Functions

For real values of $f(x)$

$ \sqrt{9-\sqrt{x^2-144}} \geq 0 $

On squaring both sides, we get

$ \begin{aligned} 9-\sqrt{x^2-144} & \geq 0 \\ 9 & \geq \sqrt{x^2-144} \end{aligned} $

Again, squaring on both sides, we get

$ \begin{aligned} 9^2 & \geq x^2-144 \\ 81+144 & \geq x^2 \\ 225 & \geq x^2 \\ x^2 & \leq 225 \end{aligned} $

On taking principal square root

$ |x| \leq 15 $

i.e. $x \leq 15$ or $x \geq-15$

$ \begin{aligned}Now,\,\,\, x^2-144 & \geq 0 \\ x^2 & \geq 144 \end{aligned} $

On taking principal square root

$ \begin{gathered} \sqrt{x^2} \geq \sqrt{12^2} \\ |x| \geq 12 \end{gathered} $

i.e. $x \leq-12$ or $x \geq 12$

We can see on number line the domain of the given function will be cross shadow area.

i.e. $[-15,-12] \cup[12,15]$

Given function $f: R \rightarrow\left[\frac{5}{2}, \infty\right]$ is defined by

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

Consider the graph

There are two points, where the expressions inside the absolute values change signs.

$ x=-\frac{1}{2} \text { or }-0.5 \text { and } x=2 $

If $x<-0.5$

$ \begin{aligned} f(x) & =-2 x-1-x+2 \\ & =-3 x+1 \end{aligned} $

If -0.5 < x < 2

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

If $x>2$

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

Each vertical line should intersect the graph at exactly one point. If not then function is not onto.

If a horizontal line intersects the graph more than once, the function is not one-one.

Clearly, from graph, through horizontal line two cut points implies that not one-one and through vertical line one cut point implies that onto.

Hence, function is onto but not one-one.

We start with the equation:

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

Differentiating both sides with respect to $ x $, we have:

$ 3 f'(x) - 2 \cdot f'\left(\frac{1}{x}\right) \times \left(-\frac{1}{x^2}\right) = 1. $

This simplifies to:

$ 3 f'(x) + \frac{2}{x^2} f'\left(\frac{1}{x}\right) = 1. $

First, set $ x = 2 $:

$ 3 f'(2) + \frac{2}{(2)^2} f'\left(\frac{1}{2}\right) = 1. $

This results in:

$ 3 f'(2) + \frac{1}{2} f'\left(\frac{1}{2}\right) = 1. $

Multiply throughout by 2 to clear the fraction:

$ 6 f'(2) + f'\left(\frac{1}{2}\right) = 2. \quad (i) $

Next, set $ x = \frac{1}{2} $:

$ 3 f'\left(\frac{1}{2}\right) + \frac{2}{\left(\frac{1}{2}\right)^2} f'(2) = 1. $

This becomes:

$ 3 f'\left(\frac{1}{2}\right) + 8 f'(2) = 1. $

Solve for $ f'\left(\frac{1}{2}\right) $:

$ f'\left(\frac{1}{2}\right) = \frac{1 - 8 f'(2)}{3}. \quad (ii) $

Substitute the expression for $ f'\left(\frac{1}{2}\right) $ from equation (ii) into equation (i):

$ 6 f'(2) + \frac{1 - 8 f'(2)}{3} = 2. $

Multiply through by 3 to eliminate the fraction:

$ 18 f'(2) + 1 - 8 f'(2) = 6. $

Simplify:

$ 10 f'(2) = 5. $

So, we find:

$ f'(2) = \frac{5}{10} = \frac{1}{2}. $

Mathematics

(d) We have,

$ f(x)=\log _2 \log _3 \log _5\left(x^2-5 x+11\right) $

$\log _5$ must be positive.

$\log _5\left(x^2-5 x+11\right)$ is defined if

$ x^2-5 x+11>0 $

Now, $D=b^2-4 a c=(-5)^2-4(1)(11)$

$ =-19 $

$\because$ If the discriminant is negative and $a>0, x^2-5 x+11>0, \forall$ real $x$.

Now, $\log _3$ must be positive.

