If $f(x) = {{(\tan 1^\circ )x + {{\log }_e}(123)} \over {x{{\log }_e}(1234) - (\tan 1^\circ )}},x > 0$, then the least value of $f(f(x)) + f\left( {f\left( {{4 \over x}} \right)} \right)$ is :
Let the sets A and B denote the domain and range respectively of the function $f(x)=\frac{1}{\sqrt{\lceil x\rceil-x}}$, where $\lceil x\rceil$ denotes the smallest integer greater than or equal to $x$. Then among the statements
(S1) : $A \cap B=(1, \infty)-\mathbb{N}$ and
(S2) : $A \cup B=(1, \infty)$
Let $f:\mathbb{R}-{0,1}\to \mathbb{R}$ be a function such that $f(x)+f\left(\frac{1}{1-x}\right)=1+x$. Then $f(2)$ is equal to
Let $f(x) = \left| {\matrix{ {1 + {{\sin }^2}x} & {{{\cos }^2}x} & {\sin 2x} \cr {{{\sin }^2}x} & {1 + {{\cos }^2}x} & {\sin 2x} \cr {{{\sin }^2}x} & {{{\cos }^2}x} & {1 + \sin 2x} \cr } } \right|,\,x \in \left[ {{\pi \over 6},{\pi \over 3}} \right]$. If $\alpha$ and $\beta$ respectively are the maximum and the minimum values of $f$, then
defined as $f(x)=\frac{x^2+2 x+1}{x^2-8 x+12}$.
Then range of $f$ is
$f(x)=\left|x^{2}-x+1\right|+\left[x^{2}-x+1\right]$,
where $[t]$ denotes the greatest integer function, in the interval $[-1,2]$, is :
Consider a function $f:\mathbb{N}\to\mathbb{R}$, satisfying $f(1)+2f(2)+3f(3)+....+xf(x)=x(x+1)f(x);x\ge2$ with $f(1)=1$. Then $\frac{1}{f(2022)}+\frac{1}{f(2028)}$ is equal to
The domain of $f(x) = {{{{\log }_{(x + 1)}}(x - 2)} \over {{e^{2{{\log }_e}x}} - (2x + 3)}},x \in \mathbb{R}$ is
Let $f:R \to R$ be a function such that $f(x) = {{{x^2} + 2x + 1} \over {{x^2} + 1}}$. Then
The number of functions
$f:\{ 1,2,3,4\} \to \{ a \in Z|a| \le 8\} $
satisfying $f(n) + {1 \over n}f(n + 1) = 1,\forall n \in \{ 1,2,3\} $ is
Let $f:\mathbb{R}\to\mathbb{R}$ be a function defined by $f(x) = {\log _{\sqrt m }}\{ \sqrt 2 (\sin x - \cos x) + m - 2\} $, for some $m$, such that the range of $f$ is [0, 2]. Then the value of $m$ is _________
Let $f(x) = 2{x^n} + \lambda ,\lambda \in R,n \in N$, and $f(4) = 133,f(5) = 255$. Then the sum of all the positive integer divisors of $(f(3) - f(2))$ is
Let $f(x)$ be a function such that $f(x+y)=f(x).f(y)$ for all $x,y\in \mathbb{N}$. If $f(1)=3$ and $\sum\limits_{k = 1}^n {f(k) = 3279} $, then the value of n is
If $f(x) = {{{2^{2x}}} \over {{2^{2x}} + 2}},x \in \mathbb{R}$, then $f\left( {{1 \over {2023}}} \right) + f\left( {{2 \over {2023}}} \right)\, + \,...\, + \,f\left( {{{2022} \over {2023}}} \right)$ is equal to
Let $\mathrm{A}=\{1,2,3,4,5\}$ and $\mathrm{B}=\{1,2,3,4,5,6\}$. Then the number of functions $f: \mathrm{A} \rightarrow \mathrm{B}$ satisfying $f(1)+f(2)=f(4)-1$ is equal to __________.
Explanation:
We want to find out how many such functions exist.
First, observe that the condition $f(1) + f(2) = f(4) - 1$ can be rewritten as $f(1) + f(2) + 1 = f(4)$. So, the sum of $f(1), f(2),$ and 1 is equal to $f(4)$. Since $f(4)$ is a value in set B, it can take values from 1 to 6.
The maximum value of $f(1) + f(2) + 1$ can be $6 + 6 + 1 = 13$, but this is more than 6 (the maximum value of $f(4)$), so it's not possible. Thus, the maximum value of $f(4)$ in this case can be 6.
Let's now analyze the number of functions for each value of $f(4)$ from 3 to 6 (we start from 3 because $f(1)$ and $f(2)$ take values from set B and their minimum sum plus 1 is 3):
1. When $f(4) = 6$, then $f(1) + f(2) = 5$. The pairs $(f(1), f(2))$ that satisfy this equation are $(1, 4), (2, 3), (3, 2), (4, 1)$. For each of these 4 cases, the functions $f(3)$ and $f(5)$ can each take 6 values from set B, resulting in $4 \times 6 \times 6 = 144$ functions.
