110 marksNumericalRange and domain of functions and their GrAnswer

Sketch the graph of $f(x) = x^2$ . Find its domain and range.

Evaluate $\lim_{x \to 1^-} \sin^{-1} \left( \frac{1-\sqrt{x}}{1-x} \right)$ [5+5]

Sketch the graph of $f(x) = x^2$. Find its domain and range.

Evaluate $\lim_{x \to 1^-} \sin^{-1} \left( \frac{1-\sqrt{x}}{1-x} \right)$ [5+5]

Model Answer

Given Data

Function: $f(x) = x^2$
Limit to evaluate: $\displaystyle \lim_{x \to 1^-} \sin^{-1}\left(\frac{1-\sqrt{x}}{1-x}\right)$
Marks: [5 + 5]

Part 1: Graph, Domain and Range of $f(x) = x^2$

Domain

$x^2$ is defined for every real number (no division or even-root restrictions).

$\text{Domain} = (-\infty, \infty) = \mathbb{R}$

Range

Since $x^2 \geq 0$ for all real $x$ , and each value $y \geq 0$ is attained at $x = \pm\sqrt{y}$ :

$\text{Range} = [0, \infty)$

Sketching Features

Feature	Value
Vertex (minimum)	$(0,0)$
Symmetry	Even function, symmetric about the $y$ -axis
Intercepts	Passes through origin only
Shape	Upward-opening parabola

Sample points:

$x$	$-3$	$-2$	$-1$	$0$	$1$	$2$	$3$
$f(x)$	$9$	$4$	$1$	$0$	$1$	$4$	$9$

Sketch

 y
 9 |*                 *
   |
 4 |   *           *
   |
 1 |      *     *
   |        * *
 0 +---------+---------> x
  -3 -2 -1   0   1  2  3

A U-shaped parabola opening upward with vertex at the origin.

Part 2: Evaluate the Limit

$\lim_{x \to 1^-} \sin^{-1}\left(\frac{1-\sqrt{x}}{1-x}\right)$

Step 1: Simplify the inner expression

Factor the denominator as a difference of squares:

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

Therefore:

$\frac{1-\sqrt{x}}{1-x} = \frac{1-\sqrt{x}}{(1-\sqrt{x})(1+\sqrt{x})} = \frac{1}{1+\sqrt{x}}, \quad x \neq 1$

Step 2: Limit of the inner expression

$\lim_{x \to 1^-} \frac{1}{1+\sqrt{x}} = \frac{1}{1+\sqrt{1}} = \frac{1}{2}$

Step 3: Continuity of $\sin^{-1}$

$\sin^{-1}$ is continuous on $[-1, 1]$ , and $\tfrac{1}{2} \in [-1,1]$ , so:

$\lim_{x \to 1^-} \sin^{-1}\left(\frac{1-\sqrt{x}}{1-x}\right) = \sin^{-1}\left(\frac{1}{2}\right)$

Step 4: Evaluate

$\sin^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{6}$

$\boxed{\lim_{x \to 1^-} \sin^{-1}\left(\frac{1-\sqrt{x}}{1-x}\right) = \frac{\pi}{6}}$

Model Answer

Given Data

Function: $f(x) = x^2$
Limit to evaluate: $\displaystyle \lim_{x \to 1^-} \sin^{-1}\left(\frac{1-\sqrt{x}}{1-x}\right)$
Marks: [5 + 5]

Part 1: Graph, Domain and Range of $f(x) = x^2$

Domain

$x^2$ is defined for every real number (no division or even-root restrictions).

$$\text{Domain} = (-\infty, \infty) = \mathbb{R}$$

Range

Since $x^2 \geq 0$ for all real $x$, and each value $y \geq 0$ is attained at $x = \pm\sqrt{y}$:

$$\text{Range} = [0, \infty)$$

Sketching Features

Feature	Value
Vertex (minimum)	$(0,0)$
Symmetry	Even function, symmetric about the $y$-axis
Intercepts	Passes through origin only
Shape	Upward-opening parabola

Sample points:

$x$	$-3$	$-2$	$-1$	$0$	$1$	$2$	$3$
$f(x)$	$9$	$4$	$1$	$0$	$1$	$4$	$9$

Sketch

 y
 9 |*                 *
   |
 4 |   *           *
   |
 1 |      *     *
   |        * *
 0 +---------+---------> x
  -3 -2 -1   0   1  2  3

A U-shaped parabola opening upward with vertex at the origin.

