Functions
If a function $f$ satisfies $f(\mathrm{~m}+\mathrm{n})=f(\mathrm{~m})+f(\mathrm{n})$ for all $\mathrm{m}, \mathrm{n} \in \mathbf{N}$ and $f(1)=1$, then the largest natural number $\lambda$ such that $\sum_\limits{\mathrm{k}=1}^{2022} f(\lambda+\mathrm{k}) \leq(2022)^2$ is equal to _________.
Explanation:
$\begin{aligned} & f(m+n)=f(m)+f(n) \\ & f(x)=k x \\ & \because f(1)=1 \\ & \Rightarrow k=1 \\ & \Rightarrow f(x)=x \end{aligned}$
$\begin{aligned} & \sum_{k=1}^{2022} f(\lambda+k)=\sum_{k=1}^{2022}(\lambda+k)=\underbrace{\lambda+\lambda+\ldots+\lambda}_{2022}+(1+2+\ldots+2022) \\ & =2022 \lambda+\frac{2022 \times 2023}{2} \leq(2022)^2 \\ & \Rightarrow \lambda \leq \frac{2021}{2} \\ & \text { largest } \lambda=1010 \end{aligned}$
If the range of $f(\theta)=\frac{\sin ^4 \theta+3 \cos ^2 \theta}{\sin ^4 \theta+\cos ^2 \theta}, \theta \in \mathbb{R}$ is $[\alpha, \beta]$, then the sum of the infinite G.P., whose first term is 64 and the common ratio is $\frac{\alpha}{\beta}$, is equal to __________.
Explanation:
To determine the range of the function $f(\theta)=\frac{\sin ^4 \theta+3 \cos ^2 \theta}{\sin ^4 \theta+\cos ^2 \theta}$, let's start by simplifying the expression. Let $\sin^2 \theta = x$, so $\cos^2 \theta = 1 - x$. The function then transforms into:
$ f(x) = \frac{x^2 + 3(1-x)}{x^2 + (1-x)} $
Simplify the numerator and denominator separately:
Numerator: $ x^2 + 3 - 3x $
Denominator: $ x^2 + 1 - x $
Thus, the function becomes:
$ f(x) = \frac{x^2 + 3 - 3x}{x^2 + 1 - x} = \frac{x^2 - 3x + 3}{x^2 - x + 1} $
Next, we need to find the range of this function. Let's analyze the function by testing specific values of $x$ in the interval $[0, 1]$ (since $\sin^2 \theta$ ranges from 0 to 1):
When $x = 0$:
$ f(0) = \frac{0^2 - 3(0) + 3}{0^2 - 0 + 1} = \frac{3}{1} = 3 $
When $x = 1$:
$ f(1) = \frac{1^2 - 3(1) + 3}{1^2 - 1 + 1} = \frac{1 - 3 + 3}{1 - 1 + 1} = \frac{1}{1} = 1 $
It appears that $f(x)$ achieves values within $[1, 3]$. To confirm this, we need to solve the quadratic inequality:
$ 1 \leq \frac{x^2 - 3x + 3}{x^2 - x + 1} \leq 3 $
By solving the inequalities, it can be confirmed that the function indeed ranges from 1 to 3 on the interval [0,1]. Hence, we have:
$ \alpha = 1 $
$ \beta = 3 $
The common ratio of the infinite geometric progression is:
$ \frac{\alpha}{\beta} = \frac{1}{3} $
Given the first term $a = 64$, the sum $S$ of the infinite geometric progression can be given as:
$ S = \frac{a}{1 - r} $
Substituting the values $a = 64$ and $r = \frac{1}{3}$, we get:
$ S = \frac{64}{1 - \frac{1}{3}} = \frac{64}{\frac{2}{3}} = 64 \times \frac{3}{2} = 96 $
Therefore, the sum of the infinite geometric progression is 96.
If $S=\{a \in \mathbf{R}:|2 a-1|=3[a]+2\{a \}\}$, where $[t]$ denotes the greatest integer less than or equal to $t$ and $\{t\}$ represents the fractional part of $t$, then $72 \sum_\limits{a \in S} a$ is equal to _________.
