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Calculation:

Given,

The function is$y = \left( 1 - \cos(x) \right)^{-1}$, where $x \neq 2n\pi$and n  is an integer.

The cosine function has a range of  [-1, 1] , so cos(x) can take values from ( -1) to ( 1 ).

The expression $1 - \cos(x)$ will take values from:

$1 - 1 = 0 \quad \text{to} \quad 1 - (-1) = 2$

So, (1 - cos(x)) takes values in the range $(0, 2]$, but $x \neq 2n\pi$ excludes the value 0.

Since $y = \frac{1}{1 - \cos(x)}$, the reciprocal function will take values in the range $\left[ \frac{1}{2}, \infty \right)$

∴ The range of the function is $[0.5, \infty)$.

The correct answer is Option (b)

Calculation:

Given,

The function is $y = \left(1 - \cos(x)\right)^{-1}$, where$x \neq 2n\pi$ and n is an integer.

The given function can be written as:

$y = \frac{1}{1 - \cos(x)}$

Using the identity $1 - \cos(x) = 2\sin^2\left(\frac{x}{2}\right)$, we rewrite the function as:

$y = \frac{1}{2\sin^2\left(\frac{x}{2}\right)}$

The integral we need to evaluate is:

$\int \frac{dx}{2\sin^2\left(\frac{x}{2}\right)}$

$\frac{1}{2} \int \csc^2\left(\frac{x}{2}\right) dx$

The integral of $\csc^2(x)$ is known to be $-\cot(x)$, so we get:

$\frac{1}{2} \left( -\cot\left(\frac{x}{2}\right) \right) + C$

$-\frac{1}{2} \cot\left(\frac{x}{2}\right) + C$

The final result is $-\cot\left(\frac{x}{2}\right) + C$

Hence, the correct answer is Option 2.

Calculation:

Given,

The function is$f(x) = \sin(\lfloor x \rfloor)$, where$\lfloor x \rfloor$is the greatest integer function, and g(x) = |x| , the absolute value function.

We are tasked with finding:

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

For $g(x) = |x|$, we know that:

$\lim_{x \to 0} g(x) = 0$

For $f(x) = \sin(\lfloor x \rfloor)$, we know that:

For $x \to 0^+$), $\lfloor x \rfloor = 0$, so f(x) = sin(0) = 0 .

For $x \to 0^-$, $\lfloor x \rfloor = -1$, so $f(x) = \sin(-1)$, which is a nonzero constant.

Evaluating the limit:

For$x \to 0^+$,$f(x)g(x) = 0 \times x = 0$

For$x \to 0^-$, $f(x)g(x) = \sin(-1) \times (-x)$, which approaches 0 as $x \to 0^- .$.

∴ The value of $\lim_{x \to 0} f(x) g(x)$is 0.

The correct answer is Option (2).

Calculation:

Given,

The function is $f(x) = \sin(\lfloor x \rfloor)$, where $\lfloor x \rfloor$ is the greatest integer function, and g(x) = |x| , the absolute value function.

We are tasked with finding:

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

For $g(x) = |x|$, we know that:

$\lim_{x \to 0} g(x) = 0$

For $f(x) = \sin(\lfloor x \rfloor)$, we know that:

For $x \to 0^+$, $\lfloor x \rfloor = 0$, so f(x) = sin(0) = 0 .

For $x \to 0^-$, $\lfloor x \rfloor = -1$, so $f(x) = \sin(-1)$, which is a nonzero constant.

Evaluating the limit:

For $x \to 0^+$, $\frac{f(x)}{g(x)} = \frac{0}{x} = 0$

For $x \to 0^-$, $\frac{f(x)}{g(x)} = \frac{\sin(-1)}{-x}$, which becomes undefined as $x \to 0^-$because the denominator approaches 0, but the numerator remains a nonzero constant.

∴ Since the left-hand and right-hand limits do not match, the limit does not exist.

The correct answer is Option (4):

Calculation:

Given,

The function is $f(x) = |x - 3|$, which is an absolute value function.

The domain of an absolute value function $f(x) = |x - a|$ is all real numbers, because the absolute value function is defined for all values of x. The function handles both positive and negative inputs for x .

Since there are no restrictions or undefined points for the absolute value function, the domain is all real numbers.

∴ The domain of the function is $(-\infty, \infty)$

Hence, the correct answer is Option 3.

Calculation:

Given,

The function is f(x) = |x - 3| , and we need to find the area bounded by the curve and the line y = 3.