$\log _3\left(\log _5\left(x^2-5 x+11\right)\right)$ is defined if

$ \log _5\left(x^2-5 x+11\right)>0 $

Since, $\log _5\left(x^2-5 x+11\right)>0$

$ \begin{aligned} & \Rightarrow \quad x^2-5 x+11>5^{\circ} \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\quad\left [\because \log _a x=b, \text { then } x=a^b\right] \\ & \Rightarrow \quad x^2-5 x+11>1 \\ & \Rightarrow \quad x^2-5 x+10>0 \\ & D=b^2-4 a c=(-5)^2-4(1)(10)=-15 \end{aligned} $

The discriminate is negative, $x^2-5 x+10$ has no real roots and is always positive. $\therefore x^2-5 x+10>0, \forall$ real $x$

Now, $\log _2$ must be positive.

$\log _2\left(\log _3\left(\log _5\left(x^2-5 x+11\right)\right)\right.$ is defined

If

$ \begin{aligned} & \log _3\left(\log _5\left(x^5-5 x+11\right)\right)>0 \\ & \log _3\left(\log _5\left(x^2-5 x+11\right)\right)>0 \\ & \log _5\left(x^2-5 x+11\right)>3^{\circ} \end{aligned} $

$ \begin{array}{lr} \Rightarrow & \log _5\left(x^2-5 x+11\right)>1 \\ \Rightarrow & x^2-5 x+11>5 \\ \Rightarrow & x^2-5 x+6>0 \\ & (x-2)(x-3)>0 \end{array} $

The quadratic $x^2-5 x+6$ changes sign at $x=2$ and $x=3$.

Thus, $x^2-5 x+6>0$ for $x \in(-\infty, 2) \cup(3, \infty)$.

Thus, the domain of $f(x)$ is $(-\infty, 2) \cup(3, \infty)$.

To find the range of the function $ f(x) = \frac{x^2 + 2x - 15}{2x^2 + 13x + 15} $, we begin by factorizing the numerator and denominator.

Denominator Factorization:

$ 2x^2 + 13x + 15 = 0 $

$ 2x^2 + 10x + 3x + 15 = 0 $

$ 2x(x + 5) + 3(x + 5) = 0 $

$ (x + 5)(2x + 3) = 0 \quad \Rightarrow \quad x = -5, \frac{-3}{2} $

Numerator Factorization:

$ x^2 + 2x - 15 = 0 $

$ x^2 + (5x - 3x) - 15 = 0 $

$ x(x + 5) - 3(x + 5) = 0 $

$ (x + 5)(x - 3) = 0 \quad \Rightarrow \quad x = -5, 3 $

Domain and Range:

The domain is $(-\infty, -5) \cup (-5, \frac{-3}{2}) \cup (\frac{-3}{2}, \infty)$.

The range is $\left(-\infty, \frac{1}{2}\right) \cup \left(\frac{1}{2}, \frac{8}{7}\right) \cup \left(\frac{8}{7}, \infty\right)$, excluding $\frac{1}{2}$ and $\frac{8}{7}$, which do not correspond to any real $x$ values for $f(x)$.

Thus, the range of the function is:

$ R - \left\{\frac{1}{2}, \frac{8}{7}\right\} $

Given, $f(\Omega)=6$

and $f(x+y)=f(x)+12 y$

Now,

$ \begin{aligned} f(2) & =f(1+1)=f(1)+12(1) \\ & =6+12=18 \\ f(3) & =f(1+2)=f(1)+12(2) \\ & =6+24=30 \\ f(4) & =f(1+3)=f(1)+12(3) \\ & =6+36=42 \end{aligned} $

and, so on

$ \begin{aligned} \therefore \quad f(n) & =6+(n-1) 12 \\ & =6+12 n-12=12 n-6 \end{aligned} $

Now, $\sum\limits_{r=1}^n f(r)=f(1)+f(2)+f(3)+\ldots+f(n)$

$ \begin{aligned} & =6+18+30+42+\ldots+(12 n-6) \\ & =6[1+3+5+7+\ldots+(2 n-1)] \\ & =6 n^2 \end{aligned} $

To determine the domain of the function $ f(x) = \sqrt{2+x} + \sqrt{3-x} $, we need to ensure that all expressions under the square roots are non-negative.