Part 2: Evaluate the Limit

$$\lim_{x \to 1^-} \sin^{-1}\left(\frac{1-\sqrt{x}}{1-x}\right)$$

Step 1: Simplify the inner expression

Factor the denominator as a difference of squares:

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

Therefore:

$$\frac{1-\sqrt{x}}{1-x} = \frac{1-\sqrt{x}}{(1-\sqrt{x})(1+\sqrt{x})} = \frac{1}{1+\sqrt{x}}, \quad x \neq 1$$

Step 2: Limit of the inner expression

$$\lim_{x \to 1^-} \frac{1}{1+\sqrt{x}} = \frac{1}{1+\sqrt{1}} = \frac{1}{2}$$

Step 3: Continuity of $\sin^{-1}$

$\sin^{-1}$ is continuous on $[-1, 1]$, and $\tfrac{1}{2} \in [-1,1]$, so:

$$\lim_{x \to 1^-} \sin^{-1}\left(\frac{1-\sqrt{x}}{1-x}\right) = \sin^{-1}\left(\frac{1}{2}\right)$$

Step 4: Evaluate

$$\sin^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{6}$$

$$\boxed{\lim_{x \to 1^-} \sin^{-1}\left(\frac{1-\sqrt{x}}{1-x}\right) = \frac{\pi}{6}}$$

210 marksNumericalDifferentiability of a functionAnswer

Where the function $f(x) = |x|$ is differentiable? Discuss.

A farmer has 1200 m. of fencing and wants to fence off a rectangular field that borders a straight river. He needs to fence along the river. What are the dimensions of the field that has the largest area? [5+5]

Where the function $f(x) = |x|$ is differentiable? Discuss.

A farmer has 1200 m. of fencing and wants to fence off a rectangular field that borders a straight river. He needs to fence along the river. What are the dimensions of the field that has the largest area? [5+5]

(a) Differentiability of f(x) = x - Function: $f(x) = x$ $f$ is differentiable at $a$ if the limit below exists: $f'(a) = \lim{h \to 0} \frac{f(a+h) - f(a)}{h}$ $f(x) = x = \begin{cases} x & x \geq 0 \\ -x & x < 0 \end{cases}$

(a) Differentiability of f(x) = x - Function: $f(x) = x$ $f$ is differentiable at $a$ if the limit below exists: $$f'(a) = \lim{h \to 0} \frac{f(a+h) - f(a)}{h}$$ $$f(x) = x = \begin{cases} x & x \geq 0 \ -x & x < 0 \end{cases}$$

310 marksNumericalLinear equationsAnswer

Find the solution of the initial value problem $x^2 y' + x y = 1$ , $y(1) = 2$ , $x > 0$ .

Find the area enclosed by the line $y = x - 1$ and the parabola $y^2 = 2x + 6$ . [5+5]

Find the solution of the initial value problem $x^2 y' + x y = 1$, $y(1) = 2$, $x > 0$.

Find the area enclosed by the line $y = x - 1$ and the parabola $y^2 = 2x + 6$. [5+5]

--- Given data: $x^2 y' + xy = 1$ , $\; y(1) = 2$ , $\; x 0$ Divide by $x^2$ : $\frac{dy}{dx} + \frac{1}{x}y = \frac{1}{x^2}$ So $P = \dfrac{1}{x}$ , $Q = \dfrac{1}{x^2}$ . $\text{I.F.} = e^{\int \frac{1}{x}\,dx} = e^{\ln x} = x$ $$y\cdot...

--- Given data: $x^2 y' + xy = 1$, $; y(1) = 2$, $; x 0$ Divide by $x^2$: $$\frac{dy}{dx} + \frac{1}{x}y = \frac{1}{x^2}$$ So $P = \dfrac{1}{x}$, $Q = \dfrac{1}{x^2}$. $$\text{I.F.} = e^{\int \frac{1}{x},dx} = e^{\ln x} = x$$ $$y\cdot...