2. When $f(4) = 5$, then $f(1) + f(2) = 4$. The pairs $(f(1), f(2))$ that satisfy this equation are $(1, 3), (2, 2), (3, 1)$. For each of these 3 cases, the functions $f(3)$ and $f(5)$ can each take 6 values from set B, resulting in $3 \times 6 \times 6 = 108$ functions.
3. When $f(4) = 4$, then $f(1) + f(2) = 3$. The pairs $(f(1), f(2))$ that satisfy this equation are $(1, 2), (2, 1)$. For each of these 2 cases, the functions $f(3)$ and $f(5)$ can each take 6 values from set B, resulting in $2 \times 6 \times 6 = 72$ functions.
4. When $f(4) = 3$, then $f(1) + f(2) = 2$. The only pair $(f(1), f(2))$ that satisfies this equation is $(1, 1)$. For this case, the functions $f(3)$ and $f(5)$ can each take 6 values from set B, resulting in $1 \times 6 \times 6 = 36$ functions.
Adding the numbers of functions from all these cases, we get a total of $144 + 108 + 72 + 36 = 360$ functions from $A$ to $B$ that satisfy the given condition.
Therefore, the number of functions $f : A \rightarrow B$ satisfying the condition $f(1) + f(2) = f(4) - 1$ is 360.
Let $\mathrm{R}=\{\mathrm{a}, \mathrm{b}, \mathrm{c}, \mathrm{d}, \mathrm{e}\}$ and $\mathrm{S}=\{1,2,3,4\}$. Total number of onto functions $f: \mathrm{R} \rightarrow \mathrm{S}$ such that $f(\mathrm{a}) \neq 1$, is equal to ______________.
Explanation:
$ \begin{aligned} & =\frac{5 !}{3 ! 2 !} \times 4 ! \\\\ & =\frac{5 \times 4}{2} \times 24=240 \end{aligned} $
When $f(a)=1$, number of onto functions
$ \begin{aligned} & =4 !+\frac{4 !}{2 ! 2 !} \times 3 ! \\\\ & =24+36=60 \end{aligned} $
So, required number of onto functions
$=240-60=180$
If domain of the function $\log _{e}\left(\frac{6 x^{2}+5 x+1}{2 x-1}\right)+\cos ^{-1}\left(\frac{2 x^{2}-3 x+4}{3 x-5}\right)$ is $(\alpha, \beta) \cup(\gamma, \delta]$, then $18\left(\alpha^{2}+\beta^{2}+\gamma^{2}+\delta^{2}\right)$ is equal to ______________.
Explanation:
So, $\frac{6 x^2+5 x+1}{2 x-1}>0$
$ \begin{aligned} & \Rightarrow \frac{(3 x+1)(2 x+1)}{2 x-1}>0 \\\\ & \Rightarrow x \in\left(\frac{-1}{2}, \frac{-1}{3}\right) \cup\left(\frac{1}{2}, \infty\right) \end{aligned} $
Domain of $ \cos ^{-1} x \rightarrow[-1,1] $
$ \text { For domain of } \cos ^{-1}\left(\frac{2 x^2-3 x+4}{3 x-5}\right) $
$ \begin{aligned} & -1 \leq \frac{2 x^2-3 x+4}{3 x-5} \leq 1 \\\\ & \frac{2 x^2-1}{3 x-5} \geq 0 \text { and } \frac{2 x^2-6 x+9}{3 x-5} \leq 0 \end{aligned} $
$ \Rightarrow x \in\left[\frac{-1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right] \cup\left(\frac{5}{3}, \infty\right) $
So, common domain is $\left(\frac{-1}{2}, \frac{-1}{3}\right) \cup\left[\frac{1}{2}, \frac{1}{\sqrt{2}}\right]$
$ \begin{aligned} & \therefore 18\left(\alpha^2+\beta^2+\gamma^2+\delta^2\right)=18\left(\frac{1}{4}+\frac{1}{9}+\frac{1}{4}+\frac{1}{2}\right) \\\\ & =18\left(\frac{9+4+9+18}{36}\right)=\frac{1}{2}(40)=20 \end{aligned} $
Explanation:
$f(1.n)=f(1).f(n)\Rightarrow f(1)=1$.
$f(3.3)=(f(3))^2$
Hence, the possibilities for $(t(3),(9))$ are $(1,1)$ and $(3,9)$.
Other three i.e. $f(2),f(5),f(8)$
Can be chosen in 6$^3$ ways.
Hence, total number of functions
$6^3\times2=432$
Let $S=\{1,2,3,4,5,6\}$. Then the number of one-one functions $f: \mathrm{S} \rightarrow \mathrm{P}(\mathrm{S})$, where $\mathrm{P}(\mathrm{S})$ denote the power set of $\mathrm{S}$, such that $f(n) \subset f(\mathrm{~m})$ where $n < m$ is ____________.