For $\sqrt{2+x}$ to be defined, the expression inside the square root must be non-negative:

$ 2 + x \geq 0 \quad \Rightarrow \quad x \geq -2 $

For $\sqrt{3-x}$ to be defined, the expression inside the square root must also be non-negative:

$ 3 - x \geq 0 \quad \Rightarrow \quad x \leq 3 $

Combining both conditions, the value of $ x $ must satisfy:

$ -2 \leq x \leq 3 $

Therefore, the domain of the function $ f(x) $ is the closed interval $[-2, 3]$.

Let $ f(x) = 3 + 2x $ and $ g_n(x) $ be the composition of $ n $ functions $ f $ with itself. If all the lines $ y = g_n(x) $ pass through a fixed point $(\alpha, \beta)$, we need to find $\alpha + \beta$.

The function $ f: \mathbb{R} \to \mathbb{R} $ is defined as:

$ f(x) = 2x + 3 = g_1(x) $

Now, find $ f \circ f(x) = g_2(x) $:

$ \begin{aligned} g_2(x) &= 2(2x + 3) + 3 \\ &= 4x + 6 + 3 \\ &= 4x + 9 \end{aligned} $

Next, find $ f \circ f \circ f(x) = g_3(x) $:

$ \begin{aligned} g_3(x) &= 2(4x + 9) + 3 \\ &= 8x + 18 + 3 \\ &= 8x + 21 \end{aligned} $

Observing a pattern, we have:

$ \begin{aligned} g_1(x) &= 2x + 3 = 2^1(x + 3) - 3 \\ g_2(x) &= 4x + 9 = 2^2(x + 3) - 3 \\ g_3(x) &= 8x + 21 = 2^3(x + 3) - 3 \\ & \vdots \\ g_n(x) &= 2^n(x + 3) - 3 \end{aligned} $

Then, the expression becomes:

$ \begin{aligned} y &= g_n(x) = 2^n(x + 3) - 3 \\ y + 3 &= 2^n(x + 3) \end{aligned} $

Now, consider the equation of a line passing through $(\alpha, \beta)$:

$ y - \beta = m(x - \alpha) $

Comparing the equations,

$ \begin{aligned} \alpha &= -3 \\ \beta &= -3 \end{aligned} $

Thus, $\alpha + \beta = -3 + (-3) = -6$.

$\text { The graph of } f(x) \text { is shown below. }$

$\therefore$ The function is not onto.

The function is many one as image of $f(x)$ is for every value of $a$ and $b$.

Hence, the function is neither one-one nor onto.

We have,

$ \begin{array}{lc} & P(x)=x^5+a x^4+b x^3+c x^2+d x+e \\ \because & P(0)=1 \\ \Rightarrow & 0+0+0+0+0+e=1 \\ \Rightarrow & e=1 \\ \because & P(1)=2 \\ \Rightarrow & 1+a+b+c+d+1=2 \\ \Rightarrow & a+b+c+d=0 ......\text { (i) } \\ \Rightarrow & 32+16 a+8 b+4 c+2 d+1=5 \\ \Rightarrow & 16 a+8 b+4 c+2 d=-28 \\ \Rightarrow & 8 a+4 b+2 c+d=-14 ......\text { (ii) }\end{array} $

$ \begin{aligned} & \because P(3)=10 \\ & \Rightarrow \quad 243+81 a+27 b+9 c+3 d+1=10 \\ & \Rightarrow 81 a+27 b+9 c+3 d=-234 \\ & \Rightarrow \quad 27 a+9 b+3 c+d=-78 ......\text { (iii) }\end{aligned} $

$ \begin{aligned} & P(4)=17 \\ & \Rightarrow 1024+256 a+64 b+16 c+4 d+1=17 \\ & \Rightarrow 256 a+64 b+16 c+4 d=-1008 \\ & \Rightarrow 64 a+16 b+4 c+d=-252 ......\text { (iv) } \end{aligned} $

