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$

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, $\beta = 3$

$(1 - \alpha )(1 - 3) = {1 \over 3}$

$\alpha = {7 \over 6}$

$f(x) = \left( {x - {7 \over 6}} \right)(x - 3)$

$f( - 3) = \left( { - 3 - {7 \over 6}} \right)(3 - 3) = 25$

$f(g(x))=8 x^{2}-2 x$ $ g(f(x))=4 x^{2}+6 x+1 $

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

$a=2 \quad b=-1$

$c=2 \quad d=3 \quad e=1$

$\therefore \quad f(x)=2 x^{2}+3 x+1$

$g(x)=2 x-1$

$f(2)+g(2)=2(2)^{2}+3(2)+1+2(2)-1$

$=18$

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$.

There are 16 ordered pairs in $S \times S$. We write all these ordered pairs in 4 sets as follows.

$A=\{(1,1)\}$

$B=\{(1,4),(2,4),(3,4)(4,4),(4,3),(4,2),(4,1)\}$

$C=\{(1,3),(2,3),(3,3),(3,2),(3,1)\}$

$D=\{(1,2),(2,2),(2,1)\}$

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 $=37$

$\because$ $f(n) = \left\{ {\matrix{ {2n,} & {n = 1,2,3,4,5} \cr {2n - 11,} & {n = 6,7,8,9,10} \cr} } \right.$

$\therefore$ 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 } $

$\therefore$ $g(10)g(1) + g(2) + g(3) + g(4) + g(5)) = 190$

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$

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$

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(mn) = f(m) . f(n)

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

Put m = n = 2

$f(4) = f(2).f(2)\left\{ \matrix{ f(2) = 1 \Rightarrow f(4) = 1 \hfill \cr or \hfill \cr f(2) = 2 \Rightarrow f(4) = 4 \hfill \cr} \right.$

Put m = 2, n = 3

$f(6) = f(2).f(3)\left\{ \matrix{ when\,f(2) = 1 \hfill \cr f(3) = 1\,to\,7 \hfill \cr \hfill \cr f(2) = 2 \hfill \cr f(3) = 1\,or\,2\,or\,3 \hfill \cr} \right.$

f(5), f(7) can take any value

Total = (1 $\times$ 1 $\times$ 7 $\times$ 1 $\times$ 7 $\times$ 1 $\times$ 7) + (1 $\times$ 1 $\times$ 3 $\times$ 1 $\times$ 7 $\times$ 1 $\times$ 7)

= 490

f(1) + f(2) = 3 $-$ f(3)

$\Rightarrow$ f(1) + f(2) = 3 + f(3) = 3

The only possibility is : 0 + 1 + 2 = 3

$\Rightarrow$ Elements 1, 2, 3 in the domain can be mapped with 0, 1, 2 only.

So number of bijective functions.

$\left| \!{\underline {\, 3 \,}} \right. $ $\times$ $\left| \!{\underline {\, 5 \,}} \right. $ = 720

Given, p(x) = f(x³) + xg(x³)

We know, x² + x + 1 = (x $-$ $\omega$) (x $-$ $\omega$²)

Given, p(x) is divisible by x² + x + 1. So, roots of p(x) is $\omega$ and $\omega$².

As root satisfy the equation,

So, put x = $\omega$

p($\omega$) = f($\omega$³) + $\omega$g($\omega$³) = 0

= f(1) + $\omega$g(1) = 0 [$\omega$³ = 1]

= f(1) + $\left( { - {1 \over 2} + {{i\sqrt 3 } \over 2}} \right)$ g(1) = 0

$ \Rightarrow $ f(1) $-$ ${{g(1)} \over 2} + i\left( {{{\sqrt 3 g(1)} \over 2}} \right)$ = 0 + i0

Comparing both sides, we get

f(1) $-$ ${{g(1)} \over 2}$ = 0

and ${{{\sqrt 3 } \over 2}g(1) = 0}$ $ \Rightarrow $ g(1) = 0

So, f(1) = 0

Now, p(1) = f(1) + 1 . g(1) = 0 + 0 = 0

$af(x) + \alpha f\left( {{1 \over x}} \right) = bx + {\beta \over x}$ ........(i)

Replace $x $ with $ {1 \over x}$

$af\left( {{1 \over x}} \right) + af(x) = {b \over x} + \beta x$ ..... (ii)

(i) + (ii)

$(a + \alpha )\left[ {f(x) + f\left( {{1 \over x}} \right)} \right] = \left( {x + {1 \over x}} \right)(b + \beta )$