Explanation:
$\because S={1,2,3,4,5,6}$ and $P(S) = \{ \phi ,\{ 1\} ,\{ 2\} ,....,\{ 1,2,3,4,5,6\} \} $
$f(n)$ corresponding a set having m elements which belongs to P(S), should be a subset of $f(n+1)$, so $f(n+1)$ should be a subset of P(S) having at least $m+1$ elements.
Now, if f(1) has one element then f(2) has 3, f(3) has 3 and so on and f(6) has 6 elements. Total number of possible functions = 6! = 720 .... (1)
If f(1) has no elements (i.e. null set $\phi$) then
Each index number represents the number of elements in respective rows
Taking every series of arrow and counting number of such possible functions (sets)
$ = {}^6{C_2} \times {}^4{C_1} \times {}^3{C_1} \times {}^2{C_1} + {}^6{C_1} \times {}^5{C_2} \times {}^3{C_1} \times {}^2{C_1} + {}^6{C_1} \times {}^5{C_1} \times {}^4{C_2} \times {}^2{C_1} + {}^6{C_1} \times {}^5{C_1} \times {}^4{C_1} \times {}^3{C_2} + {}^6{C_1} \times {}^5{C_1} \times {}^4{C_1} \times {}^3{C_1} \times {}^2{C_2} + {}^6{C_1} \times {}^5{C_1} \times {}^4{C_1} \times {}^3{C_1} \times {}^2{C_1}$
$ = 2520$ ..........(2)
From (1) and (2) : Total number of functions
= 2520 + 720 = 3240
Suppose $f$ is a function satisfying $f(x + y) = f(x) + f(y)$ for all $x,y \in N$ and $f(1) = {1 \over 5}$. If $\sum\limits_{n = 1}^m {{{f(n)} \over {n(n + 1)(n + 2)}} = {1 \over {12}}} $, then $m$ is equal to __________.
Explanation:
$f(2)=\frac{2}{5} \quad\quad f(3)=f(2)+f(1)=\frac{3}{5}$
$f(3)=\frac{3}{5}$
$\therefore \sum\limits_{n=1}^{m} \frac{f(n)}{n(n+1)(n+2)}$
$=\frac{1}{5} \sum\limits_{n=1}^{m}\left(\frac{1}{n+1}-\frac{1}{n+2}\right)$
$=\frac{1}{5}\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\ldots .+\frac{1}{m+1}-\frac{1}{m+2}\right)$
$=\frac{1}{5}\left(\frac{1}{2}-\frac{1}{m+2}\right)=\frac{m}{10(m+2)}=\frac{1}{12}$
$\therefore m=10$
For some a, b, c $\in\mathbb{N}$, let $f(x) = ax - 3$ and $\mathrm{g(x)=x^b+c,x\in\mathbb{R}}$. If ${(fog)^{ - 1}}(x) = {\left( {{{x - 7} \over 2}} \right)^{1/3}}$, then $(fog)(ac) + (gof)(b)$ is equal to ____________.
Explanation:
$g(x)=x^{b}+c$
$(fog)^{-1}=\left(\frac{x-7}{2}\right)^{\frac{1}{3}}$
$(fog)^{-1}(x)=\left(\frac{x+3-c a}{a}\right)^{\frac{1}{b}}=\left(\frac{x-7}{2}\right)^{\frac{1}{3}}$
$\Rightarrow a=2, b=3, c=5$
$fog(a c)+gof(b)$
$\because f(x)=2 x-3$
$g(x)=x^{3}+5$
$fog(10)+g o f(3)$
$=2007+32$
$=2039$
$ \text { Let } f(x)=a x^{2}+b x+c \text { be such that } f(1)=3, f(-2)=\lambda \text { and } $ $f(3)=4$. If $f(0)+f(1)+f(-2)+f(3)=14$, then $\lambda$ is equal to :
Let $\alpha, \beta$ and $\gamma$ be three positive real numbers. Let $f(x)=\alpha x^{5}+\beta x^{3}+\gamma x, x \in \mathbf{R}$ and $g: \mathbf{R} \rightarrow \mathbf{R}$ be such that $g(f(x))=x$ for all $x \in \mathbf{R}$. If $\mathrm{a}_{1}, \mathrm{a}_{2}, \mathrm{a}_{3}, \ldots, \mathrm{a}_{\mathrm{n}}$ be in arithmetic progression with mean zero, then the value of $f\left(g\left(\frac{1}{\mathrm{n}} \sum\limits_{i=1}^{\mathrm{n}} f\left(\mathrm{a}_{i}\right)\right)\right)$ is equal to :
Let $f, g: \mathbb{N}-\{1\} \rightarrow \mathbb{N}$ be functions defined by $f(a)=\alpha$, where $\alpha$ is the maximum of the powers of those primes $p$ such that $p^{\alpha}$ divides $a$, and $g(a)=a+1$, for all $a \in \mathbb{N}-\{1\}$. Then, the function $f+g$ is
The number of bijective functions $f:\{1,3,5,7, \ldots, 99\} \rightarrow\{2,4,6,8, \ldots .100\}$, such that $f(3) \geq f(9) \geq f(15) \geq f(21) \geq \ldots . . f(99)$, is ____________.