From Eqs. (i), (ii), (iii) and (iv), we get

$ \begin{aligned} & {\left[\begin{array}{cccc} 1 & 1 & 1 & 1 \\ 8 & 4 & 2 & 1 \\ 27 & 9 & 3 & 1 \\ 64 & 16 & 4 & 1 \end{array}\right]\left[\begin{array}{l} a \\ b \\ c \\ d \end{array}\right]=\left[\begin{array}{c} 0 \\ -14 \\ -78 \\ -252 \end{array}\right]} \\ & {\left[\begin{array}{cccc} 1 & 1 & 1 & 1 \\ 8 & 4 & 2 & 1 \\ 27 & 9 & 3 & 1 \\ 64 & 16 & 4 & 1 \end{array}\right]\left[\begin{array}{c} 0 \\ -14 \\ -78 \\ -252 \end{array}\right]} \\ & R_2 \rightarrow R_2-8 R_1 \\ & {\left[\begin{array}{cccc} 1 & 1 & 1 & 1 \\ 0 & -4 & -6 & -7 \\ 27 & 9 & 3 & 1 \\ 64 & 16 & 4 & 1 \end{array}\right]\left[\begin{array}{c} 0 \\ -14 \\ -78 \\ -252 \end{array}\right]} \\ & R_3 \rightarrow R_3-27 R_1 \\ & {\left[\begin{array}{cccc} 1 & 1 & 1 & 1 \\ 0 & -4 & -6 & -7 \\ 0 & -18 & -24 & -26 \\ 64 & 16 & 4 & 1 \end{array}\right]\left[\begin{array}{c} 0 \\ -14 \\ -78 \\ -252 \end{array}\right]} \\ & R_4 \rightarrow R_4-64 R_1 \\ & {\left[\begin{array}{cccc} 1 & 1 & 1 & 1 \\ 0 & -4 & -6 & -7 \\ 0 & -18 & -24 & -26 \\ 0 & -48 & -60 & -63 \end{array}\right]\left[\begin{array}{c} 0 \\ -14 \\ -78 \\ -252 \end{array}\right]} \end{aligned} $

$ \begin{aligned} & R_3 \rightarrow R_3-\frac{9}{2} R_2 \\ & {\left[\begin{array}{cccc} 1 & 1 & 1 & 1 \\ 0 & -4 & -6 & -7 \\ 0 & 0 & 3 & \frac{11}{2} \\ 0 & -48 & -60 & -63 \end{array}\right]\left[\begin{array}{c} 0 \\ -14 \\ -15 \\ -252 \end{array}\right]} \\ & R_4 \rightarrow R_4-12 R_2 \\ & {\left[\begin{array}{cccc} 1 & 1 & 1 & 1 \\ 0 & -4 & -6 & -7 \\ 0 & 0 & 3 & \frac{11}{2} \\ 0 & 0 & 12 & 21 \end{array}\right]\left[\begin{array}{c} 0 \\ -14 \\ -15 \\ -84 \end{array}\right]} \\ & R_4 \rightarrow R_4-4 R_3 \\ & {\left[\begin{array}{cccc} 1 & 1 & 1 & 1 \\ 0 & -4 & -6 & -7 \\ 0 & 0 & 3 & \frac{11}{2} \\ 0 & 0 & 0 & -1 \end{array}\right]\left[\begin{array}{c} 0 \\ -14 \\ -15 \\ -24 \end{array}\right]} \end{aligned} $

Thus,

$ \begin{aligned} a+b+c+d & =0 \\ -4 b-6 c-7 d & =-14 \\ 3 c+\frac{11}{2} d & =-15 \\ -d & =-24 \end{aligned} $

On solving, we get

$ \begin{aligned} & a=-10, b=35, c=-49, d=24 \\ & \text { Now, } P(5)=3125+625 \times(-10) \\ & \quad+125 \times 35+25 \times(-49)+24 \times 5+1 \\ & =3125-6250+4375-1225+120+1 \\ & =146 \end{aligned} $

We are given a real-valued function $ f: [a, \infty) \rightarrow [b, \infty) $ defined by $ f(x) = 2x^2 - 3x + 5 $. This function is a quadratic and is stated to be a bijection.

First, let's find the vertex of the quadratic function, as it will help determine the minimum value of the function, which is crucial for it to be a bijection.

The vertex formula for a quadratic function $ ax^2 + bx + c $ is given by:

$ \left(\frac{-b}{2a}, \frac{-D}{4a}\right) $

where $ D = b^2 - 4ac $. For our function, $ a = 2 $, $ b = -3 $, and $ c = 5 $.

Calculating the vertex:

Calculate $ x $-coordinate of the vertex:

$ \frac{-(-3)}{2 \times 2} = \frac{3}{4} $

Calculate the discriminant $ D $:

$ D = (-3)^2 - 4 \times 2 \times 5 = 9 - 40 = -31 $

Calculate $ y $-coordinate of the vertex:

$ \frac{-(-31)}{8} = \frac{31}{8} $

Thus, the vertex of the quadratic function is $ \left(\frac{3}{4}, \frac{31}{8}\right) $.

