Functions

Statement I: $f(x)=\frac{x}{1+|x|}$ is one-one

To check if a function is one-one (injective), we can check its monotonicity (whether it's always increasing or always decreasing).

1. For $x \geq 0: f(x)=\frac{x}{1+x}$. The derivative is $f^{\prime}(x)=\frac{1}{(1+x)^2}$, which is always positive.

2. For $x<0$ : $f(x)=\frac{x}{1-x}$. The derivative is $f^{\prime}(x)=\frac{1}{(1-x)^2}$, which is also always positive.

Since the derivative is positive for all $x$ in the domain, the function is strictly increasing. A strictly monotonic function is always one-one.

Statement I is True.

Statement II: $f(x)=\frac{x^2+4 x-30}{x^2-8 x+18}$ is many-one

Discriminant of Denominator: For $x^2-8 x+18, D=(-8)^2-4(1)(18)=64- 72=-8$. Since $D<0$, the denominator is always positive and the function is continuous for all $x \in \mathbb{R}$.

Behavior at Infinity: As $x \rightarrow \infty$ and $x \rightarrow-\infty$, the value of $f(x)$ approaches 1 (the ratio of the leading coefficients).

Turning Points: Since the function starts near $y=1$ at $-\infty$, moves through various values, and returns toward $y=1$ at $+\infty$, it must have at least one local maximum or minimum. This means a horizontal line will intersect the graph at more than one point.

Also, let

$ y=\frac{x^2+4 x-30}{x^2-8 x+18} $

Cross multiply:

$ y\left(x^2-8 x+18\right)=x^2+4 x-30 $

Rearrange:

$ (y-1) x^2+(-8 y-4) x+(18 y+30)=0 $

This is a quadratic in $x$.

For many values of $y$, this quadratic has two distinct real solutions. Function is MANY-ONE

$ f(x)=\operatorname{Sgn}(\sin x)+\operatorname{Sgn}(\cos x)+\operatorname{Sgn}(\tan x)+\operatorname{Sgn}(\cot x) $

Since the trig functions change signs depending on which quadrant $x$ is in, we analyze the value of $f(x)$ quadrant by quadrant.

Given that $x \neq \frac{n \pi}{2}$, we don't have to worry about any of these terms being zero. Each Sgn term will simply be either 1 (if positive) or -1 (if negative).

Range of $f(x)$ :

From table, Range $=\{-2,0,4\}$

Sum of all elements in this range $=4+(-2)+0=2$

Given

$g(x)=3 x^2+2 x-3, f(0)=-3$, and $4 g(f(x))=3 x^2-32 x+72$

Since $g(x)$ is a quadratic function (degree 2) and the composite function $4 g(f(x))$ is also a quadratic expression $\left(3 x^2-32 x+72\right)$, $f(x)$ must be a linear function.

• Let $f(x)=a x+b$.

• Using $f(0)=-3: a(0)+b=-3 \Longrightarrow b=-3$.

• So, $f(x)=a x-3$.

First, find $g(f(x))$ by substituting $f(x)$ into $g(x)$ :

$ g(f(x))=3(f(x))^2+2 f(x)-3 $

Substitute this into the given equation $4 g(f(x))=3 x^2-32 x+72$ :

$ 4\left[3(a x-3)^2+2(a x-3)-3\right]=3 x^2-32 x+72 $

Solve for the Coefficient ' $a$ '

Expand the left side of the equation:

• $4\left[3\left(a^2 x^2-6 a x+9\right)+2 a x-6-3\right]=3 x^2-32 x+72$

• $4\left[3 a^2 x^2-18 a x+27+2 a x-9\right]=3 x^2-32 x+72$

• $4\left[3 a^2 x^2-16 a x+18\right]=3 x^2-32 x+72$

• $12 a^2 x^2-64 a x+72=3 x^2-32 x+72$

Compare the coefficients:

• For $x^2: 12 a^2=3 \Longrightarrow a^2=\frac{1}{4} \Longrightarrow a= \pm \frac{1}{2}$

• For $x$ : $-64 a=-32 \Longrightarrow a=\frac{32}{64}=\frac{1}{2}$ Thus, $f(x)=\frac{1}{2} x-3$. Calculate $f(g(2))$

Find $g(2)$

$ g(2)=3(2)^2+2(2)-3=12+4-3=13 $

Step B: Find $f(g(2))$ which is $f(13)$

$ \begin{aligned} &f(13)=\frac{1}{2}(13)-3=6.5-3=3.5=\frac{7}{2} \end{aligned} $

Given $m=\sum_{i=1}^9(i)^2=1^2+2^2+\cdots 9$

We know, $1^2+2^2+3^2+\cdots n^2=\frac{n(n+1)(2 n+1)}{6}$

so, $m=\frac{9 \times 10 \times 19}{6}=3 \times 5 \times 19=285$

Given equation

$ 3 f(x)+2 f\left(\frac{m}{19 x}\right)=5 x, x \neq 0 $

put $\mathrm{m}=285,3 f(x)+2 f\left(\frac{285}{19 x}\right)=5 x$

$ 3 f(x)+2 f\left(\frac{15}{x}\right)=5 x-(1) $

put $x=\frac{15}{x}$

$ \begin{aligned} & 3 f\left(\frac{15}{x}\right)+2 f\left(\frac{15}{15 / x}\right)=5 \times \frac{15}{x} \\ & 3 f\left(\frac{15}{x}\right)+2 f(x)=\frac{75}{x}-(2) \end{aligned} $

from (1) $f\left(\frac{15}{x}\right)=\frac{5 x-3 f(x)}{2}$ substitute in equation (2)

$ \begin{aligned} & 3\left[\frac{5 x-3 f(x)}{2}\right]+2 f(x)=\frac{75}{x} \\ & 15 x-9 f(x)+4 f(x)=\frac{150}{x} \\ & 5 f(x)=15 x-\frac{150}{x} \\ & f(x)=3 x-\frac{30}{x} \end{aligned} $

to find $f(5)-f(2)$

$ \begin{aligned} & f(5)-f(2)=\left(3(5)-\frac{30}{5}\right)-\left(3 \times 2-\frac{30}{2}\right) \\ & =(15-6)-(6-15) \end{aligned} $

= $9-(-9)=18$

$ \begin{aligned} & \text { (3) } \Rightarrow f(x)=0 \Rightarrow[x]=3 \text { or }[x]=-2 \\ & \Rightarrow x \in[3,4) \cup[-2,-1) \Rightarrow f(x)=0 \text { has infinite solutions } \\ & f(x)=[x]^2-[x+3]-3 \\ & =\left[x^2\right]-[x]-6=([x]-3)([x]+2) \end{aligned} $

$ \begin{aligned} & (2) \Rightarrow \int_0^2 f(x) d x=\int_0^2\left([x]^2-[x]-6\right) d x \\ & \int_0^2 f(x) d x=\int_0^2\left([x]^2-[x]-6\right) d x=\int_0^2\left([x]^2-[x]\right) d x-12=-12 \\ & (4) \Rightarrow f(x)<0 \Rightarrow([x]-3)([x]+2)<0 \Rightarrow-2<[x]<3 \Rightarrow-1 \leq[x]<3 \\ & (1) \Rightarrow f(x)>0 \Rightarrow[x]<-2 \text { or }[x]>3 \Rightarrow x<-2 \text { or } x \geq 4 \Rightarrow x \in(-x,-2) 0[4, \infty] \end{aligned} $

$ \begin{aligned} & \text { let } x^2-10 x+85=\lambda \\ & \therefore \text { Domain for first term } \lambda>0 \ldots \ldots \ldots \ldots .(1) \\ & \& 7-\log _2 \lambda > 0 \Rightarrow \lambda < 2^7 \ldots \ldots \ldots .(2) \\ & \& \log _5\left(7-\log _2 \lambda\right) > 0 \Rightarrow \lambda < 2^6 \ldots \ldots . .(3) \\ & \therefore \text { from }(1),(2) \&(3) \\ & 0 < \lambda < 2^6 \\ & 0 < x^2-10 x+85 < 64 \Rightarrow x \in(3,7) \ldots \ldots . .(A) \\ & \& \text { domain for second term }-1 \leq \frac{3 x-7}{x-17} \leq 1 \Rightarrow x \in[-5,6] \ldots \ldots \ldots . . \end{aligned} $