The total number of functions,
$ f:\{1,2,3,4\} \rightarrow\{1,2,3,4,5,6\} $ such that $f(1)+f(2)=f(3)$, is equal to :
Let a function f : N $\to$ N be defined by
$f(n) = \left[ {\matrix{ {2n,} & {n = 2,4,6,8,......} \cr {n - 1,} & {n = 3,7,11,15,......} \cr {{{n + 1} \over 2},} & {n = 1,5,9,13,......} \cr } } \right.$
then, f is
Let f : R $\to$ R be defined as f (x) = x $-$ 1 and g : R $-$ {1, $-$1} $\to$ R be defined as $g(x) = {{{x^2}} \over {{x^2} - 1}}$.
Then the function fog is :
Let $f(x) = {{x - 1} \over {x + 1}},\,x \in R - \{ 0, - 1,1\} $. If ${f^{n + 1}}(x) = f({f^n}(x))$ for all n $\in$ N, then ${f^6}(6) + {f^7}(7)$ is equal to :
Let f : N $\to$ R be a function such that $f(x + y) = 2f(x)f(y)$ for natural numbers x and y. If f(1) = 2, then the value of $\alpha$ for which
$\sum\limits_{k = 1}^{10} {f(\alpha + k) = {{512} \over 3}({2^{20}} - 1)} $
holds, is :
Let $f:R \to R$ and $g:R \to R$ be two functions defined by $f(x) = {\log _e}({x^2} + 1) - {e^{ - x}} + 1$ and $g(x) = {{1 - 2{e^{2x}}} \over {{e^x}}}$. Then, for which of the following range of $\alpha$, the inequality $f\left( {g\left( {{{{{(\alpha - 1)}^2}} \over 3}} \right)} \right) > f\left( {g\left( {\alpha -{5 \over 3}} \right)} \right)$ holds ?
For $\mathrm{p}, \mathrm{q} \in \mathbf{R}$, consider the real valued function $f(x)=(x-\mathrm{p})^{2}-\mathrm{q}, x \in \mathbf{R}$ and $\mathrm{q}>0$. Let $\mathrm{a}_{1}$, $\mathrm{a}_{2^{\prime}}$ $\mathrm{a}_{3}$ and $\mathrm{a}_{4}$ be in an arithmetic progression with mean $\mathrm{p}$ and positive common difference. If $\left|f\left(\mathrm{a}_{i}\right)\right|=500$ for all $i=1,2,3,4$, then the absolute difference between the roots of $f(x)=0$ is ___________.
Explanation:
$\because$ ${a_1},{a_2},{a_3},{a_4}$
$\therefore$ ${a_2} = p - 3d,\,{a_2} = p - d,\,{a_3} = p + d$ and ${a_4} = p + 3d$
Where $d > 0$
$\because$ $\left| {f({a_i})} \right| = 500$
$ \Rightarrow |9{d^2} - q| = 500$
and $|{d^2} - q| = 500$ ..... (i)
either $9{d^2} - q = {d^2} - q$
$ \Rightarrow d = 0$ not acceptable
$\therefore$ $9{d^2} - q = q - {d^2}$
$\therefore$ $5{d^2} - q = 0$ ..... (ii)
Roots of $f(x) = 0$ are $p + \sqrt q $ and $p - \sqrt q $
$\therefore$ absolute difference between roots $ = \left| {2\sqrt q } \right| = 50$
The number of functions $f$, from the set $\mathrm{A}=\left\{x \in \mathbf{N}: x^{2}-10 x+9 \leq 0\right\}$ to the set $\mathrm{B}=\left\{\mathrm{n}^{2}: \mathrm{n} \in \mathbf{N}\right\}$ such that $f(x) \leq(x-3)^{2}+1$, for every $x \in \mathrm{A}$, is ___________.
Explanation:
$A = \left\{ {\matrix{ {x \in N,} & {{x^2} - 10x + 9 \le 0} \cr } } \right\}$
$ = \{ 1,2,3,\,....,\,9\} $
$B = \{ 1,4,9,16,\,.....\} $
$f(x) \le {(x - 3)^2} + 1$
$f(1) \le 5,\,f(2) \le 2,\,\,..........\,f(9) \le 37$
$x = 1$ has 2 choices
$x = 2$ has 1 choice
$x = 3$ has 1 choice
$x = 4$ has 1 choice
$x = 5$ has 2 choices
$x = 6$ has 3 choices
$x = 7$ has 4 choices
$x = 8$ has 5 choices
$x = 9$ has 6 choices
$\therefore$ Total functions = $2\times1\times1\times1\times2\times3\times4\times5\times6=1440$
Let $f(x)=2 x^{2}-x-1$ and $\mathrm{S}=\{n \in \mathbb{Z}:|f(n)| \leq 800\}$. Then, the value of $\sum\limits_{n \in S} f(n)$ is equal to ___________.