45 marksNumericalDefinite integralAnswer

Evaluate $\int_0^3 \sqrt{1 + x^2} \cdot x^3 \, dx$ [5]

Evaluate $\int_0^3 \sqrt{1 + x^2} \cdot x^3 , dx$ [5]

Evaluate $\int_0^3 \sqrt{1+x^2}\cdot x^3\, dx$

Step 1 - Given Data

Integrand: $\sqrt{1+x^2}\cdot x^3$
Limits: from $x=0$ to $x=3$

Step 2 - Substitution

Let $u = 1+x^2$ , so $du = 2x\,dx$ , and $x^2 = u-1$ .

Write $x^3\,dx = x^2\cdot x\,dx = (u-1)\cdot\dfrac{du}{2}$ .

Limits:

$x$	$u=1+x^2$
$0$	$1$
$3$	$10$

Step 3 - Transform

$\int_0^3 \sqrt{1+x^2}\cdot x^3\,dx = \frac{1}{2}\int_1^{10}\sqrt{u}\,(u-1)\,du = \frac{1}{2}\int_1^{10}\left(u^{3/2}-u^{1/2}\right)du$

Step 4 - Integrate

$= \frac{1}{2}\left[\frac{2}{5}u^{5/2} - \frac{2}{3}u^{3/2}\right]_1^{10} = \left[\frac{u^{5/2}}{5} - \frac{u^{3/2}}{3}\right]_1^{10}$

Step 5 - Apply Limits

At $u=10$ : ( $10^{5/2}=100\sqrt{10}$ , $10^{3/2}=10\sqrt{10}$ ) $\frac{100\sqrt{10}}{5} - \frac{10\sqrt{10}}{3} = 20\sqrt{10} - \frac{10\sqrt{10}}{3} = \frac{60\sqrt{10}-10\sqrt{10}}{3} = \frac{50\sqrt{10}}{3}$

At $u=1$ : $\frac{1}{5} - \frac{1}{3} = \frac{3-5}{15} = -\frac{2}{15}$

Step 6 - Combine

$= \frac{50\sqrt{10}}{3} - \left(-\frac{2}{15}\right) = \frac{50\sqrt{10}}{3} + \frac{2}{15}$

Final Answer

$\boxed{\int_0^3 \sqrt{1+x^2}\cdot x^3\,dx = \frac{50\sqrt{10}}{3} + \frac{2}{15} \approx 52.83}$

Numeric check: $\sqrt{10}\approx 3.16228$ , so $\frac{50(3.16228)}{3}\approx 52.705$ , plus $\frac{2}{15}\approx 0.133$ , giving $\approx 52.838$ . ✓

Evaluate $\int_0^3 \sqrt{1+x^2}\cdot x^3, dx$

Step 1 - Given Data

Integrand: $\sqrt{1+x^2}\cdot x^3$
Limits: from $x=0$ to $x=3$

Step 2 - Substitution

Let $u = 1+x^2$, so $du = 2x,dx$, and $x^2 = u-1$.

Write $x^3,dx = x^2\cdot x,dx = (u-1)\cdot\dfrac{du}{2}$.

Limits:

$x$	$u=1+x^2$
$0$	$1$
$3$	$10$

Step 3 - Transform

$$\int_0^3 \sqrt{1+x^2}\cdot x^3,dx = \frac{1}{2}\int_1^{10}\sqrt{u},(u-1),du = \frac{1}{2}\int_1^{10}\left(u^{3/2}-u^{1/2}\right)du$$

Step 4 - Integrate

$$= \frac{1}{2}\left[\frac{2}{5}u^{5/2} - \frac{2}{3}u^{3/2}\right]_1^{10} = \left[\frac{u^{5/2}}{5} - \frac{u^{3/2}}{3}\right]_1^{10}$$

Step 5 - Apply Limits

At $u=10$: ($10^{5/2}=100\sqrt{10}$, $10^{3/2}=10\sqrt{10}$) $$\frac{100\sqrt{10}}{5} - \frac{10\sqrt{10}}{3} = 20\sqrt{10} - \frac{10\sqrt{10}}{3} = \frac{60\sqrt{10}-10\sqrt{10}}{3} = \frac{50\sqrt{10}}{3}$$

