Functions

$f(x)=2 x^{n}+\lambda, \lambda \in \mathbb{R}, n \in \mathbb{N}$ $f(4)=2 \cdot 4^{n}+\lambda=133, f(5)=2 \cdot 5^{n}+\lambda=255$ $f(5)-f(4)=2 \cdot\left(5^{n} \cdot 4^{n}\right)=122 \Rightarrow n=3$ $\Rightarrow f(3)-f(2)=2 \cdot\left(3^{n} \cdot 2^{n}\right)=2 \cdot\left(3^{3}-2^{3}\right)=2 \times 19$ Required sum $=1+2+19+38=60$

But, if we consider the domain of the composite function $f \circ g(x)$ then in that case $f(0)$ will be not defined as $\mathrm{g}(\mathrm{x})$ cannot be equal to zero.

First, the function $f(x) = \sin^{-1}(\log_{3x}(\dfrac{6 + 2 \log_3{x}}{-5x}))$ has several restrictions : Since the arcsine function $\sin^{-1}(x)$ is only defined for $-1\leq x\leq 1$, this means that $\log_{3x}(\dfrac{6 + 2 \log _3 x}{-5 x})$ must be between -1 and 1.

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, $x \neq \dfrac{1}{3}$ .......(2) as the base cannot be 1.

Moreover, the inner function of the logarithm $\dfrac{6 + 2 \log_3{x}}{-5x}$ must be greater than 0. $\Rightarrow$ $6+2 \log _3 x0)$ $ \begin{aligned} \Rightarrow & \log _3 x 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,$\alpha^{2}+\frac{5}{\beta} \approx 135$.

For Domain $[x]^2-3[x]-10>0$

Given that $f(x)=\dfrac{\left(\tan 1^{\circ}\right) x+\log _e(123)}{x \log _e(1234)-\left(\tan 1^{\circ}\right)}$ Let us consider a similar function of $(x)$, $\therefore f(x)=\dfrac{A x+B}{C x-A}$ $\text{ Now, }$

Hence, the least value is 4 .

If $\mathrm{x} \in \mathrm{I},\lceil\mathrm{x}\rceil=[\mathrm{x}]$ (greatest integer function) If $x \notin I,\lceil x\rceil=[x]+1$

..... (1)

..... (2)

..... (3)

To find the domain of the function

we need to consider the domain conditions for both the square root function and the logarithmic function.

The square root function $\sqrt{x^2-25}$ requires that the argument of the square root be non-negative, so

This inequality is satisfied when

The denominator of the rational part of $f(x)$, $(4-x^2)$, cannot be zero, otherwise, the function will become undefined due to division by zero.

Thus, we must have

This inequality is violated when

Combining these conditions gives us the domain for the rational part of the function:

Moving on to the logarithmic function, $\log_{10}(x^2+2x-15)$, the argument must be positive:

This is a quadratic inequality, which we can factor to find the solution:

From this, we see that the inequality is satisfied for

The overall domain of $f(x)$ is the intersection of the domains for each piece.

Taking the intersection of the two sets gives us:

Since the question states that the domain is of the form $(-\infty, \alpha) \cup [\beta, \infty)$, we can infer that

We calculate $\alpha^2 + \beta^3$ as follows:

So the correct answer, representing the sum of $\alpha^2$ and $\beta^3$, is: Option D $150$.