Explanation:
$\because$ $\left| {f(n)} \right| \le 800$
$ \Rightarrow - 800 \le 2{n^2} - n - 1 \le 800$
$ \Rightarrow 2{n^2} - n - 801 \le 0$
$\therefore$ $n \in \left[ {{{ - \sqrt {6409} + 1} \over 4},{{\sqrt {6409} + 1} \over 4}} \right]$ and $n \in z$
$\therefore$ $n = - 19, - 18, - 17,\,..........,\,19,20.$
$\therefore$ $\sum {\left( {2{x^2} - x - 1} \right) = 2\sum {{x^2} - \sum {x - \sum 1 } } } $.
$ = 2\,.\,2\,.\,\left( {{1^2} + {2^2}\, + \,...\, + \,{{19}^2}} \right) + 2\,.\,{20^2} - 20 - 40$
$ = 10620$
Let $f(x)$ be a quadratic polynomial with leading coefficient 1 such that $f(0)=p, p \neq 0$, and $f(1)=\frac{1}{3}$. If the equations $f(x)=0$ and $f \circ f \circ f \circ f(x)=0$ have a common real root, then $f(-3)$ is equal to ________________.
Explanation:
Let $f(x) = (x - \alpha )(x - \beta )$
It is given that $f(0) = p \Rightarrow \alpha \beta = p$
and $f(1) = {1 \over 3} \Rightarrow (1 - \alpha )(1 - \beta ) = {1 \over 3}$
Now, let us assume that, $\alpha$ is the common root of $f(x) = 0$ and $fofofof(x) = 0$
$fofofof(x) = 0$
$ \Rightarrow fofof(0) = 0$
$ \Rightarrow fof(p) = 0$
So, $f(p)$ is either $\alpha$ or $\beta$.
$(p - \alpha )(p - \beta ) = \alpha $
$(\alpha \beta - \alpha )(\alpha \beta - \beta ) = \alpha \Rightarrow (\beta - 1)(\alpha - 1)\beta = 1$ ($\because$ $\alpha \ne 0$)
So, $\beta = 3$
$(1 - \alpha )(1 - 3) = {1 \over 3}$
$\alpha = {7 \over 6}$
$f(x) = \left( {x - {7 \over 6}} \right)(x - 3)$
$f( - 3) = \left( { - 3 - {7 \over 6}} \right)(3 - 3) = 25$
Let f(x) and g(x) be two real polynomials of degree 2 and 1 respectively. If $f(g(x)) = 8{x^2} - 2x$ and $g(f(x)) = 4{x^2} + 6x + 1$, then the value of $f(2) + g(2)$ is _________.
Explanation:
let $f(x)=c x^{2}+d x+e$
$g(x)=a x+b$
$f(g(x))=c(a x+b)^{2}+d(a x+b)+e \equiv 8 x^{2}-2 x$
$g(f(x))=a\left(c x^{2}+d x+e\right)+b \equiv 4 x^{2}+6 x+1$
$\therefore \quad a c=4 \quad a d=6 \quad a e+b=1$
$a^{2} c=8 \quad 2 a b c+a d=-2 \quad c b^{2}+b d+e=0$
By solving
$a=2 \quad b=-1$
$c=2 \quad d=3 \quad e=1$
$\therefore \quad f(x)=2 x^{2}+3 x+1$
$g(x)=2 x-1$
$f(2)+g(2)=2(2)^{2}+3(2)+1+2(2)-1$
$=18$
Let c, k $\in$ R. If $f(x) = (c + 1){x^2} + (1 - {c^2})x + 2k$ and $f(x + y) = f(x) + f(y) - xy$, for all x, y $\in$ R, then the value of $|2(f(1) + f(2) + f(3) + \,\,......\,\, + \,\,f(20))|$ is equal to ____________.
