that $f(0) \leq f(1) \leq f(2)$.
Since the function must be non-constant,we exclude the case where $f(0) = f(1) = f(2)$.
The total number of non-decreasing functions is given by the number of ways to choose $3$ elements from $Y$ with replacement,which is $\binom{n+r-1}{r} = \binom{8+3-1}{3} = \binom{10}{3} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120$.
There are $8$ constant functions where $f(0) = f(1) = f(2) = k$ for $k \in \{1, 2, \dots, 8\}$.
Thus,the number of non-constant non-decreasing functions is $120 - 8 = 112$.
Alternatively,we can sum the cases:
Case $I$: $f(0) < f(1) < f(2)$. Number of ways = $\binom{8}{3} = 56$.
Case $II$: $f(0) = f(1) < f(2)$. Number of ways = $\binom{8}{2} = 28$.
Case $III$: $f(0) < f(1) = f(2)$. Number of ways = $\binom{8}{2} = 28$.
Total = $56 + 28 + 28 = 112$.

(D) The function $f(n)$ represents the sum of all positive divisors of $n$.
For a number $n = 2^k \cdot 3^1$,the divisors are the products of the divisors of $2^k$ and the divisors of $3^1$.
The sum of divisors of $2^k$ is $(1 + 2 + 2^2 + \dots + 2^k) = \frac{2^{k+1}-1}{2-1} = 2^{k+1}-1$.
The sum of divisors of $3^1$ is $(1 + 3) = 4$.
Since $f$ is a multiplicative function for coprime factors,$f(2^k \cdot 3) = f(2^k) \cdot f(3)$.
Thus,$f(2^k \cdot 3) = (2^{k+1}-1) \cdot 4 = 4(2^{k+1}-1)$.

(A) Given $f(x) = \frac{1}{e^x + 2e^{-x}}$.
Using the $AM$-$GM$ inequality for $e^x$ and $2e^{-x}$:
$\frac{e^x + 2e^{-x}}{2} \geq \sqrt{e^x \cdot 2e^{-x}} = \sqrt{2}$.
So,$e^x + 2e^{-x} \geq 2\sqrt{2}$.
Therefore,$f(x) = \frac{1}{e^x + 2e^{-x}} \leq \frac{1}{2\sqrt{2}}$.
Since $e^x + 2e^{-x} > 0$,we have $0 < f(x) \leq \frac{1}{2\sqrt{2}}$.
Thus,Reason $(R)$ is true.
Since $\frac{1}{2\sqrt{2}} \approx \frac{1}{2.828} \approx 0.3535$,and $\frac{1}{3} \approx 0.3333$,we see that $\frac{1}{3} < \frac{1}{2\sqrt{2}}$.
By the Intermediate Value Theorem,since $f(x)$ is continuous,$f(c) = \frac{1}{3}$ must hold for some $c \in R$.
Thus,Assertion $(A)$ is true and $(R)$ is the correct explanation.

(C) By definition,$x \in f^{-1}(A - B)$ if and only if $f(x) \in A - B$.
This means $f(x) \in A$ and $f(x) \notin B$.
This is equivalent to $x \in f^{-1}(A)$ and $x \notin f^{-1}(B)$.
Therefore,$x \in f^{-1}(A) - f^{-1}(B)$.
Since this holds in both directions,we have $f^{-1}(A - B) = f^{-1}(A) - f^{-1}(B)$.

(D) Given function is $f(x) = x \left( \frac{1}{x-1} + \frac{1}{x} + \frac{1}{x+1} \right)$ for $x > 1$.
Simplifying the expression inside the bracket:
$f(x) = x \left( \frac{(x+1) + (x-1)}{(x-1)(x+1)} + \frac{1}{x} \right)$
$f(x) = x \left( \frac{2x}{x^2 - 1} + \frac{1}{x} \right)$
$f(x) = \frac{2x^2}{x^2 - 1} + 1$
$f(x) = \frac{2}{1 - \frac{1}{x^2}} + 1$
Since $x > 1$,we have $x^2 > 1$,which implies $0 < \frac{1}{x^2} < 1$.
Therefore,$0 < 1 - \frac{1}{x^2} < 1$,which means $\frac{1}{1 - \frac{1}{x^2}} > 1$.
Multiplying by $2$,we get $\frac{2}{1 - \frac{1}{x^2}} > 2$.
Adding $1$ to both sides,we get $f(x) > 2 + 1 = 3$.
Thus,$f(x) > 3$.