From (A) \& (B), domain of function will be ( 3,6 ]

$ \Rightarrow \alpha=3, \beta=6 \Rightarrow \alpha+\beta=9 $

$ \begin{aligned} & f(x+y)=f(x) \cdot f(y) \Rightarrow f(x)=a^x \\ & \left(\because f(1)=7 \Rightarrow a^1=7\right) \end{aligned} $

So $f(x)=7^x$

Now

$ g(x+y)=g(x y)(\text { put } y=1) \Rightarrow g(x+1)=g(x) $

So $g(1)=g(2)=g(3)=\ldots \ldots \ldots=g(n)=1$

$ \begin{aligned} & \text { Given } \sum_{x=1}^n \frac{f(x)}{g(x)}=19607 \\ & \sum_{x=1}^n \frac{7^x}{1}=19607 \Rightarrow 7\left(\frac{7^n-1}{7-1}\right)=19607 \\ & 7^n-1=\frac{6}{7} \times 19607 \\ & 7^n=16807 \Rightarrow n=5 \end{aligned} $

$ \begin{aligned} & f(x)=\sin ^{-1}\left[\frac{5-x}{3+2 x}\right]+\frac{1}{\log _e(10-x)} \\ & -1 \leq \frac{5-x}{3+2 x} \leq 1 \\ & \frac{2-3 x}{3+2 x} \leq 0 \\ & (3 x-2)(2 x+3) \geq 0 \\ & (3 x-2)(2 x+3) \geq 0 x \in\left(-\infty,-\frac{3}{2}\right] \cup\left[\frac{2}{3}, \infty\right) \\ & \frac{8+x}{3+2 x} \geq 0 \\ & x \in(-\infty,-8] \cup\left[\frac{-3}{2}, \infty\right) \\ & x \in(-\infty,-8] \cup\left[\frac{2}{3}, 10\right)-\{9\}=6(\alpha+\beta+\gamma+\delta)=70 \end{aligned} $

To determine the nature of the given function, let us carefully analyze it for both the one-one (injective) and onto (surjective) properties. The function is given by:

$f(x)=\frac{2 x^2-3 x+2}{3 x^2+x+3}$

where the domain is the set of real numbers $\mathbf{R}$ and the codomain is also $\mathbf{R}$.

Checking if the function is one-one

A function $f: \mathbf{R} \rightarrow \mathbf{R}$ is said to be one-one if every element in the domain has a distinct image in the codomain. That is, if $f(x_1) = f(x_2)$, then we must have $x_1 = x_2$.

Let us check if we can find two different real values of $x$ that give the same value for $f(x)$. Let us set $f(x) = 1$ and solve for $x$:

$\frac{2 x^2-3 x+2}{3 x^2+x+3} = 1$

Cross-multiplying, we obtain:

$2 x^2-3 x+2 = 3 x^2+x+3$

Bringing all terms to one side, we get:

$x^2 + 4x + 1 = 0$

Using the quadratic formula to find the roots:

$x = \frac{-4 \pm \sqrt{16 - 4(1)(1)}}{2} = \frac{-4 \pm \sqrt{12}}{2} = -2 \pm \sqrt{3}$

Since we found two distinct real values, $x_1 = -2 + \sqrt{3}$ and $x_2 = -2 - \sqrt{3}$, such that $f(x_1) = f(x_2) = 1$, the function is not one-one.

Checking if the function is onto

A function is onto if its range is equal to its codomain. Here, the codomain is $\mathbf{R}$. To find the range, let us set $f(x) = y$:

$y = \frac{2 x^2-3 x+2}{3 x^2+x+3}$

Cross-multiplying gives us:

$y(3x^2 + x + 3) = 2x^2 - 3x + 2$

$3yx^2 + yx + 3y = 2x^2 - 3x + 2$

Grouping the terms by the powers of $x$, we form a quadratic equation in $x$:

$(3y - 2)x^2 + (y + 3)x + (3y - 2) = 0$

Since $x$ is a real number, the discriminant ($D$) of this quadratic equation must be greater than or equal to zero ($D \geq 0$).

$D = (y + 3)^2 - 4(3y - 2)(3y - 2) \geq 0$

$(y + 3)^2 - 4(3y - 2)^2 \geq 0$

This is in the form of $a^2 - b^2 \geq 0$, which can be factored as $(a-b)(a+b) \geq 0$. Here, $a = y+3$ and $b = 2(3y-2) = 6y-4$:

$[(y + 3) - (6y - 4)][(y + 3) + (6y - 4)] \geq 0$

$(-5y + 7)(7y - 1) \geq 0$

To solve this inequality, we can multiply by $-1$ (which reverses the inequality sign):

$(5y - 7)(7y - 1) \leq 0$

The critical points are $y = \frac{1}{7}$ and $y = \frac{7}{5}$. Based on the inequality, $y$ must lie between these two values:

$\frac{1}{7} \leq y \leq \frac{7}{5}$

This means the range of the function is the closed interval $\left[ \frac{1}{7}, \frac{7}{5} \right]$.

Because the range is a small subset of the real numbers and not equal to the entire codomain $\mathbf{R}$, the function is not onto.

Conclusion:

Since $f(x)$ is neither one-one nor onto, the correct statement is given by Option D.

Option D

For the function

$ f(x)=\sin^{-1}\left(\frac{x+[x]}{3}\right), $

the expression inside $\sin^{-1}$ must lie between $-1$ and $1$.

So, we need

$ -1 \le \frac{x+[x]}{3} \le 1 $

Multiplying throughout by $3$,

$ -3 \le x+[x] \le 3 $

Now let

$ [x]=n $

Then by definition of greatest integer function,

$ n \le x < n+1 $

and so

$ x+[x]=x+n $

Hence the condition becomes

$ -3 \le x+n \le 3 $

But since $x \in [n,n+1)$, let us check possible integer values of $n$.

Case 1: $n=-2$

Then

$ x \in [-2,-1) $

and

$ x+[x]=x-2 $

Since $x \in [-2,-1)$,

$ x-2 \in [-4,-3) $

This does not satisfy $-3 \le x+[x]$ except at $x=-1$, but $-1 \notin [-2,-1)$.

So no value from this interval is allowed.

Case 2: $n=-1$

Then

$ x \in [-1,0) $

and

$ x+[x]=x-1 $

So

$ x-1 \in [-2,-1) $

This lies completely within $[-3,3]$.

Hence all $x \in [-1,0)$ are allowed.

Case 3: $n=0$

Then

$ x \in [0,1) $

and

$ x+[x]=x $

So

$ x \in [0,1) $

This also lies within $[-3,3]$.

Hence all $x \in [0,1)$ are allowed.

Case 4: $n=1$

Then

$ x \in [1,2) $

and

$ x+[x]=x+1 $

So

$ x+1 \in [2,3) $

This lies within $[-3,3]$.

Hence all $x \in [1,2)$ are allowed.

Case 5: $n=2$

Then

$ x \in [2,3) $

and

$ x+[x]=x+2 $

So

$ x+2 \in [4,5) $

This exceeds $3$, so no value is allowed.

Now let us also check the boundary points:

• At $x=-\dfrac{3}{2}$, actually $[x]=-2$, so this does not help.

• Better to solve more carefully near the ends.

We want the full set of $x$ satisfying

$ -3 \le x+[x] \le 3 $

Let $x \in [n,n+1)$, so $[x]=n$.