Explanation:
f(x) is polynomial
Put y = 1/x in given functional equation we get
$f\left( {x + {1 \over x}} \right) = f(x) + f\left( {{1 \over x}} \right) - 1$
$ \Rightarrow (c + 1){\left( {x + {1 \over x}} \right)^2} + (1 - {c^2})\left( {x + {1 \over x}} \right) + 2K$
$ = (c + 1){x^2} + (1 - {c^2})x + 2K + (c + 1){1 \over {{x^2}}} + (1 - {c^2}){1 \over x} + 2K - 1$
$ \Rightarrow 2(c + 1) = 2K - 1$ ..... (1)
and put $x = y = 0$ we get
$f(0) = 2 + f(0) - 0 \Rightarrow f(0) = 0 \Rightarrow k = 0$
$\therefore$ $k = 0$ and $2c = - 3 \Rightarrow c = - 3/2$
$f(x) = - {{{x^2}} \over 2} - {{5x} \over 4} = {1 \over 4}(5x + 2{x^2})$
$\left| {2\sum\limits_{i = 1}^{20} {f(i)} } \right| = \left| {{{ - 2} \over 4}\left( {{{5.20.21} \over 2} + {{2.20.21.41} \over 6}} \right)} \right|$
$ = \left| {{{ - 1} \over 2}(2730 + 5740)} \right|$
$ = \left| { - {{6790} \over 2}} \right| = 3395$.
Let S = {1, 2, 3, 4}. Then the number of elements in the set { f : S $\times$ S $\to$ S : f is onto and f (a, b) = f (b, a) $\ge$ a $\forall$ (a, b) $\in$ S $\times$ S } is ______________.
Explanation:
$A=\{(1,1)\}$
$B=\{(1,4),(2,4),(3,4)(4,4),(4,3),(4,2),(4,1)\}$
$C=\{(1,3),(2,3),(3,3),(3,2),(3,1)\}$
$D=\{(1,2),(2,2),(2,1)\}$
All elements of set $B$ have image 4 and only element of $A$ has image 1.
All elements of set $C$ have image 3 or 4 and all elements of set $D$ have image 2 or 3 or 4 .
We will solve this question in two cases.
Case I: When no element of set $C$ has image 3.
Number of onto functions $=2$ (when elements of set $D$ have images 2 or 3$)$
Case II: When atleast one element of set $C$ has image 3.
Number of onto functions $=\left(2^{3}-1\right)(1+2+2)$
$ =35 $
Total number of functions $=37$
Let S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}. Define f : S $\to$ S as
$f(n) = \left\{ {\matrix{ {2n} & , & {if\,n = 1,2,3,4,5} \cr {2n - 11} & , & {if\,n = 6,7,8,9,10} \cr } } \right.$.
Let g : S $\to$ S be a function such that $fog(n) = \left\{ {\matrix{ {n + 1} & , & {if\,n\,\,is\,odd} \cr {n - 1} & , & {if\,n\,\,is\,even} \cr } } \right.$.
Then $g(10)g(1) + g(2) + g(3) + g(4) + g(5))$ is equal to _____________.
Explanation:
$\because$ $f(n) = \left\{ {\matrix{ {2n,} & {n = 1,2,3,4,5} \cr {2n - 11,} & {n = 6,7,8,9,10} \cr} } \right.$
$\therefore$ f(1) = 2, f(2) = 4, ......, f(5) = 10
and f(6) = 1, f(7) = 3, f(8) = 5, ......, f(10) = 9
Now, $f(g(n)) = \left\{ {\matrix{ {n + 1,} & {if\,n\,is\,odd} \cr {n - 1,} & {if\,n\,is\,even} \cr } } \right.$
$\therefore$ $\matrix{ {f(g(10)) = 9} & { \Rightarrow g(10) = 10} \cr {f(g(1)) = 2} & { \Rightarrow g(1) = 1} \cr {f(g(2)) = 1} & { \Rightarrow g(2) = 6} \cr {f(g(3)) = 4} & { \Rightarrow g(3) = 2} \cr {f(g(4)) = 3} & { \Rightarrow g(4) = 7} \cr {f(g(5)) = 6} & { \Rightarrow g(5) = 3} \cr } $
$\therefore$ $g(10)g(1) + g(2) + g(3) + g(4) + g(5)) = 190$
Let f : R $\to$ R be a function defined by $f(x) = {{2{e^{2x}}} \over {{e^{2x}} + e}}$. Then $f\left( {{1 \over {100}}} \right) + f\left( {{2 \over {100}}} \right) + f\left( {{3 \over {100}}} \right) + \,\,\,.....\,\,\, + \,\,\,f\left( {{{99} \over {100}}} \right)$ is equal to ______________.