${{f(x) + f\left( {{1 \over x}} \right)} \over {x + {1 \over x}}} = {{\beta + b} \over {a + \alpha }} = {2 \over 1} = 2$

f(x + y) = f(x) f(y)

put x = y = 1
$ \therefore $ f(2) = (ƒ(1))² = 3²

put x = 2, y = 1
$ \therefore $ f(3) = (ƒ(1))³ = 3³

Similarly f(x) = 3^x
$ \Rightarrow $ f(i) = 3ⁱ

Given, $\sum\limits_{i = 1}^n {f(i)} = 363$

$ \Rightarrow $ 3 + 3² + 3³ +.... + 3ⁿ = 363

$ \Rightarrow $ ${{3\left( {{3^n} - 1} \right)} \over {3 - 1}}$ = 363

$ \Rightarrow $ 3ⁿ - 1 = ${{363 \times 2} \over 3}$ = 242

$ \Rightarrow $ 3ⁿ = 243 = 3⁵

$ \Rightarrow $ n = 5

The desired functions will contain either one element or two elements in its codomain of which '2' always belongs to f(A).

Case 1 : When 2 is the image of all element of set A.

Number of ways this is possible = 1

Case 2 : When one image is 2 and other one image is one of {1, 3, 4}.

Number of ways we can choose one of {1, 3, 4} is = ³C₁.

Now divide 3 elements {a, b, c} of set A into two parts.
We can do this ${{3!} \over {2!1!}}$ ways.

Now map one part of set A into the element 2 of set B and map other part of set A into one of {1, 3, 4} of set B.
We can do that 2! ways.

So number of functions in this case
= ³C₁ $ \times $ ${{3!} \over {2!1!}}$ $ \times $ 2! = 18

$ \therefore $ Total number of functions = 1 + 18 = 19

Function $ f $:

$ f(x) = \log_e\left(x^2 + 2x + 4\right) = \log_e\left[(x+1)^2 + 3\right] $

Function $ g $:

$ g(x) = \frac{4}{1 + e^{-2x}} $

To find the inverse of $ g $, $ g^{-1} $, we solve for $ x $:

Start with $ y = \frac{4}{1 + e^{-2x}} $.

Rearranging gives $ 1 + e^{-2x} = \frac{4}{y} $.

Solving for $ e^{-2x} $, we have $ e^{-2x} = \frac{4}{y} - 1 $.

Taking the natural logarithm, we find $ -2x = \ln\left(\frac{4}{y} - 1\right) $.

Thus, we solve for $ x $:

$ x = -\frac{1}{2} \ln\left(\frac{4-y}{y}\right) = \frac{1}{2} \ln\left(\frac{y}{4-y}\right) $

So, the inverse function:

$ g^{-1}(x) = \frac{1}{2} \ln\left(\frac{x}{4-x}\right) $

Now, let's evaluate $ g^{-1}(2) $:

$ g^{-1}(2) = \frac{1}{2} \ln\left(\frac{2}{4-2}\right) = \frac{1}{2} \ln(1) = 0 $

Thus, $ f(g^{-1}(x)) = f\left(\frac{1}{2} \ln\left(\frac{x}{4-x}\right)\right) $.

Next, we differentiate $ f \circ g^{-1} $:

$ \frac{d}{dx}\left(f(g^{-1}(x))\right) = \frac{d}{dx}\left[\log_e\left(\left(g^{-1}(x) + 1\right)^2 + 3\right)\right] $

Using the chain rule, the derivative is:

$ \frac{1}{\left[\left(g^{-1}(x) + 1\right)^2 + 3\right]} \cdot 2 \left(g^{-1}(x) + 1\right) \cdot \frac{d}{dx}\left(g^{-1}(x)\right) $

Calculate the derivative of $ g^{-1}(x) $:

$ \frac{d}{dx}\left(\frac{1}{2} \ln\left(\frac{x}{4-x}\right)\right) = \frac{1}{2} \cdot \frac{1}{\frac{x}{4-x}} \cdot \left(\frac{(4-x)x' + x(4-x)'}{(4-x)^2}\right) $

Evaluate at $ x = 2 $:

$ g^{-1}(2) = 0 $ means $ \left(g^{-1}(2) + 1\right)^2 + 3 = 4 $.

At $ x = 2 $:

$ \frac{d}{dx} f(g^{-1}(x))\big|_{x=2} = \frac{2}{4} \cdot \frac{2}{4} = \frac{1}{4} = 0.25 $

Therefore, the value of the derivative of the composite function $ f \circ g^{-1} $ at $ x = 2 $ is $ 0.25 $.

