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Calculation:

Given,

The function is $y = \sin^{-1} \left( x - \frac{4x^3}{27} \right)$, and we are tasked with simplifying it.

$y = \sin^{-1} \left( x - \frac{4x^3}{27} \right)$

We recognize that we can factor the expression as:

$x - \frac{4x^3}{27} = \left( \frac{x}{3} \right) \left( 3 - 4 \left( \frac{x}{3} \right)^3 \right)$

Thus, the function becomes:

$y = \sin^{-1} \left( \frac{x}{3} \left( 3 - 4 \left( \frac{x}{3} \right)^3 \right) \right)$

Recognizing that this fits a known identity for inverse trigonometric functions:

$\sin^{-1}(x) = 3 \sin^{-1} \left( \frac{x}{3} \right)$

Thus, we can simplify the expression to:

$y = 3 \sin^{-1} \left( \frac{x}{3} \right)$

Hence, the correct answer is Option 4.

Calculation:

Given,

The function is $y = 3 \sin^{-1} \left( \frac{x}{3} \right)$, and we are tasked with finding the derivative of y , i.e., $\frac{dy}{dx}$.

The general derivative of $\sin^{-1}(u)$ is:

$\frac{d}{dx} \sin^{-1}(u) = \frac{1}{\sqrt{1 - u^2}} \cdot \frac{du}{dx}$, where$u = \frac{x}{3}$

$\frac{du}{dx} = \frac{1}{3}$

Apply the chain rule to find the derivative of y :

$\frac{dy}{dx} = 3 \times \frac{1}{\sqrt{1 - \left( \frac{x}{3} \right)^2}} \times \frac{1}{3}$

$\frac{dy}{dx} = \frac{1}{\sqrt{1 - \frac{x^2}{9}}} = \frac{3}{\sqrt{9 - x^2}}$

Hence, the correct answer is Option 3.

Calculation:

Given,

The function is $f(x) = \sqrt{x^2 + 9} - 3$ and $g(x) = \sqrt{x^2 + 16} - 4$.

We are tasked with finding:

$\lim_{x \to 0} \frac{f(x)}{g(x)}$

Multiply both the numerator and denominator by their respective conjugates:

$\frac{\sqrt{x^2 + 9} - 3}{\sqrt{x^2 + 16} - 4} \times \frac{\sqrt{x^2 + 9} + 3}{\sqrt{x^2 + 9} + 3} \times \frac{\sqrt{x^2 + 16} + 4}{\sqrt{x^2 + 16} + 4}$

Simplify the numerator:

$(\sqrt{x^2 + 9} - 3)(\sqrt{x^2 + 9} + 3) = x^2$

Simplify the denominator:

$(\sqrt{x^2 + 16} - 4)(\sqrt{x^2 + 16} + 4) = x^2$

Now, the expression becomes:

$\frac{x^2}{x^2} \times \frac{\sqrt{x^2 + 16} + 4}{\sqrt{x^2 + 9} + 3}$

Simplify and evaluate the limit:

$\frac{\sqrt{x^2 + 16} + 4}{\sqrt{x^2 + 9} + 3}$ becomes:

$\frac{\sqrt{16} + 4}{\sqrt{9} + 3} = \frac{4 + 4}{3 + 3} = \frac{8}{6} = \frac{4}{3}$

Hence, the correct answer is Option 3.

Calculation:

Given,

The function is $f(x) = x^2 + 9$.

Statement I: f(x) is an increasing function.

The derivative of f(x) is:

$f'(x) = \frac{d}{dx}(x^2 + 9) = 2x$

When $( x > 0 )$, $( f'(x) > 0 )$, so (f(x) is increasing.

When (x < 0), f'(x) < 0 ), so f(x)  is decreasing.

At (x = 0), ( f'(x) = 0 ), meaning the function is neither increasing nor decreasing at this point.

Hence, f(x)  is not entirely increasing. It is increasing for (x > 0) and decreasing for ( x < 0).

Statement II: f(x) has local maximum at x = 0

Since the function$f(x) = x^2 + 9$ is a parabola opening upwards (because the coefficient of x² is positive), it has a global minimum at x = 0, not a local maximum.

Conclusion:

• Statement I is incorrect because the function is not entirely increasing. It is increasing for x > 0  and decreasing for  x < 0 .
• Statement II is incorrect because the function has a global minimum at x = 0, not a local maximum.

Hence, the correct answer is Option 4.

Calculation:

Given,

The function satisfies the equation $\frac{f(x)}{f(y)} = \frac{x}{y}$ for all positive real values of x and y, and $f(2) = 3$.

We are tasked with finding $f(16)$.

Using the functional equation, for $f(4)$, we have:

$\frac{f(4)}{f(2)} = f\left( \frac{4}{2} \right) = f(2)$

Since $f(2) = 3$, we can calculate $f(4)$

$\frac{f(4)}{3} = 3 \quad \Rightarrow \quad f(4) = 9$

Next, to find $f(16)$, we use the functional equation again:

$\frac{f(16)}{f(4)} = f\left( \frac{16}{4} \right) = f(4)$

Since $f(4) = 9$, we can calculate $f(16)$

$\frac{f(16)}{9} = 9 \quad \Rightarrow \quad f(16) = 81$

Hence, the correct answer is Option 4.