Explanation:
Given,
$f(x) = {{2{e^{2x}}} \over {{e^{2x}} + e}}$
$\therefore$ $f(1 - x) = {{2{e^{2(1 - x)}}} \over {{e^{2(1 - x)}} + e}}$
$ = {{2\,.\,{{{e^2}} \over {{e^{2x}}}}} \over {{{{e^2}} \over {{e^{2x}}}} + e}}$
$ = {{2{e^2}} \over {{e^2} + {e^{2x}}\,.\,e}}$
$ = {{2{e^2}} \over {e(e + {e^{2x}})}}$
$ = {{2e} \over {e + {e^{2x}}}}$
$\therefore$ $f(x) + f(1 - x) = {{2{e^{2x}}} \over {{e^{2x}} + e}} + {{2e} \over {{e^{2x}} + e}}$
$ = {{2({e^{2x}} + e)} \over {{e^{2x}} + e}}$
$ = 2$ ...... (1)
Now,
$f\left( {{1 \over {100}}} \right) + f\left( {{{99} \over {100}}} \right)$
$ = f\left( {{1 \over {100}}} \right) + f\left( {1 - {1 \over {100}}} \right)$
$ = 2$ [as $f(x) + f(1 - x) = 2$]
Similarly,
$f\left( {{2 \over {100}}} \right) + f\left( {1 - {2 \over {100}}} \right) = 2$
$ \vdots $
$f\left( {{{49} \over {100}}} \right) + f\left( {1 - {{49} \over {100}}} \right) = 2$
$\therefore$ Total sum $ = 49 \times 2$
Remaining term $ = f\left( {{{50} \over {100}}} \right) = f\left( {{1 \over 2}} \right)$
Put $x = {1 \over 2}$ in equation (1), we get
$f\left( {{1 \over 2}} \right) + f\left( {1 - {1 \over 2}} \right) = 2$
$ \Rightarrow 2f\left( {{1 \over 2}} \right) = 2$
$ \Rightarrow f\left( {{1 \over 2}} \right) = 1$
$\therefore$ Sum $ = 49 \times 2 + 1 = 99$
Let $f:R \to R$ be a function defined by
$f(x) = {\left( {2\left( {1 - {{{x^{25}}} \over 2}} \right)(2 + {x^{25}})} \right)^{{1 \over {50}}}}$. If the function $g(x) = f(f(f(x))) + f(f(x))$, then the greatest integer less than or equal to g(1) is ____________.
Explanation:
Given,
$f(x) = {\left( {2\left( {1 - {{{x^{25}}} \over 2}} \right)\left( {2 + {x^{25}}} \right)} \right)^{{1 \over {50}}}}$
and $g(x) = f\left( {f\left( {f\left( x \right)} \right)} \right) + f\left( {f\left( x \right)} \right)$
$\therefore$ $g(1) = f\left( {f\left( {f\left( 1 \right)} \right)} \right) + f\left( {f\left( 1 \right)} \right)$
Now, $f(1) = {\left( {2\left( {1 - {{{1^{25}}} \over 2}} \right)\left( {2 + {1^{25}}} \right)} \right)^{{1 \over {50}}}}$
$ = {\left( {2\left( {1 - {1 \over 2}} \right)\left( {2 + 1} \right)} \right)^{{1 \over {50}}}}$
$ = {\left( 3 \right)^{{1 \over {50}}}}$
$\therefore$ $f\left( {f\left( 1 \right)} \right) = f\left( {{3^{{1 \over {50}}}}} \right)$
$ = {\left( {2\left( {1 - {{{{\left( {{3^{{1 \over {50}}}}} \right)}^{25}}} \over 2}} \right)\left( {2 + {{\left( {{3^{{1 \over {50}}}}} \right)}^{25}}} \right)} \right)^{{1 \over {50}}}}$
$ = {\left( {2\left( {1 - {{{3^{{1 \over 2}}}} \over 2}} \right)\left( {2 + {3^{{1 \over 2}}}} \right)} \right)^{{1 \over {50}}}}$
$ = {\left( {2 \times \left( {{{2 - \sqrt 3 } \over 2}} \right)\left( {2 + \sqrt 3 } \right)} \right)^{{1 \over {50}}}}$
$ = {\left[ {\left( {2 - \sqrt 3 } \right)\left( {2 + \sqrt 3 } \right)} \right]^{{1 \over {50}}}}$
$ = {\left( {4 - 3} \right)^{{1 \over {50}}}}$
$ = {1^{{1 \over {50}}}} = 1$
Now, $f\left( {f\left( {f\left( 1 \right)} \right)} \right) = f(1) = {3^{{1 \over {50}}}}$
$\therefore$ $g(1) = f\left( {f\left( {f\left( 1 \right)} \right)} \right) + f\left( {f\left( 1 \right)} \right)$
$ = {3^{{1 \over {50}}}} + 1$
Now, greatest integer less than or equal to $g(1)$
$ = \left[ {g(1)} \right]$
$ = \left[ {{3^{{1 \over {50}}}} + 1} \right]$
$ = \left[ {{3^{{1 \over {50}}}}} \right] + \left[ 1 \right]$
$ = [1.02] + 1$
$ = 1 + 1 = 2$
The number of one-one functions f : {a, b, c, d} $\to$ {0, 1, 2, ......, 10} such
that 2f(a) $-$ f(b) + 3f(c) + f(d) = 0 is ___________.