$\begin{aligned} & \mathrm{f}(\mathrm{x}+\mathrm{y})=\mathrm{f}(\mathrm{x})+\mathrm{f}(\mathrm{y}) \quad \text{... (1)}\\ & \Rightarrow \quad \mathrm{f}(\mathrm{nx})=\operatorname{nf}(\mathrm{x}) \forall \mathrm{n} \in \mathrm{N} \quad \text{... (2)} \end{aligned}$

$ \begin{array}{ll} \text { Now } & \text { put } \mathrm{y}=-\mathrm{x} \text { in eq.(1) } \\ & \mathrm{f}(\mathrm{x})+\mathrm{f}(-\mathrm{x})=\mathrm{f}(0) \quad\{\mathrm{f}(0)=0\} \\ \Rightarrow \quad & \mathrm{f}(-\mathrm{x})=-\mathrm{f}(\mathrm{x}) \\ \Rightarrow \quad & \mathrm{f} \text { is odd function } \\ & \text { from eq. (2) } \\ & \mathrm{f}(-\mathrm{nx})=\mathrm{nf}(-\mathrm{x}) \\ \Rightarrow \quad & \mathrm{f}(-\mathrm{nx})=-\mathrm{nf}(\mathrm{x}) \\ \Rightarrow \quad & \mathrm{f}(\mathrm{mx})=\operatorname{mf}(\mathrm{x}) \forall \mathrm{m} \in \mathrm{Z}^{-} \quad \text{... (3)}\\ & \text {from eq. (2) and eq. (3) } \\ & \mathrm{f}(\mathrm{nx})=\mathrm{nf}(\mathrm{x}) \forall \mathrm{n} \in \mathrm{Z} \quad \text{... (4)} \end{array}$

$\begin{aligned} \text { Now } & \text { put } x=\frac{p}{q} \text { where } p, q \in Z, q \neq 0 \\ f\left(\frac{n p}{q}\right) & =n f\left(\frac{p}{q}\right) \forall n \in Z \\ \text { put } n & =q \\ & f(p)=q f\left(\frac{p}{q}\right) \end{aligned}$

$\Rightarrow \quad \mathrm{pf}(1)=\mathrm{qf}\left(\frac{\mathrm{p}}{\mathrm{q}}\right) \quad \{\text{from eq.(4)}\}$

Let $f(1)=a$

$\begin{array}{ll} \text { then } \quad & \mathrm{pa}=\mathrm{qf}\left(\frac{\mathrm{p}}{\mathrm{q}}\right) \\ & \mathrm{f}\left(\frac{\mathrm{p}}{\mathrm{q}}\right)=\frac{\mathrm{ap}}{\mathrm{q}} \\ \Rightarrow \quad & \mathrm{f}(\mathrm{x})=\mathrm{ax} \forall \mathrm{x} \in \mathbb{Q} \end{array}$

Now, $f\left(\frac{-3}{5}\right)=a\left(\frac{-3}{5}\right)=12 \Rightarrow a=-20$

$\Rightarrow \quad \mathrm{f}(\mathrm{x})=-20 \mathrm{x} \forall \mathrm{x} \in \mathbb{Q} \quad \text{... (5)}$

From the given functional equation it is not possible to find a unique function for irrational values of '$x$', there are infinitely many such functions satisfying given functional equation for irrational values of $x$, but in this problem we finally need the function at rational values of '$x$' only. So, for rational values of $x$ we are getting a unique function mentioned in (5).

Now, $g(x+y)=g(x) \cdot g(y)$

$\begin{array}{ll} \Rightarrow & \ln (\mathrm{g}(\mathrm{x}+\mathrm{y})=\ln (\mathrm{g}(\mathrm{x}))+\ln (\mathrm{g}(\mathrm{y})) \\ \text { Let } & \ln (\mathrm{g}(\mathrm{x}))=\mathrm{h}(\mathrm{x}) \\ \Rightarrow & \mathrm{h}(\mathrm{x}+\mathrm{y})=\mathrm{h}(\mathrm{x})+\mathrm{h}(\mathrm{y}) \\ \Rightarrow & \mathrm{h}(\mathrm{x})=\mathrm{kx} \forall \mathrm{x} \in \mathbb{Q} \\ \Rightarrow & \mathrm{g}(\mathrm{x})=\mathrm{e}^{\mathrm{kx}} \forall \mathrm{x} \in \mathbb{Q} \quad \text{... (6)} \end{array}$