Then

$ -3 \le x+n \le 3 $

which gives

$ -3-n \le x \le 3-n $

Together with $n \le x < n+1$, we need intersection of

$ [n,n+1) \cap [-3-n,\,3-n]. $

Now check relevant integers:

For $n=-2$

$ x \in [-2,-1) \cap [-1,5] = \{-1\} $

but $-1$ is not included in $[-2,-1)$, so no solution.

For $n=-1$

$ x \in [-1,0) \cap [-2,4] = [-1,0) $

For $n=0$

$ x \in [0,1) \cap [-3,3] = [0,1) $

For $n=1$

$ x \in [1,2) \cap [-4,2] = [1,2) $

For $n=2$

$ x \in [2,3) \cap [-5,1] = \varnothing $

Thus domain is

$ [-1,0)\cup [0,1)\cup [1,2)=[-1,2) $

So,

$ \alpha=-1,\qquad \beta=2 $

Therefore,

$ \alpha^2+\beta^2=(-1)^2+2^2=1+4=5 $

Hence the correct answer is

$ \boxed{5} $

So, Option B is correct.

For

$ f:[1,\infty)\to [1,\infty),\qquad f(x)=(x-1)^4+1, $

we first find the inverse function.

Since for $x\ge 1$, the function $(x-1)^4+1$ is strictly increasing, $f$ is one-one and onto from $[1,\infty)$ to $[1,\infty)$.

Let

$ y=(x-1)^4+1. $

Then

$ y-1=(x-1)^4. $

Since $x\ge 1$, we take the positive fourth root:

$ x-1=(y-1)^{1/4}. $

Hence

$ f^{-1}(y)=1+(y-1)^{1/4}. $

So

$ f^{-1}(x)=1+(x-1)^{1/4}. $

Now check the two statements one by one.

Statement (I)

We need to solve

$ f(x)=f^{-1}(x). $

That is,

$ (x-1)^4+1=1+(x-1)^{1/4}. $

So

$ (x-1)^4=(x-1)^{1/4}. $

Put

$ t=x-1 \quad \text{where } t\ge 0. $

Then the equation becomes

$ t^4=t^{1/4}. $

Now solve:

$ t^4-t^{1/4}=0. $

This gives

$ t^{1/4}\left(t^{15/4}-1\right)=0. $

Hence

$ t=0 \quad \text{or} \quad t^{15/4}=1. $

Since $t\ge 0$, from $t^{15/4}=1$ we get

$ t=1. $

Therefore

$ x-1=0 \Rightarrow x=1, $

or

$ x-1=1 \Rightarrow x=2. $

So the set

$ S=\{x\in[1,\infty): f(x)=f^{-1}(x)\} $

contains exactly two elements:

$ S=\{1,2\}. $

Thus, Statement (I) is TRUE.

Statement (II)

We need to solve

$ f(x)=f^{-1}(x+1). $

Now

$ f(x)=(x-1)^4+1, $

and

$ f^{-1}(x+1)=1+\big((x+1)-1\big)^{1/4}=1+x^{1/4}. $

So the equation becomes

$ (x-1)^4+1=1+x^{1/4}. $

Hence

$ (x-1)^4=x^{1/4}. $

Again put

$ t=x^{1/4}, \quad t\ge 1 \text{ since } x\ge 1. $

Then

$ x=t^4. $

So the equation becomes

$ (t^4-1)^4=t. $

Instead of solving this directly, test simple values.

For $x=1$:

$ f(1)=(1-1)^4+1=1, $

and

$ f^{-1}(1+1)=f^{-1}(2)=1+(2-1)^{1/4}=2. $

So $x=1$ is not a solution.

For $x=2$:

$ f(2)=(2-1)^4+1=2, $

and

$ f^{-1}(2+1)=f^{-1}(3)=1+(3-1)^{1/4}=1+2^{1/4}\neq 2. $

So $x=2$ is not a solution.

Now define

$ g(x)=(x-1)^4-x^{1/4}, \qquad x\ge 1. $

We need to check whether $g(x)=0$ has any solution.

Observe:

$ g(1)=0-1=-1<0, $

and

$ g(2)=1-2^{1/4}<0, $

while

$ g(3)=16-3^{1/4}>0. $

Since $g$ is continuous, there is at least one root in $(2,3)$.

So the set

$ S=\{x\in[1,\infty): f(x)=f^{-1}(x+1)\} $

is not empty.

Thus, Statement (II) is FALSE.

Therefore, the correct option is

$ \boxed{\text{Option A: only (I) is TRUE}} $

Given

$ f(x+y)=f(x)+2y^2+y+\alpha xy \quad \text{for all } x,y\in \mathbb{R} $

with

$ f(0)=-1,\qquad f(1)=2. $

We have to find

$ \sum_{n=1}^5 (\alpha+f(n)). $

Step 1: Use the given relation to find $\alpha$

Put $x=0$ in

$ f(x+y)=f(x)+2y^2+y+\alpha xy. $

Then

$ f(y)=f(0)+2y^2+y+\alpha\cdot 0\cdot y $

so

$ f(y)=-1+2y^2+y. $

Thus, for every real $y$,

$ f(y)=2y^2+y-1. $

Now use $f(1)=2$:

$ f(1)=2(1)^2+1-1=2. $

This is already satisfied, so now we must use the original equation to determine $\alpha$.

Substitute

$ f(x+y)=2(x+y)^2+(x+y)-1 $

and

$ f(x)=2x^2+x-1 $

into the given equation:

$ 2(x+y)^2+(x+y)-1 = 2x^2+x-1+2y^2+y+\alpha xy. $

Expand the left side:

$ 2x^2+4xy+2y^2+x+y-1. $

Right side is:

$ 2x^2+x-1+2y^2+y+\alpha xy. $

Comparing both sides,

$ 4xy=\alpha xy. $

Hence,

$ \alpha=4. $

Step 2: Find $f(n)$ for $n=1$ to $5$

Since

$ f(n)=2n^2+n-1, $

we get:

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

$ f(2)=2(2)^2+2-1=8+2-1=9 $

$ f(3)=2(3)^2+3-1=18+3-1=20 $

$ f(4)=2(4)^2+4-1=32+4-1=35 $

$ f(5)=2(5)^2+5-1=50+5-1=54 $

Step 3: Compute the required sum

We need

$ \sum_{n=1}^5 (\alpha+f(n)). $

Since $\alpha=4$,

$ \alpha+f(1)=4+2=6 $

$ \alpha+f(2)=4+9=13 $

$ \alpha+f(3)=4+20=24 $

$ \alpha+f(4)=4+35=39 $

$ \alpha+f(5)=4+54=58 $

Therefore,

$ 6+13+24+39+58=140. $

So the required value is

$ \boxed{140}. $

Hence, the correct option is B.

For the function

$ f(x)=\cos^{-1}\left(\frac{4x+2[x]}{3}\right), $

the expression inside $\cos^{-1}$ must lie between $-1$ and $1$.

So, for domain:

$ -1 \le \frac{4x+2[x]}{3} \le 1 $

Multiplying by $3$:

$ -3 \le 4x+2[x] \le 3 $

Now let

$ [x]=n $

where $n$ is an integer and

$ n \le x < n+1. $

Then the inequality becomes

$ -3 \le 4x+2n \le 3. $

Subtracting $2n$:

$ -3-2n \le 4x \le 3-2n $

Dividing by $4$:

$ \frac{-3-2n}{4} \le x \le \frac{3-2n}{4} $

But also,

$ n \le x < n+1. $

So we find possible integer values of $n$ for which these intervals overlap.

Case 1: $n=-1$

Then

$ [x]=-1,\quad -1 \le x < 0 $

and

$ \frac{-3-2(-1)}{4} \le x \le \frac{3-2(-1)}{4} $

$ \frac{-1}{4} \le x \le \frac{5}{4} $

Intersection with $[-1,0)$ gives

$ -\frac14 \le x < 0. $

Case 2: $n=0$

Then

$ 0 \le x < 1 $

and

$ \frac{-3}{4} \le x \le \frac{3}{4} $

Intersection gives

$ 0 \le x \le \frac34. $

Case 3: $n=-2$

Then

$ -2 \le x < -1 $

and

$ \frac{-3-2(-2)}{4} \le x \le \frac{3-2(-2)}{4} $

$ \frac14 \le x \le \frac74 $

No intersection.