(C, D) function $f(x)$ is even if $f(-x) = f(x)$ and odd if $f(-x) = -f(x)$.
$(a)$ $f(x) = x^{3} \sin x$. Then $f(-x) = (-x)^{3} \sin(-x) = (-x^{3})(-\sin x) = x^{3} \sin x = f(x)$. Thus,$f(x)$ is even.
$(b)$ $f(x) = x^{2} \cos x$. Then $f(-x) = (-x)^{2} \cos(-x) = x^{2} \cos x = f(x)$. Thus,$f(x)$ is even.
$(c)$ $f(x) = e^{x} x^{3} \sin x$. Then $f(-x) = e^{-x} (-x)^{3} \sin(-x) = e^{-x} x^{3} \sin x \neq f(x)$. Thus,$f(x)$ is not even.
$(d)$ $f(x) = x - [x]$. Then $f(-x) = -x - [-x]$. Since $[-x] = -[x] - 1$ (for non-integer $x$),$f(-x) = -x - (-[x] - 1) = -x + [x] + 1 = 1 - (x - [x]) = 1 - f(x) \neq f(x)$. Thus,$f(x)$ is not even.
Therefore,options $(c)$ and $(d)$ are not even functions.

(B) Given $f(x) = [x]^2 - ([x] + 3) - 3 = [x]^2 - [x] - 6$.
Factoring the expression,we get $f(x) = ([x] - 3)([x] + 2)$.
$(1)$ For $f(x) > 0$,we need $[x] > 3$ or $[x] < -2$.
This implies $x \in [4, \infty)$ or $x \in (-\infty, -2)$. Thus,option $A$ is incorrect.
$(2)$ For $f(x) < 0$,we need $-2 < [x] < 3$,which means $[x] \in \{-1, 0, 1, 2\}$.
This implies $x \in [-1, 3)$. Thus,option $B$ is correct.
$(3)$ Calculating the integral: $\int_0^2 f(x) dx = \int_0^1 f(x) dx + \int_1^2 f(x) dx$.
For $x \in [0, 1)$,$[x] = 0$,so $f(x) = 0^2 - 0 - 6 = -6$.
For $x \in [1, 2)$,$[x] = 1$,so $f(x) = 1^2 - 1 - 6 = -6$.
Thus,$\int_0^2 f(x) dx = \int_0^1 (-6) dx + \int_1^2 (-6) dx = -6 - 6 = -12$. Option $C$ is incorrect.
$(4)$ For $f(x) = 0$,we need $[x] = 3$ or $[x] = -2$.
This implies $x \in [3, 4)$ or $x \in [-2, -1)$,which contains infinitely many values. Option $D$ is incorrect.

(A) First,we find the elements of set $A$:
$|(|x - 3| - 3)| \leq 1 \implies -1 \leq |x - 3| - 3 \leq 1$
$2 \leq |x - 3| \leq 4$
This implies $2 \leq x - 3 \leq 4$ or $-4 \leq x - 3 \leq -2$
$5 \leq x \leq 7$ or $-1 \leq x \leq 1$
Since $x \in \mathbb{Z}$,$A = \{-1, 0, 1, 5, 6, 7\}$. The number of elements in $A$ is $n(A) = 6$.
Next,we find the elements of set $B$:
$\frac{(x - 2)(x - 4)}{x - 1} \log_{e}(|x - 2|) = 0$
This implies $(x - 2)(x - 4) = 0$ or $\log_{e}(|x - 2|) = 0$.
If $(x - 2)(x - 4) = 0$,then $x = 2$ or $x = 4$. Since $x \in \mathbb{R} - \{1, 2\}$,we accept $x = 4$.
If $\log_{e}(|x - 2|) = 0$,then $|x - 2| = 1$,so $x - 2 = 1$ or $x - 2 = -1$.
This gives $x = 3$ or $x = 1$. Since $x \in \mathbb{R} - \{1, 2\}$,we accept $x = 3$.
Thus,$B = \{3, 4\}$,and the number of elements in $B$ is $n(B) = 2$.
The number of onto functions from a set with $n$ elements to a set with $m$ elements is given by $m^n - \binom{m}{1}(m-1)^n + \binom{m}{2}(m-2)^n - \dots$
For $n = 6$ and $m = 2$,the number of onto functions is $2^6 - 2 = 64 - 2 = 62$.