Explanation:
Given one-one function
$f:\{ a,b,c,d\} \to \{ 0,1,2,\,\,....\,\,10\} $
and $2f(a) - f(b) + 3f(c) + f(d) = 0$
$ \Rightarrow 3f(c) + 2f(a) + f(d) = f(b)$
Case I:
(1) Now let $f(c) = 0$ and $f(a) = 1$ then
$3 \times 0 + 2 \times 1 + f(d) = f(b)$
$ \Rightarrow 2 + f(d) = f(b)$
Now possible value of $f(d) = 2,3,4,5,6,7,$ and $8$.
f(d) can't be 9 and 10 as if $f(d) = 9$ or 10 then $f(b) = 2 + 9 = 11$ or $f(b) = 2 + 10 = 12$, which is not possible as here any function's maximum value can be 10.
$\therefore$ Total possible functions when $f(c) = 0$ and $f(a) = 1$ are = 7
(2) When $f(c) = 0$ and $f(a) = 2$ then
$3 \times 0 + 2 \times 2 + f(d) = f(b)$
$ \Rightarrow 4 + f(d) = f(b)$
$\therefore$ possible value of $f(d) = 1,3,4,5,6$
$\therefore$ Total possible functions in this case = 5
(3) When $f(c) = 0$ and $f(a) = 3$ then
$3 \times 0 + 2 \times 3 + f(d) = f(b)$
$ \Rightarrow 6 + f(d) = f(b)$
$\therefore$ Possible value of $f(d) = 1,2,4$
$\therefore$ Total possible functions in this case = 3
(4) When $f(c) = 0$ and $f(a) = 4$ then
$3 \times 0 + 2 \times 4 + f(d) = f(b)$
$ \Rightarrow 8 + f(d) = f(b)$
$\therefore$ Possible value of $f(d) = 1,2$
$\therefore$ Total possible functions in this case = 2
(5) When $f(c) = 0$ and $f(a) = 5$ then
$3 \times 0 + 2 \times 5 + f(d) = f(b)$
$ \Rightarrow 10 + f(d) = f(b)$
Possible value of f(d) can be 0 but f(c) is already zero. So, no value to f(d) can satisfy.
$\therefore$ No function is possible in this case.
$\therefore$ Total possible functions when $f(c) = 0$ and $f(a) = 1,2,3$ and $4$ are $ = 7 + 5 + 3 + 2 = 17$
Case II:
(1) When $f(c) = 1$ and $f(a) = 0$ then
$3 \times 1 + 2 \times 0 + f(d) = f(b)$
$ \Rightarrow 3 + f(d) = f(b)$
$\therefore$ Possible value of $f(d) = 2,3,4,5,6,7$
$\therefore$ Total possible functions in this case = 6
(2) When $f(c) = 1$ and $f(a) = 2$ then
$3 \times 1 + 2 \times 2 + f(d) = f(b)$
$ \Rightarrow 7 + f(d) = f(b)$
$\therefore$ Possible value of $f(d) = 0,3$
$\therefore$ Total possible functions in this case = 2
(3) When $f(c) = 1$ and $f(a) = 3$ then
$3 \times 1 + 2 \times 3 + f(d) = f(b)$
$ \Rightarrow 9 + f(d) = f(b)$
$\therefore$ Possible value of $f(d) = 0$
$\therefore$ Total possible functions in this case = 1
$\therefore$ Total possible functions when $f(c) = 1$ and $f(a) = 0,2$ and $3$ are
$ = 6 + 2 + 1 = 9$
Case III:
(1) When $f(c) = 2$ and $f(a) = 0$ then
$3 \times 2 + 2 \times 0 + f(d) = f(b)$
$ \Rightarrow 6 + f(d) = f(b)$
$\therefore$ Possible values of $f(d) = 1,3,4$
$\therefore$ Total possible functions in this case = 3
(2) When $f(c) = 2$ and $f(a) = 1$ then,
$3 \times 2 + 2 \times 1 + f(d) = f(b)$
$ \Rightarrow 8 + f(d) = f(b)$
$\therefore$ Possible values of $f(d) = 0$
$\therefore$ Total possible function in this case = 1
$\therefore$ Total possible functions when $f(c) = 2$ and $f(a) = 0,1$ are
$ = 3 + 1 = 4$
Case IV:
(1) When $f(c) = 3$ and $f(a) = 0$ then
$3 \times 3 + 2 \times 0 + f(d) = f(b)$
$ \Rightarrow 9 + f(d) = f(b)$
$\therefore$ Possible values of $f(d) = 1$
$\therefore$ Total one-one functions from four cases
$ = 17 + 9 + 4 + 1 = 31$
$f(x) = {\log _{\sqrt 5 }}\left( {3 + \cos \left( {{{3\pi } \over 4} + x} \right) + \cos \left( {{\pi \over 4} + x} \right) + \cos \left( {{\pi \over 4} - x} \right) - \cos \left( {{{3\pi } \over 4} - x} \right)} \right)$ is :
g(3n + 1) = 3n + 2,
g(3n + 2) = 3n + 3,
g(3n + 3) = 3n + 1, for all n $\ge$ 0.
Then which of the following statements is true?