At $u=1$: $$\frac{1}{5} - \frac{1}{3} = \frac{3-5}{15} = -\frac{2}{15}$$

Step 6 - Combine

$$= \frac{50\sqrt{10}}{3} - \left(-\frac{2}{15}\right) = \frac{50\sqrt{10}}{3} + \frac{2}{15}$$

Final Answer

$$\boxed{\int_0^3 \sqrt{1+x^2}\cdot x^3,dx = \frac{50\sqrt{10}}{3} + \frac{2}{15} \approx 52.83}$$

Numeric check: $\sqrt{10}\approx 3.16228$, so $\frac{50(3.16228)}{3}\approx 52.705$, plus $\frac{2}{15}\approx 0.133$, giving $\approx 52.838$. ✓

55 marksTaylor's and Maclaurin's seriesAnswer

Find the Maclaurin series expansion of $f(x) = \sin x$ for all x. [5]

The Maclaurin series of a function f(x) is the Taylor series expanded about x = 0, given by: $f(x) = f(0) + f'(0)\cdot x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots + \frac{f^{(n)}(0)}{n!}x^n + \cdots$ --- Let f(x) = sin x....

The Maclaurin series of a function f(x) is the Taylor series expanded about x = 0, given by: $$f(x) = f(0) + f'(0)\cdot x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots + \frac{f^{(n)}(0)}{n!}x^n + \cdots$$ --- Let f(x) = sin x....

65 marksNumericalNormal and binormal vectorsAnswer

Find the unit normal and binormal vectors for the circular helix $\mathbf{r}(t) = \cos t \hat{i} + \sin t \hat{j} + t \hat{k}$ . [5]

Find the unit normal and binormal vectors for the circular helix $\mathbf{r}(t) = \cos t \hat{i} + \sin t \hat{j} + t \hat{k}$. [5]

$\mathbf{r}(t) = \cos t\,\hat{i} + \sin t\,\hat{j} + t\,\hat{k}$ All data present and readable. --- $\mathbf{r}'(t) = -\sin t\,\hat{i} + \cos t\,\hat{j} + \hat{k}$ $\mathbf{r}'(t) = \sqrt{\sin^2 t + \cos^2 t + 1} = \sqrt{2}$ $$\mat...

$$\mathbf{r}(t) = \cos t,\hat{i} + \sin t,\hat{j} + t,\hat{k}$$ All data present and readable. --- $$\mathbf{r}'(t) = -\sin t,\hat{i} + \cos t,\hat{j} + \hat{k}$$ $$\mathbf{r}'(t) = \sqrt{\sin^2 t + \cos^2 t + 1} = \sqrt{2}$$ $$\mat...

75 marksNumericalLimit and continuityAnswer

If $f(x,y) = \frac{xy}{x^2 + y^2}$ , does $\lim_{(x, y) \to (0, 0)} f(x, y)$ exist? Justify. [5]

If $f(x,y) = \frac{xy}{x^2 + y^2}$, does $\lim_{(x, y) \to (0, 0)} f(x, y)$ exist? Justify. [5]

Given data: - Function: $f(x,y) = \dfrac{xy}{x^2+y^2}$ - Point of interest: $(x,y) \to (0,0)$ - Task: determine whether the limit exists. All data present. To test a two-variable limit, approach $(0,0)$ along different paths. If limits d...

85 marksNumericalInfinite sequence and seriesAnswer

Determine whether the sequence $a_n = (-1)^n$ is convergent or divergent. [5]

Given: - Sequence: $an = (-1)^n$ No other numeric data required. This is a proof-based analysis problem. --- $a1 = -1,\quad a2 = +1,\quad a3 = -1,\quad a4 = +1,\quad \dots$ The sequence oscillates permanently between $+1$ and $-1$ . A s...

Given: - Sequence: $an = (-1)^n$ No other numeric data required. This is a proof-based analysis problem. --- $$a1 = -1,\quad a2 = +1,\quad a3 = -1,\quad a4 = +1,\quad \dots$$ The sequence oscillates permanently between $+1$ and $-1$. A s...