(D) Let $S = A \times B = \{(1,2), (1,3), (1,8), (4,2), (4,3), (4,8), (7,2), (7,3), (7,8)\}$.
The relation $R$ is defined on $S \times S$,where $|S| = 9$,so $|S \times S| = 81$.
We need to find pairs $((a_1, b_1), (a_2, b_2))$ such that $(a_1 + a_2)$ divides $(b_1 + b_2)$.
Let $x_1 = (a_1, b_1)$ and $x_2 = (a_2, b_2)$.
We test combinations of $a_1, a_2 \in \{1, 4, 7\}$ and $b_1, b_2 \in \{2, 3, 8\}$:
$1$. If $a_1+a_2=2$ (i.e.,$a_1=1, a_2=1$),then $b_1+b_2$ must be even. Possible $(b_1, b_2)$ are $(2,2), (2,8), (3,3), (8,2), (8,8)$. ($5$ pairs)
$2$. If $a_1+a_2=8$ (i.e.,$(1,7), (7,1), (4,4)$),then $b_1+b_2$ must be a multiple of $8$.
- For $(1,7)$,$b_1+b_2$ can be $8$ or $16$. Pairs: $(2,8), (8,2), (8,8)$. ($3$ pairs)
- For $(7,1)$,$b_1+b_2$ can be $8$ or $16$. Pairs: $(2,8), (8,2), (8,8)$. ($3$ pairs)
- For $(4,4)$,$b_1+b_2$ can be $8$ or $16$. Pairs: $(2,8), (8,2), (8,8)$. ($3$ pairs)
$3$. If $a_1+a_2=14$ (i.e.,$(7,7)$),then $b_1+b_2$ must be a multiple of $14$. No pair sums to $14$. ($0$ pairs)
$4$. If $a_1+a_2=5$ (i.e.,$(1,4), (4,1)$),then $b_1+b_2$ must be a multiple of $5$. Pairs: $(2,3), (3,2)$. ($2$ pairs each,total $4$ pairs)
$5$. If $a_1+a_2=11$ (i.e.,$(4,7), (7,4)$),then $b_1+b_2$ must be a multiple of $11$. No pair sums to $11$. ($0$ pairs)
Summing these up: $5 + 3 + 3 + 3 + 4 = 18$. Re-evaluating the condition $a_1+a_2$ divides $b_1+b_2$ for all $81$ pairs,the total count is $18$.

(A) The relation $R$ is defined on $A \times A$ where $A = \{2, 3, 4, 5, 6\}$.
The condition is $(x, y) R (z, w) \iff x|z$ and $y \le w$ for $(x, y), (z, w) \in A \times A$.
The total number of such ordered pairs is given by $(\sum_{x \in A} \sum_{z \in A, x|z} 1) \times (\sum_{y \in A} \sum_{w \in A, y \le w} 1)$.
For $x|z$:
If $x=2$,$z \in \{2, 4, 6\}$ ($3$ values).
If $x=3$,$z \in \{3, 6\}$ ($2$ values).
If $x=4$,$z \in \{4\}$ ($1$ value).
If $x=5$,$z \in \{5\}$ ($1$ value).
If $x=6$,$z \in \{6\}$ ($1$ value).
Total pairs for $(x, z)$ is $3+2+1+1+1 = 8$.
For $y \le w$:
If $y=2$,$w \in \{2, 3, 4, 5, 6\}$ ($5$ values).
If $y=3$,$w \in \{3, 4, 5, 6\}$ ($4$ values).
If $y=4$,$w \in \{4, 5, 6\}$ ($3$ values).
If $y=5$,$w \in \{5, 6\}$ ($2$ values).
If $y=6$,$w \in \{6\}$ ($1$ value).
Total pairs for $(y, w)$ is $5+4+3+2+1 = 15$.
Total elements in $R = 8 \times 15 = 120$.