Functions

$\begin{aligned} & f(x)=\cos ^{-1}\left(\frac{4 x+5}{3 x-7}\right) \\ & \Rightarrow-1 \leq\left(\frac{4 x+5}{3 x-7}\right) \leq 1 \\ & \left(\frac{4 x+5}{3 x-7}\right) \geq-1 \\ & \frac{4 x+5+3 x-7}{3 x-7} \geq 0 \\ & \Rightarrow \frac{7 x-2}{3 x-7} \geq 0 \end{aligned}$

$\begin{aligned} & x \in\left(-\infty, \frac{2}{7}\right] \cup\left(\frac{7}{3}, \infty\right) \\ & \& \frac{4 x+5}{3 x-7} \leq 1 \Rightarrow \frac{x+12}{3 x-7} \leq 0 \end{aligned}$

$\therefore$ Domain of $\mathrm{f}(\mathrm{x})$ is

$\left[-12, \frac{2}{7}\right] \alpha=-12, \beta=\frac{2}{7}$

$g(x)=\log _2\left(2-6 \log _{27}(2 x+5)\right)$

Domain

$2-6 \log _{27}(2 x+5)>0$

$\begin{array}{ll} \Rightarrow & 6 \log _{27}(2 \mathrm{x}+5)<2 \\ \Rightarrow & \log _{27}(2 \mathrm{x}+5)<\frac{1}{3} \\ \Rightarrow & 2 \mathrm{x}+5<3 \\ \Rightarrow & \mathrm{x}<-1 \end{array}$

$\& 2 x+5>0 \Rightarrow x>-\frac{5}{2}$

Domain is $\mathrm{x} \in\left(-\frac{5}{2},-1\right)$

$\begin{aligned} &\gamma=-\frac{5}{2}, \delta=-1\\ &\begin{aligned} & |7(\alpha+\beta)+4(\gamma+\delta)|=\left\lvert\, 7\left(\left.-12+\frac{2}{7}+4\left(-\frac{5}{2}-1\right) \right\rvert\,\right.\right. \\ & |-82-14|=96 \end{aligned} \end{aligned}$

$\begin{aligned} & A=\{(x, y) ; 2 x+3 y=23, x, y \in N\} \\ & A=\{(1,7),(4,5),(7,3),(10,1)\} \\ & B=\{x:(x, y) \in A\} \\ & B=\{1,4,7,10\} \end{aligned}$

So, total number of one-one functions from A to B is $4!=24$

$\begin{aligned} & f(m+n)=f(m)+f(n) \\ & f(x)=k x \\ & \because f(1)=1 \\ & \Rightarrow k=1 \\ & \Rightarrow f(x)=x \end{aligned}$

$\begin{aligned} & \sum_{k=1}^{2022} f(\lambda+k)=\sum_{k=1}^{2022}(\lambda+k)=\underbrace{\lambda+\lambda+\ldots+\lambda}_{2022}+(1+2+\ldots+2022) \\ & =2022 \lambda+\frac{2022 \times 2023}{2} \leq(2022)^2 \\ & \Rightarrow \lambda \leq \frac{2021}{2} \\ & \text { largest } \lambda=1010 \end{aligned}$

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.

$\begin{aligned} & S:\{a \in R:|2 a-1|=3[a]+2\{a\}\} \\ & |2 a-1|=3[a]+2(a-[a]) \\ & |2 a-1|=[a]+2 a \end{aligned}$

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 $a=\frac{1}{4}$

$72 \sum_\limits{a \in S} a=72 \times \frac{1}{4}=18$

$f(x)=\frac{2 x}{\sqrt{1+9 x^2}}$

$(f \circ f)(x)=\frac{2 f(x)}{\sqrt{1+9(f(x))^2}}=\frac{\frac{4 x}{\sqrt{1+9 x^2}}}{\sqrt{1+9 \times \frac{4 x^2}{1+9 x^2}}}=\frac{4 x}{\sqrt{1+45 x^2}}$