To find the points of intersection, we set the function equal to 3:

$|x - 3| = 3$

Solving for x :

• For$x \geq 3$, $x - 3 = 3$, which gives x = 6 .

• For ( x < 3 ), 3 - x = 3 , which gives x = 0 .

Therefore, the points of intersection are x = 0  and x = 6 .

The area can be calculated by integrating the difference between the curve and the line from x = 0 to x = 6. The integral is split into two parts due to the absolute value function:

$A = \int_{0}^{3} (3 - x) , dx + \int_{3}^{6} (x - 3) , dx$

For x in [0, 3] , ( f(x) = 3 - x ), and for ( x in [3, 6] ), ( f(x) = x - 3 ).

Compute both integrals:

• For x in [0, 3] :

$\int_{0}^{3} (3 - x) , dx = \left[ 3x - \frac{x^2}{2} \right]_{0}^{3} = 9 - 4.5 = 4.5$

• For x in [3, 6] :

$\int_{3}^{6} (x - 3) , dx = \left[ \frac{x^2}{2} - 3x \right]_{3}^{6} = 4.5$

Step 4: The total area is the sum of the two areas:

$A = 4.5 + 4.5 = 9 , \text{square units}$

∴ The total area bounded by the curve and the line is 9 square units.

The correct answer is Option (4):

Calculation:

Given,

The function is $f = {(1, 1), (2, 4), (3, 7), (4, 10)}$, and we are also given that the function is $f(x) = px + q$, where p and q  are constants.

Using the points f(1) = 1  and f(2) = 4 , we create the following system of equations:

• From f(1) = 1 , we have $p(1) + q = 1$, so p + q = 1 .......(1).
• From f(2) = 4 , we have p(2) + q = 4 , so 2p + q = 4  ..........( 2).

Subtract Equation 1 from Equation 2:

$(2p + q) - (p + q) = 4 - 1$

$p = 3$

Now, substitute p = 3 into Equation 1:

$3 + q = 1$

$q = -2$

$p + q = 3 + (-2) = 1$

∴ The value of p + q  is 1.

The correct answer is Option (3).

Calculation:

Given,

The function is $f(x)=px+q$, a linear polynomial.

Since $f(x)=px+q$ is linear with $p\neq0$, it is one-one (injective): different $x$ give different $f(x)$.

The actual outputs given are ${1,4,7,10}$, so the range of $f$ is ${1,4,7,10}$.

The codomain is the set of natural numbers $\mathbb{N}$. For $f$ to be onto, its range must equal its codomain, but here:

${1,4,7,10}\neq\mathbb{N}$

∴ f is one-one but not onto.

Hence, the correct answer is Option 1.

Calculation:

Given,

The function is $f(x) = x^2 - 1$ and we are tasked with finding $\lim_{x \to 1} (f \circ f(x))$.

$f \circ f(x) = f(f(x))$

Given $f(x) = x^2 - 1$, we substitute $f(x)$ into itself:

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

Now, apply the function $f$ to $(x^2 - 1)$:

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

This simplifies to:

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

Thus, $f \circ f(x) = x^4 - 2x^2$.

Now, we need to find $\lim_{x \to 1} (x^4 - 2x^2)$.

Substitute x = 1 into the function:

$f \circ f(1) = 1^4 - 2(1^2) = 1 - 2 = -1$

∴ The value of $\lim_{x \to 1} (f \circ f(x))$ is -1.

The correct answer is Option (1).

Calculation:

The function is $y = x^2 - 1$, and we need to find the area bounded by the curve and the x-axis between x = -1  and x = 1 .

The required area is given by the definite integral of the function from  -1  to 1 :

$\text{Area} = \int_{-1}^{1} (x^2 - 1) , dx$

$\int (x^2 - 1) , dx = \frac{x^3}{3} - x$

Evaluate the integral from  -1 to  1 :

$\left[\frac{x^3}{3} - x\right]_{-1}^{1} = \left(\frac{1^3}{3} - 1\right) - \left(\frac{(-1)^3}{3} - (-1)\right)$

$= \left(\frac{1}{3} - 1\right) - \left(\frac{-1}{3} + 1\right) = \left(\frac{1}{3} - \frac{3}{3}\right) - \left(\frac{-1}{3} + \frac{3}{3}\right) = \frac{2}{3} - \frac{2}{3} = \frac{4}{3}$

∴ The area is $\frac{4}{3}$ square units.

Hence, the correct answer is Option 3.