Case 4: $n=1$

Then

$ 1 \le x < 2 $

and

$ \frac{-5}{4} \le x \le \frac14 $

No intersection.

So the domain is

$ \left[-\frac14,0\right)\cup\left[0,\frac34\right] = \left[-\frac14,\frac34\right]. $

Thus,

$ \alpha=-\frac14,\quad \beta=\frac34. $

Therefore,

$ \alpha+\beta=\frac12 $

and

$ 12(\alpha+\beta)=12\cdot \frac12=6. $

Hence the correct answer is

$ \boxed{6} $

So, Option A is correct.

Total number of functions from $\{1,2,3,4\}$ to $\{a,b,c\}$ is

$ 3^4=81 $

Now we find the number of onto functions.

For a function to be onto, all the three elements $a,b,c$ must appear as images of elements of $\{1,2,3,4\}$.

Using inclusion-exclusion principle:

• Total functions: $3^4=81$

Let

• $A$ = functions missing $a$

• $B$ = functions missing $b$

• $C$ = functions missing $c$

If one element is missing, then the function goes into only $2$ elements.

So,

$ |A|=|B|=|C|=2^4=16 $

Hence,

$ |A|+|B|+|C|=3\times 16=48 $

If two elements are missing, then all $4$ elements of domain must go to only one element of codomain.

So,

$ |A\cap B|=|B\cap C|=|C\cap A|=1^4=1 $

Thus,

$ |A\cap B|+|B\cap C|+|C\cap A|=3 $

If all three are missing, that is impossible, so

$ |A\cap B\cap C|=0 $

Therefore, number of functions which are not onto is

$ |A\cup B\cup C|=48-3=45 $

So the correct answer is

$ \boxed{45} $

Hence, Option B is correct.

$\begin{aligned} & y=\frac{5-x}{x^2-3 x+2} \\ & y x^2-3 x y+2 y+x-5=0 \\ & y z^2+(-3 y+1) x+(2 y-5)=0 \end{aligned}$

Case I : If $y=0$ (Accepted)

$\Rightarrow x=5$

Case II : If $y \neq 0$

$\begin{aligned} & \mathrm{D} \geq 0 \\ & (-3 y+1)^2-4(y)(2 y-5) \geq 0 \\ & 9 y^2+1-6 y-8 y^2+20 y \geq 0 \\ & y^2+14 y+1 \geq 0 \\ & (y+7)^2-48 \geq 0 \\ & |y+7| \geq 4 \sqrt{3} \\ & \Rightarrow y+7 \geq 4 \sqrt{3} \text { or } \mathrm{y}+7 \leq-4 \sqrt{3} \\ & \Rightarrow \mathrm{y} \geq 4 \sqrt{3}-7 \text { or } \mathrm{y} \leq-4 \sqrt{3}-7 \end{aligned}$

From Case I and Case II

$y \in(-\infty,-4 \sqrt{3}-7] \cup[4 \sqrt{3}-7, \infty)$

$\begin{aligned} & \text { So } \alpha=-4 \sqrt{3}-7 \\ & \quad \beta=4 \sqrt{3}-7 \\ & \begin{aligned} \Rightarrow a^2+b^2 & =(-4 \sqrt{3}-7)^2+(4 \sqrt{3}-7)^2 \\ & =2(48+49) \\ & =194 \end{aligned} \end{aligned}$

$f(x)=\log _4\left(\log _3\left(\log _7\left(8-\log _2\left(x^2+4 x+5\right)\right)\right)\right.$

$\begin{aligned} & \log _3\left(\log _1\left(8-\log _2\left(x^2+4 x+5\right)\right)\right)>0 \\ & \log _7\left(8-\log _2\left(x^2+4 x+5\right)\right)>1 \\ & 8-\log _2\left(x^2+4 x+5\right)>7 \\ & -\log _2\left(x^2+4 x+5\right)>-1 \\ & \log _2\left(x^2+4 x+5\right)<1 \\ & x^2+4 x+5<2 \\ & x^2+4 x+3<0 \\ & \Rightarrow(x+3)(x+1)<0 \quad \ldots(1) \\ & \log _7\left(8-\log _2\left(x^2+4 x+5\right)\right)>0 \\ & 8-\log _2\left(x^2+4 x+5\right)>1 \\ & \log _2\left(x^2+4 x+5\right)<9 \\ & x^2+4 x+5<2^9 \\ & x^2+4 x+5<512 \\ & \Rightarrow x^2+4 x-507<0 \\ & \Rightarrow x=-4 \pm \sqrt{16+2028} \end{aligned}$

$\begin{aligned} & x=\frac{-4 \pm \sqrt{2044}}{2} \quad\text{..... (2)}\\ & \Rightarrow\left(x-\left(\frac{-4+\sqrt{2044}}{2}\right)\right)\left(x-\left(\frac{-4-\sqrt{2044}}{2}\right)\right)<0 \\ & x^2+4 x+5>0 \\ & D>0 \\ & x \in R \\ & \text { Also, } 8-\log _2\left(x^2+4 x+5\right)>0 \\ & \log _2\left(x^2+4 x+5\right)<8 \\ & x^2+4 x+5<256 \\ & \Rightarrow x^2+4 x-251<0 \\ & \Rightarrow x=-4 \pm \sqrt{16+1004} \\ & \Rightarrow x=\frac{-4 \pm \sqrt{1020}}{2} \end{aligned}$

$\begin{aligned} &\Rightarrow\left(x-\left(\frac{-4+\sqrt{1020}}{2}\right)\right)\left(x-\left(\frac{-4-\sqrt{1020}}{2}\right)\right)<0\\ &\therefore \text { Intersection of (1), (2) and (3) } \end{aligned}$

$\begin{aligned} & \therefore x \in(-3,-1) \\ & -1 \leq \frac{7 x+10}{x-2} \leq 1 \\ & \Rightarrow x \in[-2,-1] \\ & \therefore \alpha^2+\beta^2+\gamma^2+\delta^2=(-3)^2+(-1)^2+(-2)^{-2}+(-1)^2 \\ & =9+1+4+1 \\ & =15 \end{aligned}$

$\begin{aligned} & g(2)=4, g(4)=\frac{10}{3} \\ & f \text { is decreasing in }\left(\frac{10}{3}, 4\right) \\ & \therefore \quad \alpha=f(4)=\frac{1}{2} \\ & \beta=f\left(\frac{10}{3}\right)=\frac{29}{56} \\ & \frac{1}{\beta-\alpha}=\frac{1}{\frac{29}{56}-\frac{1}{2}}=56 \end{aligned}$

$\begin{aligned} & f(x)+3 f\left(\frac{24}{x}\right)=4 x, x \neq 0 \quad \ldots(1)\\ & \text { replace } x \text { by } \frac{24}{x} \\ & f\left(\frac{24}{x}\right)+3 f\left(\frac{24}{24}\right)=4\left(\frac{24}{x}\right)=\frac{96}{x} \quad \ldots(2) \\ & 3 \times(2)-(1) \\ & \Rightarrow 8 f(x)=\frac{96.3}{x}-4 x \Rightarrow f(x)=\frac{36}{x}-\frac{x}{2} \\ & f(3)+f(8)=\left(12-\frac{3}{2}\right)+\left(\frac{36}{8}-4\right) \end{aligned}$

$=8+\frac{36}{8}-\frac{12}{8}=11$

$\begin{aligned} & 1-\log _4\left(x^2-9 x+18\right)>0 \\ & \log _4\left(x^2-9 x+18\right)<1 \\ & x^2-9 x+18<4 \\ & x^2-9 x+14<0 \\ & x \in(2,7) \\ & x^2-9 x+18>0 \\ & x \in(-\infty, 3) \cup(6, \infty) \end{aligned}$