Since $ f(x) = 2x^2 - 3x + 5 $ is a parabola opening upwards (as the coefficient of $ x^2 $ is positive), the minimum value of $ f(x) $ occurs at the vertex. Therefore, to achieve a bijective function, the domain should start from the $ x $-coordinate of the vertex, and the range should start from the $ y $-coordinate of the vertex.

This implies:

$ a = \frac{3}{4} $

$ b = \frac{31}{8} $

To find the value of $ 3a + 2b $:

$ 3 \times \frac{3}{4} + 2 \times \frac{31}{8} = \frac{9}{4} + \frac{62}{8} = \frac{9}{4} + \frac{31}{4} = \frac{40}{4} = 10 $

Thus, $ 3a + 2b = 10 $.

To find the domain of the real-valued function $ f(x) = \frac{1}{\sqrt{\log_{0.5}(2x-3)}} + \sqrt{4-9x^2} $, we analyze the conditions under which each part of the function is defined.

First Part: $ \frac{1}{\sqrt{\log_{0.5}(2x-3)}} $

For the square root in the denominator to be defined and non-zero, the expression inside must be positive: $\log_{0.5}(2x-3) > 0$.

The base of the logarithm is less than 1. This means:

$ 2x - 3 < 1 \quad \Rightarrow \quad x < 2 $

Additionally, the expression inside the logarithm must be positive:

$ 2x - 3 > 0 \quad \Rightarrow \quad x > \frac{3}{2} $

Second Part: $ \sqrt{4-9x^2} $

The expression under the square root must be non-negative:

$ 4 - 9x^2 > 0 \quad \Rightarrow \quad x^2 < \frac{4}{9} $

Solving for $ x $, we derive:

$ -\frac{2}{3} < x < \frac{2}{3} $

Intersection of Conditions:

From the first part, we need $ x $ to satisfy both conditions: $ \frac{3}{2} < x < 2 $.

From the second part, $ x $ must be in the interval $ -\frac{2}{3} < x < \frac{2}{3} $.

The intersection of these conditions is empty, as there is no number that simultaneously satisfies $ \frac{3}{2} < x < 2 $ and $ -\frac{2}{3} < x < \frac{2}{3} $.

Hence, the domain of the function is a null set.

$f(x)=x^3-x$ is not one-one function as $f(A)=f(0)=f(-1)=0$ $f(x)=x^3-x$ is a polynomial function and every polynomial fanction is continnous and differentiable in its domala.

If $x \rightarrow \infty$, then $f(x) \rightarrow \infty$

$[\because f(x)$ is a cubic function]

If $x \rightarrow-\infty$, then $f(x) \rightarrow-\infty$

So, $\rightarrow \infty

$f(x) \in R$ and $x \in R$

$\therefore f(x)$ is onto function.

To determine $ g(x) $, let's first understand the given functions:

We have $ f(x) = \sqrt{x - 1} $. This means $ f(x) $ is a function that transforms $ x $ into $ \sqrt{x - 1} $.

Next, the composition $ g(f(x)) = x + 2x^2 + 1 $ can be rewritten using substitution and simplification steps:

Begin by rewriting the expression for $ g(f(x)) $:

$ g(f(x)) = x + 2\sqrt{x-1} + 1 $

Recognize a perfect square identity in:

$ x + 2\sqrt{x-1} + 1 = (\sqrt{x-1} + 1)^2 $

Here, $ (\sqrt{x-1} + 1)^2 = \sqrt{x-1}^2 + 2\sqrt{x-1} + 1 $, which simplifies to $ x - 1 + 2\sqrt{x-1} + 1 = x + 2\sqrt{x-1} $.

Notice then that:

$ g(f(x)) = (\sqrt{x-1} + 1)^2 = ((\sqrt{x-1} - 1) + 2)^2 = (f(x) + 2)^2 $

Thus, we can deduce that $ g(x) $ must match the transformed output when $ x = f(x) $ plus an added constant:

$ g(x) = (x + 2)^2 $

Therefore, $ g(x) $ is the function that squares the input, transformed by adding $ 2 $.

Let's solve the problem such that the given expression assumes all real values.