$\begin{aligned} & \text { and } \quad \mathrm{g}\left(\frac{-1}{3}\right)=\mathrm{e}^{-\frac{\mathrm{K}}{3}}=2 \quad \Rightarrow \quad \mathrm{K}=-3 \ln 2 \\ & \Rightarrow \quad \mathrm{K}=\ln \left(\frac{1}{8}\right) \\ & \Rightarrow \quad \mathrm{g}(\mathrm{x})=\mathrm{e}^{\ln \left(\frac{1}{8}\right) \cdot \mathrm{x}}=\left(\frac{1}{8}\right)^x=2^{-3 \mathrm{x}} \forall \mathrm{x} \in \mathbb{Q} \end{aligned}$

Now, $f\left(\frac{1}{4}\right)=-5, g(-2)=2^6=64$

$\mathrm{g}(0)=1$

$\begin{aligned} \text{So} \quad & \left(\mathrm{f}\left(\frac{1}{4}\right)+\mathrm{g}(-2)-(8) \mathrm{g}(0)\right) \\ & =(-5+64-8)(1)=51 \end{aligned}$

$\begin{aligned} & \mathrm{f}(\mathrm{x})=\frac{\left(\mathrm{x}^{2023}+2024 \mathrm{x}+2025\right)}{\mathrm{e}^{\pi \mathrm{x}}\left(\mathrm{x}^2-\mathrm{x}+3\right)}(\sin \mathrm{x}+2) \\ & \because(\sin \mathrm{x}+2) \text { is never zero } \\ & \therefore \text { for } \mathrm{x}^{2223}+2024 \mathrm{x}+2025=0 \\ & \text { let } \phi(\mathrm{x})=\mathrm{x}^{2023}+2024 \mathrm{x}+2025 \\ & \phi^{\prime}(\mathrm{x})=2023 \mathrm{x}^{2022}+2024>0 \forall \mathrm{x} \in \mathrm{R} \\ & \therefore \phi(\mathrm{x}) \text { is an Strictly Increasing function } \\ & \therefore \phi(\mathrm{x})=0 \text { for exactly one value of } \mathrm{x} \\ & \therefore \mathrm{f}(\mathrm{x})=0 \text { has one solution } \end{aligned}$

The given function f : [0, 1] $ \to $ R be define by

$f(x) = {{{4^x}} \over {{4^x} + 2}}$

$ \because $ $f(1 - x) = {{{4^{1 - x}}} \over {{4^{1 - x}} + 2}} = {2 \over {2 + {4^x}}}$

$ \therefore $ $f(x) + f(1 - x) = {{{4^x}} \over {{4^x} + 2}} + {2 \over {2 + {4^x}}}$

$ = {{{4^x} + 2} \over {{4^x} + 2}}$

So, f(x) + f(1 $-$ x) = 1 .....(i)

$ \therefore $ $f\left( {{1 \over {40}}} \right) + f\left( {{2 \over {40}}} \right) + f\left( {{3 \over {40}}} \right) + ... + f\left( {{{39} \over {40}}} \right) - f\left( {{1 \over 2}} \right)$

$ = \left[ {f\left( {{1 \over {40}}} \right) + f\left( {{{39} \over {40}}} \right)} \right] + \left[ {f\left( {{2 \over {40}}} \right) + f\left( {{{38} \over {40}}} \right)} \right] + ... + \left[ {f\left( {{{18} \over {40}}} \right) + f\left( {{{22} \over {40}}} \right)} \right] + \left[ {f\left( {{{19} \over {40}}} \right) + f\left( {{{21} \over {40}}} \right)} \right] + \left[ {f\left( {{{20} \over {40}}} \right) - f\left( {{1 \over 2}} \right)} \right]$

$ = \{ 1 + 1 + ... + 1 + 1\} + f\left( {{1 \over 2}} \right) - f\left( {{1 \over 2}} \right)$

$ = \{ 1 + 1 + ... + 1 + 1\}$(19 times) {from Eq. (i)}

= 19.

The given function f : R $ \to $ R be defined by

$f(\theta ) = {(\sin \theta + \cos \theta )^2} + {(\sin \theta - \cos \theta )^4}$

$ = 1 + \sin 2\theta + {(1 - \sin 2\theta )^2}$

$ = 1 + \sin 2\theta + 1 + {\sin ^2}2\theta - 2\sin 2\theta $

$ = {\sin ^2}2\theta - \sin 2\theta + 2$

$ = {\left( {\sin 2\theta - {1 \over 2}} \right)^2} + {7 \over 4}$

The local minimum of function 'f' occurs when

$\sin 2\theta = {1 \over 2}$

$ \Rightarrow 2\theta = {\pi \over 6},\,{{5\pi } \over 6},\,{{13\pi } \over 6},\,...$

$ \Rightarrow \theta = {\pi \over {12}},\,{{5\pi } \over {12}},\,{{13\pi } \over {12}},\,...$

but $\theta \in \{ {\lambda _1}\pi ,\,{\lambda _2}\pi ,\,...,\,{\lambda _r}\pi \} $,

where $0 < {\lambda _1} < .... < {\lambda _r} < 1$.