$\begin{aligned} & x \in(2,3) \cup(6,7) \\ & \alpha+\beta+\gamma+\delta=18 \end{aligned}$

$\frac{2 x-3}{4 x+5}>0$

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

$\begin{aligned} & \Rightarrow \quad \frac{3 x+4-2+x}{x-2} \geq 0 \\ & \Rightarrow \quad \frac{4 x+2}{x-2} \geq 0 \\ & \Rightarrow \quad x \in\left(-\infty,-\frac{1}{2}\right] \cup(2, \infty)........(ii)\end{aligned}$

$ \begin{aligned} & \frac{3 x+4}{2-x} \geq-1 \\ \Rightarrow & \frac{3 x+4}{2-x}+1 \geq 0 \\ \Rightarrow & \frac{3 x+4+2-x}{2-x} \geq 0 \\ \Rightarrow & \frac{2 x+6}{x-2} \leq 0 \\ \therefore & x \in[-3,2)........(iii) \end{aligned} $

Taking intersection of (i), (ii) and (iii)

$ \begin{aligned} & x \in\left[-3,-\frac{5}{4}\right) \\ & \alpha=-3, \beta=-\frac{5}{4} \\ & \alpha^2+4 \beta=4 \end{aligned} $

$x+|x|= \begin{cases}2 x, & x \geq 0 \\ 0, & x<0\end{cases}$

$\Rightarrow \frac{1}{\sqrt{x+|x|}}$, domain is $x>0$, as $2 x \neq 0$

Similarly,

$\begin{aligned} &\frac{1}{\sqrt{3 x+10-x^2}} \text { is defined when } 3 x+10-x^2>0\\ &\begin{aligned} \Rightarrow & x^2-3 x-10<0 \\ & (x-5)(x+2)<0 \\ \Rightarrow & x \in(-2,5) \\ \Rightarrow & \text { Domain will be }(0, \infty) \cap(-2,5)=(0,5) \\ \Rightarrow & (1+a)^2+b^2=1+25=26 \end{aligned} \end{aligned}$

$\begin{aligned} & f_1(x)=\log _5\left(18 x-x^2-77\right) \\ & \therefore 18 x-x^2-77>0 \\ & \quad x^2-18 x+77<0 \\ & \quad x \in(7,11) \alpha=7, \beta=11 \\ & f_2(x)=\log _{(x-1)}\left(\frac{2 x^2+3 x-2}{x^2-3 x-4}\right) \\ & \therefore \quad x-1>0, x-1 \neq 1, \frac{2 x^2+3 x-2}{x^2-3 x-4}>0 \\ & \quad x>1, x \neq 2, \frac{(2 x-1)(x+2)}{(x-4)(x+1)}>0 \\ & \quad x>1, x \neq 2, \end{aligned}$

$\begin{aligned} & \therefore \quad x \in(4, \infty) \\ & \therefore \gamma=4 \\ & \therefore \alpha^2+\beta^2+\gamma^2=49+121+16 \\ & =186 \end{aligned}$

as $f(x)$ is onto hence $A$ is range of $f(x)$

$\text { now } \begin{aligned} f^{\prime}(x) & =6 x^2-30 x+36 \\ & =6(x-2)(x-3) \end{aligned}$

$f(2)=16-60+72+7=35$

$\begin{aligned} & \mathrm{f}(3)=54-135+108+7=34 \\ & \mathrm{f}(0)=7 \end{aligned}$

hence range $\in[7,35]=\mathrm{A}$

also for range of $g(x)$

$\begin{aligned} & g(x)=1-\frac{1}{x^{2025}+1} \in[0,1)=B \\ & s=\{0,7,8, \ldots . .35\} \text { hence } n(s)=30 \end{aligned}$

$\begin{aligned} & f(x)=\frac{2^x}{2^x+\sqrt{2}} \\ & f(x)+f(1-x)=\frac{2^x}{2^x+\sqrt{2}}+\frac{2^{1-x}}{2^{1-x}+\sqrt{2}} \\ & =\frac{2^x}{2^x+\sqrt{2}}+\frac{2}{2+\sqrt{2} 2^x}=\frac{2^x+\sqrt{2}}{2^x+\sqrt{2}}=1 \end{aligned}$

$\begin{array}{r} \text { Now, } \sum_{\mathrm{k}=1}^{81} \mathrm{f}\left(\frac{\mathrm{k}}{82}\right)=\mathrm{f}\left(\frac{1}{82}\right)+\mathrm{f}\left(\frac{2}{82}\right)+\ldots \ldots+\mathrm{f}\left(\frac{81}{82}\right) \\ =\mathrm{f}\left(\frac{1}{82}\right)+\mathrm{f}\left(\frac{2}{82}\right)+\ldots \ldots .+\mathrm{f}\left(1-\frac{2}{82}\right)+\mathrm{f}\left(1-\frac{1}{82}\right) \\ =\left[\mathrm{f}\left(\frac{1}{82}\right)+\mathrm{f}\left(1-\frac{1}{82}\right)\right]+\left[\mathrm{f}\left(\frac{2}{82}\right)+\mathrm{f}\left(1-\frac{2}{82}\right)\right]+\ldots . .40 \text { cases }+\mathrm{f}\left(\frac{41}{82}\right) \end{array}$

$\begin{aligned} & =(1+1+\ldots 40 \text { times })+\frac{2^{1 / 2}}{2^{1 / 2}+2^{1 / 2}} \\ & 40+\frac{1}{2}=\frac{81}{2} \end{aligned}$

$\begin{aligned} & f(x)=(3 a+2) x^2+\left(\frac{a+2}{a-1}\right) x+b \\ & f\left(x+\frac{1}{2}\right)=f(x)+f(y)+1-\frac{2}{7} x y\quad\text{.... (1)} \end{aligned}$

In (1) Put $x=y=0 \Rightarrow f(0)=2 f(0)+1 \Rightarrow f(0)=-1$

So, $\mathrm{f}(0)=0+0+\mathrm{b}=-1 \Rightarrow \mathrm{~b}=-1$

$\begin{aligned} & \text { In (1) Put } y=-x \Rightarrow f(0)=f(x)+f(-x)+1+\frac{2}{7} x^2 \\ & -1=2(3 a+2) x^2+2 b+1+\frac{2}{7} x^2 \end{aligned}$

$\begin{aligned} & -1=\left(2(3 a+2)+\frac{2}{7}\right) x^2+1-2 \\ & \Rightarrow 6 a+4+\frac{2}{7}=0 \\ & a=-\frac{5}{7} \end{aligned}$

So $\mathrm{f}(\mathrm{x})=-\frac{1}{7} \mathrm{x}^2-\frac{3}{4} \mathrm{x}-1$

$\Rightarrow|\mathrm{f}(\mathrm{x})|=\frac{1}{28}\left|4 \mathrm{x}^2+21 \mathrm{x}+28\right|$

Now, $28 \sum_{i=1}^5|f(i)|=28(|f(1)|+|f(2)|+\ldots+|f(5)|)$

$28 \cdot \frac{1}{28} \cdot 675=675$

$\begin{aligned} & f(x)=\frac{2^{2 \mathrm{x}}-1}{2^{2 \mathrm{x}}+1} \\ & =1-\frac{2}{2^{2 \mathrm{x}}+1} \\ & \mathrm{f}^{\prime}(\mathrm{x})=\frac{2}{\left(2^{2 \mathrm{x}}+1\right)^2} \cdot 2 \cdot 2^{2 \mathrm{x}} \cdot \ln 2 \text { i.e always }+\mathrm{ve} \end{aligned}$

so $f(x)$ is $\uparrow$ function

$\begin{aligned} & \therefore \mathrm{f}(-\infty)=-1 \\ & \mathrm{f}(\infty)=1 \\ & \therefore \mathrm{f}(\mathrm{x}) \in(-1,1) \neq \text { co-domain } \end{aligned}$

so function is one-one but not onto

$\begin{aligned} & f(x)=\frac{42^x+16}{2.2^{2 x}+16.2^x+32} \\ & f(x)=\frac{2\left(2^x+4\right)}{2^{2 x}+8.2^x+16} \\ & f(x)=\frac{2}{2^x+4} \\ & f(4-x)=\frac{2^x}{2\left(2^x+4\right)} \\ & f(x)+f(4-x)=\frac{1}{2} \end{aligned}$