We're looking at the expression:

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

To ensure that $ y $ can take all real values, the denominator must not be zero:

$ 2x^2 - 3x + 1 \neq 0 $

This can be factored as:

$ (2x - 1)(x - 1) \neq 0 $

Thus, $ x \neq 1 $ and $ x \neq \frac{1}{2} $.

For $ y $ to take all real values for real $ x $, the quadratic expression $ (3y+1)^2 - 4y(y-2) \geq 0 $ must hold:

$ (3y+1)^2 - 4[y(y-2) - a] \geq 0 $

Simplifying the terms:

$ 9y^2 + 6y + 1 - (4y^2 - 8y - 4a) \geq 0 $

This simplifies to:

$ 5y^2 + 14y + 1 + 4a \geq 0 $

As a quadratic in terms of $ y $, for it to be non-positive (invertible for any $ x $), the discriminant must be negative:

The discriminant:

$ 6^2 - 4 \times 5 \times (1 + 4a) < 0 $

$ 36 - 20(1 + 4a) < 0 $

$ 36 - 20 - 80a < 0 $

$ 16 < 80a $

Thus:

$ a > \frac{1}{2} $

Therefore, the condition for $ a $ is:

$ -1 < a < -\frac{1}{2} $

The correct understanding implies the feasible range for $ a $.

To find the value of $ h $ that removes the $ x^3 $ term from the polynomial $ f(x) = x^4 + 2x^3 - 19x^2 - 8x + 60 = 0 $ when transformed to $ f(x + h) = 0 $, follow these steps:

Polynomial Transformation: The transformation affects each term in the original polynomial. For the $ x^3 $ term in the polynomial, apply the transformation:

$ f(x+h) = (x+h)^4 + 2(x+h)^3 - 19(x+h)^2 - 8(x+h) + 60 $

Focus on the $ x^3 $ Terms: When expanding, consider only the terms involving $ x^3 $:

From $(x+h)^4$:

$ ^4C_1 \cdot x^3 \cdot h = 4hx^3 $

From $2(x+h)^3$:

$ 2 \times 1 \times x^3 = 2x^3 $

Collect $ x^3 $ Terms: Adding these terms together gives:

$ 4hx^3 + 2x^3 = (4h + 2)x^3 $

Set the Coefficient to Zero: To eliminate the $ x^3 $ term from the polynomial, the coefficient must be zero:

$ 4h + 2 = 0 $

Solve for $ h $:

$ 4h = -2 \quad \implies \quad h = \frac{-2}{4} = \frac{-1}{2} $

Thus, the value of $ h $ that removes the $ x^3 $ term is $ h = -\frac{1}{2} $.

First, the function $f(x) = \sin^{-1}(\log_{3x}(\frac{6 + 2 \log_3{x}}{-5x}))$ has several restrictions :

1. Since the arcsine function $\sin^{-1}(x)$ is only defined for $-1\leq x\leq 1$, this means that $\log_{3x}(\frac{6 + 2 \log _3 x}{-5 x})$ must be between -1 and 1.

2. For the logarithm to be defined, $3x > 0 \Rightarrow x > 0$ .......(1)
   
   because the base of a logarithm must be greater than 0 and not equal to 1. Also, $x \neq \frac{1}{3}$ .......(2)
   
   as the base cannot be 1.

3. Moreover, the inner function of the logarithm $\frac{6 + 2 \log_3{x}}{-5x}$ must be greater than 0.
   
   $ \Rightarrow $ $6+2 \log _3 x<0 \quad(\because x>0)$
   
   $ \begin{aligned} \Rightarrow & \log _3 x<-3 \\\\ \Rightarrow & x<3^{-3} \\\\ \Rightarrow & x<\frac{1}{27} .........(3) \end{aligned} $
   
   $ \Rightarrow \text { From (1), (2), (3), } 0 < x < \frac{1}{27} $

The next step is to solve the inequality for $-1 \leq \log_{3x}(\frac{6+2 \log _3 x}{-5 x}) \leq 1$. To do this, we make the observation that for the logarithmic part to be within $[-1,1]$, it must be true that $3x \leq \frac{6+2 \log_3 x}{-5x} \leq \frac{1}{3x}$.