Calculation:

Given,

The function satisfies the functional equation:

$f\left(\frac{x}{y}\right) = \frac{f(x)}{f(y)}$

Also, we are given that:

$f(2) = 3$

We use the functional equation to calculate f(4) . Substituting x = 4 and y = 2 , we get:

$\frac{f(4)}{f(2)} = f(2)$

$\frac{f(4)}{3} = 3 \quad \Rightarrow f(4) = 3 \times 3 = 9$

$\frac{f(2)}{f(2)} = f(1) \quad \Rightarrow f(1) = 1$

$f(1) \times f(4) = 1 \times 9 = 9$

Hence, the correct answer is Option 3.

Calculation:

Given,

The function is such that: $f(xy) = f(x + y)$ for all real values of x  and y , and $f(5) = 10$.

We need to find the value of $f(0)$.

Substitute $x = 5$ and $y = 0$ in the given functional equation $f(xy) = f(x + y)$:

$f(5 \cdot 0) = f(5 + 0)$

$f(0) = f(5)$

Since $f(5) = 10$, we conclude:

$f(0) = 10$

Hence, the correct answer is Option 4.

Calculation:

Given,

The function is $f(xy) = f(x + y)$ for all real values of x  and y, and f(5) = 10 .

We are tasked with finding:

$f(20) + f(-20)$

Using the given functional equation, we have:

For $f(0 \cdot 5) = f(0 + 5)$, we get:

$f(0) = f(5) = 10$

For $f(0 \cdot 20) = f(0 + 20)$, we get:

$f(0) = f(20) = 10$

For $f(0 \cdot -20) = f(0 + (-20))$, we get:

$f(0) = f(-20) = 10$

Thus,

$f(20) + f(-20) = 10 + 10 = 20$

Hence, the correct answert is Option 3.

Calculation:

Given,

The function is $f(x) = \left\lfloor x^2 \right\rfloor$, where $\left\lfloor x^2 \right\rfloor$ is the greatest integer function.

We are tasked with finding:

$\int_{\frac{\sqrt{3}}{2}}^{\sqrt{3}} \left\lfloor x^2 \right\rfloor , dx$

Decomposing the integral based on the function:

For $\frac{\sqrt{3}}{2} \leq x < 1$, x² lies between $\frac{3}{4}$ and 1, so $\left\lfloor x^2 \right\rfloor = 0$. Therefore,

$\int_{\frac{\sqrt{3}}{2}}^1 0 , dx = 0$

For$1 \leq x < \sqrt{2}$, x²  lies between 1 and 2, so $\left\lfloor x^2 \right\rfloor = 1 .$ Therefore,

$\int_1^{\sqrt{2}} 1 , dx = \sqrt{2} - 1$

For $\sqrt{2} \leq x < \sqrt{3}$, x²  lies between 2 and 3, so $\left\lfloor x^2 \right\rfloor = 2$ . Therefore,

$\int_{\sqrt{2}}^{\sqrt{3}} 2 , dx = 2(\sqrt{3} - \sqrt{2})$

Summing up all the results:

$\int_{\frac{\sqrt{3}}{2}}^{\sqrt{3}} \left\lfloor x^2 \right\rfloor , dx = 0 + (\sqrt{2} - 1) + 2(\sqrt{3} - \sqrt{2})$

= $2(\sqrt{3} - \sqrt{2})$.

Hence, the correct answer is Option 2.

Calculation:

Given,

The function is $f(x) = \lfloor x^2 \rfloor$.

We are tasked with finding the value of $\int_{\sqrt{2}}^{2} f(x) dx$.

We can break the integral into two parts as follows:

$\int_{\sqrt{2}}^{2} f(x) dx = \int_{\sqrt{2}}^{\sqrt{3}} 2 dx + \int_{\sqrt{3}}^{2} 3 dx$

For the range $\sqrt{2} \leq x \leq \sqrt{3}$, $\lfloor x^2 \rfloor = 2$. So the first part of the integral is:

$\int_{\sqrt{2}}^{\sqrt{3}} 2 dx = 2 \times (\sqrt{3} - \sqrt{2})$

For the range $\sqrt{3} \leq x \leq 2$, $\lfloor x^2 \rfloor = 3$. So the second part of the integral is:

$\int_{\sqrt{3}}^{2} 3 dx = 3 \times (2 - \sqrt{3})$

Now, we calculate the values:

$2 \times (\sqrt{3} - \sqrt{2}) = 2\sqrt{3} - 2\sqrt{2}$

$3 \times (2 - \sqrt{3}) = 6 - 3\sqrt{3}$

Combining the two parts:

$2\sqrt{3} - 2\sqrt{2} + 6 - 3\sqrt{3} = 6 - 2\sqrt{2} - \sqrt{3}$

Hence, the correct answer is Option 1.