So, $\quad \mathrm{f}\left(\frac{1}{15}\right)+\mathrm{f}\left(\frac{59}{15}\right)=\frac{1}{2}$

$\begin{aligned} & \text { Similarly }=f\left(\frac{29}{15}\right)+f\left(\frac{31}{15}\right)=\frac{1}{2} \\ & f\left(\frac{30}{15}\right)=f(2)=\frac{2}{2^2+4}=\frac{2}{8}=\frac{1}{4} \\ & \Rightarrow 8\left(29 \times \frac{1}{2}+\frac{1}{4}\right) \end{aligned}$

Ans. 118

$\begin{aligned} & f(x)=\ln x \\ & g(x)=\frac{x^4-2 x^3+3 x^2-2 x+2}{2 x^2-2 x+1} \\ & D_g \in R \\ & D_f \in(0, \infty) \end{aligned}$

For $D_{f o g} \Rightarrow g(x)>0$

$\begin{aligned} & \frac{x^4-2 x^3+3 x^2-2 x+2}{2 x^2-2 x+1}>0 \\ & \Rightarrow x^4-2 x^3+3 x^2-2 x+2>0 \end{aligned}$

Clearly $\mathrm{x}<0$ satisfies which are included in option (1) only.

$ \textbf{Step 1: Total Functions with } 1 \in f(A) $

Any function $ f: A \to B $ where $ A = \{1, 2, 3, 4\} $ and $ B = \{1, 4, 9, 16\} $ is defined by choosing one of the four elements of $ B $ for each element of $ A $. Thus, the total number of functions is

$ 4^4 = 256. $

To count those functions where $ 1 $ appears at least once in the set $ f(A) $, we can use the complementary counting method: subtract the functions that never use $ 1 $. If $ 1 $ is excluded, each element of $ A $ has only 3 choices (namely, $ \{4, 9, 16\} $), so the number of such functions is

$ 3^4 = 81. $

Thus, the number of functions such that $ 1 \in f(A) $ is

$ 256 - 81 = 175. $

$ \textbf{Step 2: Counting Many-One Functions} $

In this context, "many-one functions" are understood to be non-injective functions. Since an injective (one-to-one) function from $ A $ to $ B $ must be a permutation (because both sets have 4 elements), the number of one-to-one functions is

$ 4! = 24. $

It is important to note that every injective function $ f: A \to B $ has $ f(A) = B $ (a full permutation) which automatically means $ 1 \in f(A) $.

Thus, the number of many-one (non-injective) functions $ f: A \to B $ with $ 1 \in f(A) $ is found by subtracting the one-to-one functions from the total functions that include $ 1 $:

$ 175 - 24 = 151. $

$ \boxed{151} $

This detailed explanation shows that the number of many-one functions $ f: A \rightarrow B $ such that $ 1 \in f(A) $ is indeed $ 151 $.

$f(x)=\frac{1}{\sqrt{|x|-x^2}}$

For $f(x)$ to be defined $|x|-x^2>0$

$\operatorname{Case} \mathbf{I}|x|=x, x>0$

$ \begin{array}{lr} \Rightarrow & x-x^2>0 \\ \Rightarrow & x(1-x) > 0 \\ \Rightarrow & 0 < x < 1 \end{array} $

Case II $|x|=-x, x<0$

$ \begin{aligned} & \Rightarrow \quad-x-x^2 > 0 \\ & \Rightarrow x(1+x) < 0 \\ & \Rightarrow \quad-1 < x < 0 \end{aligned} $

So, $A=(-1,1)-\{0\}$

Let $y=\frac{1}{\sqrt{|x|-x^2}}$

When $0 < x < 1$

$ y=\frac{1}{\sqrt{x-x^2}} $

As, $x \in(0,1)$

$ \begin{array}{ll} \Rightarrow & 0 < x-x^2 \leq \frac{1}{4} \\ \Rightarrow & 0<\sqrt{x-x^2} \leq \frac{1}{2} \\ \Rightarrow & 2 \leq \frac{1}{\sqrt{x-x^2}}<\infty \end{array} $

So, $y \in[2, \infty)$ for $x \in(0,1)$

$ y=\frac{1}{\sqrt{-x-x^2}} $

$ y \in[2, \infty) \text { for } x \in(-1,0) $

So, $B=[2, \infty)$

So, $A \cup B$

$ \begin{aligned} & (-1,1)-\{0\} \cup[2, \infty) \\ & (-1,0) \cup(0,1) \cup[2, \infty) \end{aligned} $

$f(x)= \begin{cases}2 x+3, & x \leq \frac{4}{3} \\ -3 x^2+8 x, & x>\frac{4}{3}\end{cases}$

$ \Rightarrow f^{\prime}(x)= \begin{cases}2 & x<\frac{4}{3} \\ -6 x+8, & x>\frac{4}{3}\end{cases} $

$ f^{\prime}(x)>0, \forall x<\frac{4}{3} \text { and } f^{\prime}(x)<0, \forall x>\frac{4}{3} $

So, $f$ is many-one.

Sign scheme for $f^{\prime}(x)$

$ \begin{aligned} & f(-\infty) \rightarrow-\infty \\ & f(\infty) \rightarrow-\infty \\ & \qquad f\left(\frac{4}{3}\right)=2 \times \frac{4}{3}+3 \end{aligned} $

So, range $=\left(-\infty, \frac{17}{3}\right]$

Range is subset of codomain.

So, $f(x)$ is not onto.

$ \begin{aligned} & \text { Let } f(x)=2^{4 n+3}+3^{3 n+1} \\ & \Rightarrow P \text { divides } f(1), f(2), f(3), \ldots, f(n) \\ & \qquad \begin{array}{l} f(1)=2^7+3^4=128+81 \\ f(1)=209 \end{array} \end{aligned} $

$f(1)$ is divisible by 11 .

11 is an odd prime integer.

$ \begin{aligned} &\\ &\begin{aligned} & \cosh (x)=\frac{e^x+e^{-x}}{2} \\ & \tanh (x)=\frac{e^x-e^{-x}}{e^x+e^{-x}} \\ & x=\frac{e^{\cosh ^{-1}(x)}+e^{-\cosh ^{-1}(x)}}{2} \\ & x=\frac{y+\frac{1}{y}}{2} \\ & y^2-2 x y+1=0 \\ & y=\frac{2 x \pm \sqrt{4 x^2-4}}{2} \\ & e^{\cosh ^{-1}(x)}=x+\sqrt{x^2-1} \\ & \cosh ^{-1}(x)=\ln \left(x+\sqrt{x^2-1}\right) \end{aligned} \end{aligned} $

Again, $x=\frac{e^{\tanh ^{-1}(x)}-e^{-\tanh ^{-1}(x)}}{e^{\tanh ^{-1}(x)}+e^{-\tanh ^{-1}(x)}}$

Let $y=e^{\tanh ^{-1}(x)}$

$ \begin{array}{ll} \Rightarrow & x=\frac{y^2-1}{y^2+1} \\ \Rightarrow & x y^2+x=y^2-1 \\ \Rightarrow & \frac{x+1}{1-x}=y^2 \\ \Rightarrow & e^{\tanh ^{-1}(x)}=\sqrt{\frac{1+x}{1-x}} \\ \Rightarrow & \tanh ^{-1}(x)=\frac{1}{2} \ln \left(\frac{1+x}{1-x}\right) \end{array} $

Now, $\cosh ^{-1}(x)=\tanh ^{-1}(x)$

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

No solution

Aas $x \geq 1$ for $\ln \left(x+\sqrt{x^2-1}\right)$ to be real.