Solving the inequality $15x^2 + 6 + 2 \log_3 x \geq 0$, we find that $x \in(0, \frac{1}{27})$ ........(4)

Likewise, solving the inequality $6+2 \log_3 x + \frac{5}{3} \geq 0$, we find that $x \geq 3^{-\frac{23}{6}}$ ........(5)

Combining Equations (3), (4), and (5), the intersection of all these intervals is $x \in[3^{-\frac{23}{6}}, \frac{1}{27})$.

Now, consider the function $g(x) = x - [x]$, where $[x]$ is the greatest integer function. For this function, the range is the fractional part of $x$. In this case, the range $(\alpha, \beta)$ is given by the minimum and maximum possible values of $x$ in its domain. Hence, $\alpha = 3^{-\frac{23}{6}}$ and $\beta = \frac{1}{27}$.

Finally, substitute these values into the equation $\alpha^{2}+\frac{5}{\beta}$:

$\alpha^{2}+\frac{5}{\beta} = (3^{-\frac{23}{6}})^2 + \frac{5}{\frac{1}{27}} = 3^{-\frac{23}{3}} + 135$ = 135.0002198.

Since $3^{-\frac{23}{3}}$ = 0.0002198 is an extremely small number, it's approximately 0. So, $\alpha^{2}+\frac{5}{\beta} \approx 135$.

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

$y' = {{ - 1} \over {2\sqrt 3 - x}} + {1 \over {2\sqrt {x + 2} }} = 0$

$ \Rightarrow \sqrt x + 2 = \sqrt 3 - x$

$ \Rightarrow x = {1 \over 2}$

$y\left( {{1 \over 2}} \right) = \sqrt {{5 \over 2}} + \sqrt {{5 \over 2}} = \sqrt {10} $

${y_{\min }}$ at $x = - 2$ or $x = 3$, i.e., $\sqrt 5 $

$\therefore$ $f(x) \in [\sqrt 5 ,\sqrt {10} ]$

$f(1) + 2f(2) + 3f(3)\, + \,...\, + \,nf(n) = n(n + 1) + (n)$ ..... (i)

$n \to n + 1$

$f(1) + 2f(2)\, + \,...\, + \,(n + 1)f(n + 1) = (n + 1)(n + 2)f(n + 1)$ ...... (ii)

(i) and (ii) gives

$3f(3) - 2f(2) = 0$

$4f(4) - 3f(3) = 0$

$ \vdots $

$(n + 1)f(n + 1) - nf(n) = 0$

$ \Rightarrow f(n + 1) = {{2f(2)} \over {n + 1}}$

$f(n) = {1 \over {2n}}$

${1 \over {f(2022)}} + {1 \over {f(2028)}} = 8100$

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

$ \Rightarrow f'(x) = {{({x^2} + 1)(2x + 2) - ({x^2} + 2x + 1)(2x)} \over {{{({x^2} + 1)}^2}}}$

$ \Rightarrow f'(x) = {{2 - 2{x^2}} \over {{{({x^2} + 1)}^2}}}$

$ = {{2\left( {1 + x} \right)\left( {1 - x} \right)} \over {{{\left( {{x^2} + 1} \right)}^2}}}$

$ \therefore $ $f'(x) = 0$ at x = 1 and x = -1. So, x = 1 and x = -1 are point of maxima/minima.

Option A : $f(x)$ is many-one in $( - \infty , - 1)$.

As x = -1 is a point of maxima/minima. So it is a boundary point in the range $( - \infty , - 1)$. So, $f(x)$ is one-one in $( - \infty , - 1)$.

$ \therefore $ Option A is incorrect.

Option B : $f(x)$ is one-one in $( - \infty ,\infty )$.

As, x = 1 and x = -1 are point of maxima/minima which is present inside the range $( - \infty ,\infty )$. So, $f(x)$ is many-one function in $( - \infty ,\infty )$.

$ \therefore $ Option B is incorrect.

Option C : $f(x)$ is one-one in $[1,\infty )$ but not in $( - \infty ,\infty )$.

As x = 1 is a point of maxima/minima. So it is a boundary point in the range $[1,\infty )$. So, $f(x)$ is one-one in $[1,\infty )$.

As, x = 1 and x = -1 are point of maxima/minima which is present inside the range $( - \infty ,\infty )$. So, $f(x)$ is many-one function in $( - \infty ,\infty )$ .

$ \therefore $ Option C is correct.

Option D : $f(x)$ is many-one in $(1,\infty )$.

As x = 1 is a point of maxima/minima. So it is a boundary point in the range $[1,\infty )$. So, $f(x)$ is one-one in $[1,\infty )$.