$ \therefore $ $\theta = {\pi \over {12}},\,{{5\pi } \over {12}}$

So, ${\lambda _1} + ... + {\lambda _r} = {1 \over {12}} + {5 \over {12}} = 0.50$

The given function f : [0, 2] $ \to $ R defined by

$f(x) = (3 - \sin (2\pi x))\sin \left( {\pi x - {\pi \over 4}} \right) - \sin \left( {3\pi x + {\pi \over 4}} \right)$

$ = (3 - \sin (2\pi x))\left[ {{{\sin \pi x} \over {\sqrt 2 }} - {{\cos \pi x} \over {\sqrt 2 }}} \right] - \left\{ {{{\sin 3\pi x} \over {\sqrt 2 }} + {{\cos (3\pi x)} \over {\sqrt 2 }}} \right\}$

$ = (3 - \sin (2\pi x)){{[\sin (\pi x) - \cos (\pi x)} \over {\sqrt 2 }} - {1 \over {\sqrt 2 }}[3\sin (\pi x) - 4{\sin ^3}(\pi x) + 4{\cos ^3}(\pi x) - 3\cos (\pi x)]$

$ = {{\sin (\pi x) - \cos (\pi x)} \over {\sqrt 2 }}[3 - \sin (2\pi x) - 3 + 4\{ {\sin ^2}(\pi x) + {\cos ^2}(\pi x) + \sin (\pi x)\cos (\pi x)\} ]$

$ = {{\sin (\pi x) - \cos (\pi x)} \over {\sqrt 2 }}[4 + \sin (2\pi x)]$

As, $f(x) \ge 0\forall \in [\alpha ,\beta ]$, where $\alpha ,\beta \in [0,2]$, so

$\sin (\pi x) - \cos (\pi x) \ge 0$

as $4 + \sin (2\pi x) > 0\,\forall x \in R$.

$ \Rightarrow \pi x \in \left[ {{\pi \over 4},{{5\pi } \over 4}} \right] \Rightarrow x \in \left[ {{1 \over 4},{5 \over 4}} \right]$

$ \therefore $ $\alpha = {1 \over 4}$ and $\beta = {5 \over 4}$

Therefore the value of $(\beta - \alpha ) = 1$

Given set S of polynomials with real coefficients

$S = \{ {({x^2} - 1)^2}({a_0} + {a_1}x + {a_2}{x^2} + {a_3}{x^3}):{a_0},{a_1},{a_2},{a_3} \in R\} $

and for a polynomial $f \in S$, Let

$f(x) = {({x^2} - 1)^2}({a_0} + {a_1}x + {a_2}{x^2} + {a_3}{x^3})$

it have $-$1 and 1 as repeated roots twice, so graph of f(x) touches the X-axis at x = $-$1 and x = 1, so f'(x) having at least three roots x = $-$1, 1 and $\alpha $. Where $\alpha $$ \in $($-$1, 1) and f''(x) having at least two roots in interval ($-$1, 1)

So, m_f' = 3 and m_f'' = 2

$ \therefore $ Minimum possible value of (m_f' + m_f'') = 5

Given, X has exactly 5 elements and Y has exactly 7 elements.

$ \therefore $ n(X) = 5

and n(Y) = 7

Now, number of one-one functions from X to Y is

$\alpha = {}^7{P_5} = {}^7{C_5} \times 5!$

Number of onto functions from Y to X is $\beta $


1, 1, 1, 1, 3 or 1, 1, 1, 2, 2

$ \therefore $ $\beta = {{7!} \over {3!4!}} \times 5! + {{7!} \over {{{(2!)}^3}3!}} \times 5!$

$ = ({}^7{C_3} + 3{}^7{C_3})5! = 4 \times {}^7{C_3} \times 5!$

$ \therefore $ ${{\beta - \alpha } \over {5!}} = {{(4 \times {}^7{C_3} - {}^7{C_5})5!} \over {5!}}$

$ = 4 \times 35 - 21 = 140 - 21 = 119$

We have $f(0) = 1,f'(x) = 3{x^2} + {1 \over 2}{e^{x/2}}$

$ \Rightarrow f'(g(x))g'(x) = 1$

Substituting $x = 0 \Rightarrow g'(1) = {1 \over {f'(0)}} = 2$.