And for $\ln \left(\frac{1+x}{1-x}\right)$ to be defined

$ -1

Now, $\cosh ^{-1}(x)=\operatorname{coth}^{-1}(x)$

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

Now, $\ln \left(x+\sqrt{x^2-1}\right)$ is strictly increasing function, $\forall x \geq 1$ and $\frac{1}{2} \ln \left(\frac{x+1}{x-1}\right)$ is strictly decreasing function for $x>1$.

⇒ Both the graph will cut and one point exactly.

For $f(x)$ to be defined.

$ \begin{aligned} & \sqrt{x^2+x}+\sqrt{x^2-x}>0 \\ & \Rightarrow \quad x^2+x \geq 0 \text { and } x^2-x \geq 0 \\ & \Rightarrow \quad x(x+1) \geq 0 \text { and } x(x-1) \geq 0 \\ & \Rightarrow \quad(-\infty,-1] \cup[0, \infty) \end{aligned} $

and $(-\infty, 0] \cup[1, \infty)$

Domain

$ \begin{aligned} & =(-\infty,-1] \cup[0, \infty) \cap(-\infty, 0] \cup[1, \infty)-\{0\} \\ & =(-\infty,-1] \cup[1, \infty) \end{aligned} $

$ \begin{aligned} & \text { } \frac{x+1}{x^3(x-1)}=\frac{a}{x}+\frac{b}{x^2}+\frac{c}{x^3}+\frac{d}{x-1} \\ & =\frac{a x^2(x-1)+b x(x-1)+c(x-1)+d x^3}{x^3(x-1)} \\ & \Rightarrow x+1=x^3(a+d)+x^2(b-a)+x(c-b)-c \\ & \Rightarrow a+d=0, b-a=0 \end{aligned} $

$ \begin{aligned} & c-b=1,-c=1 \\ & \text { So, } c=-1, b=-2 a=-2 \\ & \Rightarrow a=b=2 c=-d \end{aligned} $

Given, $f(x)=5^{|x|}+\operatorname{Sgn}\left(5^x\right)$

$ \begin{array}{ll} \because & 5^x>0 \forall x \in R \\ \therefore & \operatorname{sgn}\left(5^x\right)=1 \\ \text { And } & |x| \geq 0 \Rightarrow-|x| \leq 0 \\ \Rightarrow & 5^x \leq 5^0 \Rightarrow 5^x \leq 1 \\ & 5^{|x|} \leq 5^0 \Rightarrow 5^{|x|} \leq 1 \\ \therefore & f(x)=5^{|x|}+1 \leq 1+1 \leq 2 \\ \because & f(1)=f(-1) \end{array} $

$\therefore f(x)$ is neither one-one nor onto.

$f(x)=\frac{x^2+x+k}{x^2-x+k}$

$ \text { ∵ Range }=\left[\frac{1}{3}, 3\right] $

For range let $f(x)=y$

$ \begin{aligned} & y=\frac{x^2+x+k}{x^2-x+k} \\ & \Rightarrow \quad x^2(y-1)-x(y+1)+k y-k=0 \\ & \therefore \quad D \geq 0 \\ & \Rightarrow \quad(y+1)^2-4 k(y-1)(y-1) \geq 0 \\ & \Rightarrow \quad(y+1)^2-(2 \sqrt{k}(y-1))^2 \geq 0 \\ & \Rightarrow \quad(y+1+2 \sqrt{k}(y-1)) \\ & \quad \quad(y+1-2 \sqrt{k}(y-1) \geq 0 \\ & \Rightarrow \quad(y(2 \sqrt{k}+1)+1-2 \sqrt{k}) \\ & \quad(y(2 \sqrt{k}-1)-2 \sqrt{k}-1) \leq 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) \\\because \,\,\,\, & y \in\left[\frac{1}{3}, 3\right] \\ &(3 y-1)(y-3) \leq 0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

Comparing Eqs. (i) and (ii),

$ 2 \sqrt{k}+1=3 \Rightarrow 2 \sqrt{k}=2 \Rightarrow k=1 $

We have,

$ \begin{aligned} & f(x)=4 x^3+a x^2+3 x-2 \\ & f^{\prime}(x)=12 x^2+2 a x+3 \end{aligned} $

For monotonic $f^{\prime}(x)>0$

$ \begin{aligned} & 12 x^2+2 a x+3>0 \\ & D<0 \\ \Rightarrow \quad & (2 a)^2-4(12)(3)<0 \Rightarrow a^2-36<0 \\ \Rightarrow \quad & (a-6)(a+6)<0 \Rightarrow a \in(-6,6) \end{aligned} $

$f(x)=\frac{x^2+x+a}{x^2-x+a}$

$\because f(x)$ is surjection, then range $=$ Codomain

Range $=R$

Now for range, put $f(x)=y$.

$ \begin{gathered} y=\frac{x^2+x+a}{x^2-x+a} \\ x^2(y-1)-x(y+1)+a y-a=0 \end{gathered} $

Now $x \in R \Rightarrow D \geq 0$

$ \begin{aligned} & (y+1)^2-4(y-1)(a y-a) \geq 0 \\ & \Rightarrow(y+1)^2-4 a(y-1)^2 \geq 0, \forall y \in R \\ & \Rightarrow y^2(1-4 a)+y(2+8 a)+1-4 a \geq 0, \forall y \in R\end{aligned} $

$ \begin{aligned} & \therefore 1-4 a \geq 0 \text { and } D \leq 0 \\ & \qquad a \leq \frac{1}{4} \text { and }(2+8 a)^2-4(1-4 a)^2 \leq 0 \\ & \Rightarrow(1+4 a)^2-(1-4 a)^2 \leq 0 \\ & \Rightarrow(1+4 a+1-4 a)(1+4 a-1+4 a) \leq 0 \\ & \Rightarrow 2(8 a) \leq 0 \\ & \Rightarrow a \leq 0 \\ & \therefore a \in(-\infty, 0) . \end{aligned} $

Given, $f(x)=\frac{1}{\sqrt{\log _{1 / 3}\left(\frac{x-1}{2-x}\right)}}$

∴ Domain

$ \begin{aligned} & \log _{1 / 3} \frac{x-1}{2-x}>0 \text { and } \frac{x-1}{2-x}>0 \\ & \Rightarrow \quad \frac{x-1}{2-x}<\left(\frac{1}{3}\right)^0 \Rightarrow \frac{x-1}{2-x}<1 \\ & \Rightarrow \quad \frac{x-1}{2-x}-1<0 \\ & \Rightarrow \quad \frac{x-1-2+x}{2-x}<0 \Rightarrow \frac{2 x-3}{2-x}<0 \end{aligned} $

Now,

Now, $x \in\left(1, \frac{3}{2}\right)$

$ \therefore \quad a=1, b=\frac{3}{2} $

Now, $2 b=3=a+2$

$f(x)=\left(x^2+x+1\right)^{x^2-3 x-4}$

Let $g(x)=x^2+x+1, \quad h(x)=x^2-3 x-4$

$ \begin{aligned} & g^{\prime}(x)=2 x+1, \quad h^{\prime}(x)=2 x-3 \\ & \therefore \quad x>4 \\ & \quad g^{\prime}(x)>0 \text { and } h^{\prime}(x)>0 \end{aligned} $

∴ both $g(x)$ and $h(x)$ are increasing

$\therefore f(x)$ is monotonically increasing function.