$ \therefore $ Option D is incorrect.

Note :

Methods to Check One-One Function :

(i) If a function is one-one, any line parallel to $x$-axis cuts the graph of the function maximum at one point.

(ii) Any continuous function which is entirely increasing or decreasing (no maxima/minima is present) in whole domain will always be one-one function.

$ \text { Here, } f(x)=\left\{\begin{array}{cc} x^2-4 x+3 & , x<2 \\ x-3 & , x \geq 2 \end{array}\right. $

$ \text { Number of solutions }=2 $

$ \begin{aligned} & \text { Here, } f(x)=3 x-2, g(x)=x^2+2 \\ & g o f(x)=(3 x-2)^2+2=9 x^2-12 x+6 \\ & f og(x)=3\left(x^2+2\right)-2=3 x^2+4 \\ & \text { Now, }[(g \circ f)+(f \circ g)](x)=12 x^2-12 x+10 \\ & =12\left(x^2+2\right)-4(f(x))-22 \\ & =12 g(x)-4 f(x)-22 \end{aligned} $

$ \begin{array}{r} f(x)=a x^9+b x^7+c x^5+d x^3 \\ +e x+\frac{15}{x}+12 \end{array} $

Let $g(x)=a x^9+b x^7+c x^5+d x^3+e x+\frac{15}{x}$ which is an odd function odd function, let $g(x)$ We have,

$ \begin{aligned} \Rightarrow \quad g(-x) & =-g(x) \\ -4 & =f(4)=g(4)+12 \end{aligned} $

$ \begin{array}{llrl} \Rightarrow g(4) =-16 \\ \Rightarrow g(-4) =16 \\ f(-4)=g(-4)+12 & =16+12=28 \end{array} $

As, we know in ${ }^n C_r, n \geq r$ and $n \in I$, $r \geq 0$

So, to define, ${ }^{(16-x)} C_{(2 x-1)}$

$ \begin{array}{ll} & 16-x \geq 2 x-1 \\ \Rightarrow & 3 x \leq 17 \\ \Rightarrow & x \leq \frac{17}{3} \text { and } 2 x-1 \geq 0 \\ \Rightarrow & x \geq \frac{1}{2} \end{array} $

So, $\quad \frac{1}{2} \leq x \leq \frac{17}{3}$

Hence, $x=\{1,2,3,4,5\}$

$ \begin{aligned} & X=\left\{\left.\left[\begin{array}{ll} a & b \\ c & d \end{array}\right] \right\rvert\, a, b, c, d \in R\right\} \\ & f: X \rightarrow R \\ & f(A)=\operatorname{det}(A), \forall A \in X \end{aligned} $

Now, by taking examples

$ \begin{aligned} & {\left[\begin{array}{ll} 1 & 1 \\ 1 & 1 \end{array}\right],\left[\begin{array}{ll} 2 & 2 \\ 2 & 2 \end{array}\right] \in X} \\ & \operatorname{det}\left[\begin{array}{ll} 1 & 1 \\ 1 & 1 \end{array}\right]=\operatorname{det}\left[\begin{array}{ll} 2 & 2 \\ 2 & 2 \end{array}\right]=0 \\ & f\left(\left[\begin{array}{ll} 1 & 1 \\ 1 & 1 \end{array}\right]\right)=f\left(\left[\begin{array}{ll} 2 & 2 \\ 2 & 2 \end{array}\right]\right) \\ & \Rightarrow\left[\begin{array}{ll} 1 & 1 \\ 1 & 1 \end{array}\right] \neq\left[\begin{array}{ll} 2 & 2 \\ 2 & 2 \end{array}\right] \end{aligned} $

So, this function is not one-one.

Let $f: X \rightarrow Y$

to show onto for each such that $f(x)=y_0$; $y_0 \in Y$

there exists $x \in X$

$ \begin{aligned} & y_0 \in R \\ & {\left[\begin{array}{cc} y_0 & 0 \\ 0 & 1 \end{array}\right] \text { such that }\left|\begin{array}{cc} y_0 & 0 \\ 0 & 1 \end{array}\right|=y_0} \\ & f\left(\left[\begin{array}{cc} y_0 & 0 \\ 0 & 1 \end{array}\right]\right)=y_0 \end{aligned} $

This function is onto function.