Explanation:
$\begin{aligned} & S:\{a \in R:|2 a-1|=3[a]+2\{a\}\} \\ & |2 a-1|=3[a]+2(a-[a]) \\ & |2 a-1|=[a]+2 a \end{aligned}$
Case I: If $0 < a < \frac{1}{2}$
$\begin{aligned} & 1-2 a=0+2 a \\ & \Rightarrow a=\frac{1}{4} \end{aligned}$
Case II: If $\frac{1}{2} < a < 1$
$2 a-1=0+2 a$
No solution
Case III: If $1 \leq a<2$
$2 a-1=1+2 a$
$\Rightarrow$ No solution
$\therefore$ only solution is $a=\frac{1}{4}$
$72 \sum_\limits{a \in S} a=72 \times \frac{1}{4}=18$
Consider the function $f: \mathbb{R} \rightarrow \mathbb{R}$ defined by $f(x)=\frac{2 x}{\sqrt{1+9 x^2}}$. If the composition of $f, \underbrace{(f \circ f \circ f \circ \cdots \circ f)}_{10 \text { times }}(x)=\frac{2^{10} x}{\sqrt{1+9 \alpha x^2}}$, then the value of $\sqrt{3 \alpha+1}$ is equal to _______.
Explanation:
$f(x)=\frac{2 x}{\sqrt{1+9 x^2}}$
$(f \circ f)(x)=\frac{2 f(x)}{\sqrt{1+9(f(x))^2}}=\frac{\frac{4 x}{\sqrt{1+9 x^2}}}{\sqrt{1+9 \times \frac{4 x^2}{1+9 x^2}}}=\frac{4 x}{\sqrt{1+45 x^2}}$
$(f \circ f \circ f)(x)=\frac{4 \times \frac{2 x}{\sqrt{1+9 x^2}}}{\sqrt{1+45 \times \frac{4 x^2}{1+9 x^2}}}=\frac{8 x}{\sqrt{1+21 \times 9 x^2}}$
$(f \circ f \circ f \circ f)(x)=\frac{16 x}{\sqrt{1+85 \times 9 x^2}}$
$\Rightarrow \alpha$ is $10^{\text {th }}$ term of $1,5,21,85, \ldots \alpha$ is $10^{\text {th }}$ term of
$\begin{aligned} & \frac{\left(2^1\right)^2-1}{3}, \frac{\left(2^2\right)^2-1}{3}, \frac{\left(2^3\right)^2-1}{3}, \frac{\left(2^4\right)^2-1}{3}, \ldots \\ \Rightarrow \quad & \alpha=\frac{\left(2^{10}\right)^2-1}{3} \\ \Rightarrow \quad & \sqrt{3 \alpha+1}=2^{10}=1024 \end{aligned}$
Let $\mathrm{A}=\{1,2,3, \ldots, 7\}$ and let $\mathrm{P}(\mathrm{A})$ denote the power set of $\mathrm{A}$. If the number of functions $f: \mathrm{A} \rightarrow \mathrm{P}(\mathrm{A})$ such that $\mathrm{a} \in f(\mathrm{a}), \forall \mathrm{a} \in \mathrm{A}$ is $\mathrm{m}^{\mathrm{n}}, \mathrm{m}$ and $\mathrm{n} \in \mathrm{N}$ and $\mathrm{m}$ is least, then $\mathrm{m}+\mathrm{n}$ is equal to _________.
Explanation:
$\begin{aligned} & f: A \rightarrow P(A) \\ & a \in f(a) \end{aligned}$
That means '$a$' will connect with subset which contain element '$a$'.
Total options for 1 will be $2^6$. (Because $2^6$ subsets contains 1)
Similarly, for every other element
Hence, total is $2^6 \times 2^6 \times 2^6 \times 2^6 \times 2^6 \times 2^6 \times 2^6=2^{42}$
Ans. $2+42=44$
The range of $f(x)=4 \sin ^{-1}\left(\frac{x^{2}}{x^{2}+1}\right)$ is
For $x \in \mathbb{R}$, two real valued functions $f(x)$ and $g(x)$ are such that, $g(x)=\sqrt{x}+1$ and $f \circ g(x)=x+3-\sqrt{x}$. Then $f(0)$ is equal to
Let $\mathrm{D}$ be the domain of the function $f(x)=\sin ^{-1}\left(\log _{3 x}\left(\frac{6+2 \log _{3} x}{-5 x}\right)\right)$. If the range of the function $\mathrm{g}: \mathrm{D} \rightarrow \mathbb{R}$ defined by $\mathrm{g}(x)=x-[x],([x]$ is the greatest integer function), is $(\alpha, \beta)$, then $\alpha^{2}+\frac{5}{\beta}$ is equal to
The domain of the function $f(x)=\frac{1}{\sqrt{[x]^{2}-3[x]-10}}$ is : ( where $[\mathrm{x}]$ denotes the greatest integer less than or equal to $x$ )
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 } $
$\therefore$ $g(10)g(1) + g(2) + g(3) + g(4) + g(5)) = 190$