We have, $3 f(x)+4 f\left(\frac{1}{x}\right)=\frac{2}{x}-1$

$ \therefore \quad 3 f(3)+4 f\left(\frac{1}{3}\right)=\frac{2}{3}-1=\frac{-1}{3} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

Also, $3 f\left(\frac{1}{3}\right)+4 f(3)=6-1=5\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

On solving Eqs. (i) and (ii), we get

$ f(3)=3 $

$ \begin{aligned} & \text { We have, } y=\frac{10^x-10^{-x}}{10^x+10^{-x}}+1 \\ & \Rightarrow \quad y-1=\frac{10^x-10^{-x}}{10^x+10^{-x}} \\ & \text { Let } \quad a=10^x \\ & \therefore \quad 10^{-x}=\frac{1}{a} \\ & \therefore \quad y-1=\frac{a-\frac{1}{a}}{a+\frac{1}{a}}=\frac{a^2-1}{a^2+1} \\ & \Rightarrow \quad(y-1)\left(a^2+1\right)=a^2-1 \Rightarrow(y-1) a^2+(y-1)=a^2-1 \\ & \Rightarrow \quad a^2(y-1-1)=-y \Rightarrow a^2(y-2)=-y \\ & \Rightarrow \quad a^2=\frac{-y}{y-2} \Rightarrow 10^{2 x}=\frac{-y}{y-2} \\ & \Rightarrow \quad 2 x=\log _{10}\left(\frac{-y}{y-2}\right) \Rightarrow 2 x=\log _{10}\left(\frac{y}{2-y}\right) \\ & \Rightarrow \quad x=\frac{1}{2} \log _{10}\left(\frac{y}{2-y}\right) \end{aligned} $

Given, $f(x)=\tan \left(\frac{\pi}{\sqrt{x+1}+4}\right)$

For $\sqrt{x+1}$ to be defined,

$ \begin{array}{ll} \therefore & \sqrt{x+1}+4 \geq 0+4=4 \\ \Rightarrow & 0<\frac{1}{\sqrt{x+1}+4} \leq \frac{1}{4} \\ \Rightarrow & 0<\frac{\pi}{\sqrt{x+1}+4} \leq \frac{\pi}{4} \\ \therefore & f(x)=\tan \left(\frac{\pi}{\sqrt{x+1}}+4\right), \text { where } \\ & 0<\frac{\pi}{\sqrt{x+1}+4} \leq \frac{\pi}{4} \end{array} $

So, the function $f(x)$ is strictly increasing in the interval $\left(0, \frac{\pi}{4}\right]$

And the range of the given function is $\left(\tan (0), \tan \left(\frac{\pi}{4}\right)\right]$, which is $(0,1]$

$\therefore$ The range of $f(x)$ is $(0,1]$

Given,

$ \frac{x^3+3}{(x-3)^3}=a+\frac{b}{x-3}+\frac{c}{(x-3)^2}+\frac{d}{(x-3)^3} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

Let $y=x-3$

$ \Rightarrow x=y+3 $

Then, $\frac{x^3+3}{(x-3)^3}=\frac{(y+3)^3+3}{y^3}$

$ \begin{aligned} & =\frac{y^3+9 y^2+27 y+27+3}{y^3} \\ & =\frac{y^3+9 y^2+27 y+30}{y^3} \\ & =1+\frac{9}{y}+\frac{27}{y^2}+\frac{30}{y^3} \\ & =1+\frac{9}{x-3}+\frac{27}{(x-3)^2}+\frac{30}{(x-3)^3} \end{aligned} $

Comparing this with Eq. (i), we get

$ a=1, b=9, c=27, d=30 $

So, $(a+d)-(b+c)=(1+30)-(9+27)$

$ =31-36=-5 $

$ \begin{aligned} &\begin{aligned} & \text { } f(x)=\frac{3}{4-x^2}+\log _{10}\left(x^3-x\right) \\ & \text { for } \frac{3}{4-x^2}, 4-x^2 \neq 0 \\ & \Rightarrow \quad x \neq \pm 2 \end{aligned}\\ &\text { and for } \log _{10}\left(x^3-x\right) \text {, }\\ &x^3-x>0 \Rightarrow x\left(x^2-1\right)>0 \end{aligned} $

and $x \neq 2$

Here, $x \in(-1,0) \cup(1,2) \cup(2, \infty)$

$\because f(x)=\frac{4-x^2}{4+x^2}, \forall x \in A$ and $f(x)$ is bijective So, Domain of $f(x) 4+x^2 \neq 0$ which is always true. therefore, Domain of $f(x) \in R$ Lets check if $f(x)$ be injective

$ \begin{aligned} & f\left(x_1\right)=f\left(x_2\right) \\ & \Rightarrow \frac{4-x_1^2}{4+x_1^2}=\frac{4-x_2^2}{4+x_2^2} \\ & \Rightarrow 16+4 x_2^2-4 x_1^2-x_1^2 x_2^2 \\ & \quad=16+4 x_1^2-4 x_2^2-x_1^2 x_2^2 \\ & \Rightarrow x_1^2=x_2^2 \Rightarrow x_2= \pm x_1 \end{aligned} $

for $f(x)$ to be injective.

We must have $x_1=x_2$ and $-4 \in A$ therefore, domain must be ( $-\infty, 0$ ] Now, determine the range

Let, $y=\frac{4-x^2}{4+x^2}$

$ \begin{aligned} & \Rightarrow y\left(4+x^2\right)=4-x^2 \Rightarrow 4 y+y x^2=4-x^2 \\ & \Rightarrow x^2=\frac{4(1-y)}{1+y} \\ & \because x^2 \geq 0 \end{aligned} $

So, $\frac{4(1-y)}{1+y} \geq 0$

So, $1-y \geq 0$ and $1+y>0$

therefore, range of $f(x) \in(-1,1]$

that means, $A \subseteq(-\infty, 0]$

and $B=(-1,1]$

Hence, $A \cap B=(-1,0]$

$f(x)=x^2+2 b x+2 c^2$ is a quadratic, opening upward

So, minmum at $(x)=\frac{-2 b}{2}=-b$

So, maximum of $g(x)$ at $x=\frac{-(-2 c)}{2(-1)}=-c$

then, $g(-c)=-c^2+2 c^2+b^2=c^2+b^2$

and given that

$\min f(x)>\max g(x)$

$\Rightarrow \quad-b^2+2 c^2>c^2+b^2$

$\Rightarrow \quad c^2>2 b^2 \Rightarrow\left(\frac{c}{b}\right)^2>2$

$\Rightarrow \quad\left|\frac{c}{b}\right|>\sqrt{2}$

Hence, $\left|\frac{c}{b}\right| \in(\sqrt{2}, \infty)$

We have, $f: R \rightarrow A$

$ f(x)=\cos x+\sqrt{3} \sin x-1 $

is onto.

∴ Range $=$ codomain

$ \begin{aligned} & \text { Now, range of } f(x) \\ & \begin{aligned} \because \quad & -\sqrt{1^2+(\sqrt{3})^2} \leq \cos x \\ & +\sqrt{3} \sin x \leq \sqrt{1^2+(\sqrt{3})^2} \\ & -2 \leq \cos x+\sqrt{3} \sin x \leq 2 \\ & -3 \leq \cos x+\sqrt{3} \sin x-1 \leq 1 \\ & -3 \leq f(x) \leq 1 \end{aligned} \\ & \therefore A=[-3,1] \end{aligned} $

$ \begin{aligned} & \text { Given, } g(x)=1+x-[x] \\ &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, =1+\{x\} \\ & \begin{aligned} \because \quad 0 & \{x\}<1 \\ \Rightarrow \quad 1 & \leq g(x)<2 \end{aligned} \\ & f(x)= \begin{cases}-1, & x<0 \\ 0, & x=0 \\ 1, & x>0\end{cases} \\ & \text { Now, } f\{g(x)\}= \begin{cases}-1, & g(x)<0 \\ 0, & g(x)=0 \\ 1, & g(x)>0\end{cases} \end{aligned} $

$\therefore g(x)$ is always $[1,2)$

$ \therefore \quad f\{g(x)\}=1 $