$(f \circ f \circ f)(x)=\frac{4 \times \frac{2 x}{\sqrt{1+9 x^2}}}{\sqrt{1+45 \times \frac{4 x^2}{1+9 x^2}}}=\frac{8 x}{\sqrt{1+21 \times 9 x^2}}$

$(f \circ f \circ f \circ f)(x)=\frac{16 x}{\sqrt{1+85 \times 9 x^2}}$

$\Rightarrow \alpha$ is $10^{\text {th }}$ term of $1,5,21,85, \ldots \alpha$ is $10^{\text {th }}$ term of

$\begin{aligned} & \frac{\left(2^1\right)^2-1}{3}, \frac{\left(2^2\right)^2-1}{3}, \frac{\left(2^3\right)^2-1}{3}, \frac{\left(2^4\right)^2-1}{3}, \ldots \\ \Rightarrow \quad & \alpha=\frac{\left(2^{10}\right)^2-1}{3} \\ \Rightarrow \quad & \sqrt{3 \alpha+1}=2^{10}=1024 \end{aligned}$

$\begin{aligned} & f: A \rightarrow P(A) \\ & a \in f(a) \end{aligned}$

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. $2+42=44$

Given that the function $f : A \rightarrow B$ satisfies the condition $f(1) + f(2) = f(4) - 1$, where the set $A = \{1, 2, 3, 4, 5\}$ and the set $B = \{1, 2, 3, 4, 5, 6\}$.

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.

Total number of onto functions

$ \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

$=240-60=180$

Domain of $\log _e\left(\frac{6 x^2+5 x+1}{2 x-1}\right)$

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} $

$f(1.n)=f(1).f(n)\Rightarrow f(1)=1$.

$f(3.3)=(f(3))^2$

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

$6^3\times2=432$

$\because S={1,2,3,4,5,6}$ 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

$\because f(1)=\frac{1}{5} ~\therefore f(2)=f(1)+f(1)=\frac{2}{5}$

$f(2)=\frac{2}{5} \quad\quad f(3)=f(2)+f(1)=\frac{3}{5}$

$f(3)=\frac{3}{5}$

$\therefore \sum\limits_{n=1}^{m} \frac{f(n)}{n(n+1)(n+2)}$

$=\frac{1}{5} \sum\limits_{n=1}^{m}\left(\frac{1}{n+1}-\frac{1}{n+2}\right)$

$=\frac{1}{5}\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\ldots .+\frac{1}{m+1}-\frac{1}{m+2}\right)$

$=\frac{1}{5}\left(\frac{1}{2}-\frac{1}{m+2}\right)=\frac{m}{10(m+2)}=\frac{1}{12}$

$\therefore m=10$

$f(x)=a x-3$

$g(x)=x^{b}+c$

$(fog)^{-1}=\left(\frac{x-7}{2}\right)^{\frac{1}{3}}$

$(fog)^{-1}(x)=\left(\frac{x+3-c a}{a}\right)^{\frac{1}{b}}=\left(\frac{x-7}{2}\right)^{\frac{1}{3}}$

$\Rightarrow a=2, b=3, c=5$

$fog(a c)+gof(b)$

$\because f(x)=2 x-3$

$g(x)=x^{3}+5$

$fog(10)+g o f(3)$

$=2007+32$

$=2039$

$\because$ ${a_1},{a_2},{a_3},{a_4}$

$\therefore$ ${a_2} = p - 3d,\,{a_2} = p - d,\,{a_3} = p + d$ 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$

$A = \left\{ {\matrix{ {x \in N,} & {{x^2} - 10x + 9 \le 0} \cr } } \right\}$

$ = \{ 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 = $2\times1\times1\times1\times2\times3\times4\times5\times6=1440$

$\because$ $\left| {f(n)} \right| \le 800$

$ \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$