95 marksNumericalMotion in spaceAnswer

The position vector of an object moving in a plane is given by $\mathbf{r}(t) = t^2 \hat{i} + t^2 \hat{j}$ . Find its velocity, speed, and acceleration when $t = 1$ and illustrate geometrically. [5]

The position vector of an object moving in a plane is given by $\mathbf{r}(t) = t^2 \hat{i} + t^2 \hat{j}$. Find its velocity, speed, and acceleration when $t = 1$ and illustrate geometrically. [5]

$\vec{r}(t) = t^2\hat{i} + t^2\hat{j}, \qquad t = 1$ --- $\vec{v}(t) = \frac{d\vec{r}}{dt} = \frac{d}{dt}(t^2)\hat{i} + \frac{d}{dt}(t^2)\hat{j} = 2t\hat{i} + 2t\hat{j}$ At $t = 1$ : $\vec{v}(1) = 2\hat{i} + 2\hat{j}$ --- $$\vec{v}(...

$$\vec{r}(t) = t^2\hat{i} + t^2\hat{j}, \qquad t = 1$$ --- $$\vec{v}(t) = \frac{d\vec{r}}{dt} = \frac{d}{dt}(t^2)\hat{i} + \frac{d}{dt}(t^2)\hat{j} = 2t\hat{i} + 2t\hat{j}$$ At $t = 1$: $$\vec{v}(1) = 2\hat{i} + 2\hat{j}$$ --- $$\vec{v}(...

105 marksNumericalIntroduction to first order equations SepaAnswer

Show that every member of the family of function $y = \frac{1 + ce^t}{1 - ce^t}$ is a solution of the differential equation $y' = \frac{1}{2}(y^2 - 1)$ . [5]

Show that every member of the family of function $y = \frac{1 + ce^t}{1 - ce^t}$ is a solution of the differential equation $y' = \frac{1}{2}(y^2 - 1)$. [5]

Solution

Given Data

Family of functions: $y = \dfrac{1 + ce^t}{1 - ce^t}$
Differential equation: $y' = \dfrac{1}{2}(y^2 - 1)$

Step 1: Compute the Left Hand Side, $y'$

Using the quotient rule with $u = 1 + ce^t$ , $v = 1 - ce^t$ :

$u' = ce^t, \qquad v' = -ce^t$

$y' = \frac{u'v - uv'}{v^2} = \frac{ce^t(1 - ce^t) - (1 + ce^t)(-ce^t)}{(1 - ce^t)^2}$

Simplify the numerator:

$ce^t(1 - ce^t) + ce^t(1 + ce^t) = ce^t\left[(1 - ce^t) + (1 + ce^t)\right] = ce^t(2) = 2ce^t$

Therefore:

$y' = \frac{2ce^t}{(1 - ce^t)^2}$

Step 2: Compute the Right Hand Side, $\tfrac{1}{2}(y^2 - 1)$

$y^2 = \frac{(1 + ce^t)^2}{(1 - ce^t)^2}$

$y^2 - 1 = \frac{(1 + ce^t)^2 - (1 - ce^t)^2}{(1 - ce^t)^2}$

Using $A^2 - B^2 = (A+B)(A-B)$ with $A = 1 + ce^t$ , $B = 1 - ce^t$ :

$A + B = 2, \qquad A - B = 2ce^t$

$(1 + ce^t)^2 - (1 - ce^t)^2 = (2)(2ce^t) = 4ce^t$

So:

$y^2 - 1 = \frac{4ce^t}{(1 - ce^t)^2}$

Multiply by $\tfrac{1}{2}$ :

$\frac{1}{2}(y^2 - 1) = \frac{1}{2}\cdot\frac{4ce^t}{(1 - ce^t)^2} = \frac{2ce^t}{(1 - ce^t)^2}$

Step 3: Compare

$y' = \frac{2ce^t}{(1 - ce^t)^2} = \frac{1}{2}(y^2 - 1)$

The two sides are identical for every value of the arbitrary constant $c$ .

Conclusion

Since $y'$ equals $\tfrac{1}{2}(y^2 - 1)$ for all $c$ , every member of the family $y = \dfrac{1 + ce^t}{1 - ce^t}$ is a solution of the differential equation $y' = \tfrac{1}{2}(y^2 - 1)$ . $\blacksquare$

Solution

Given Data

Family of functions: $y = \dfrac{1 + ce^t}{1 - ce^t}$
Differential equation: $y' = \dfrac{1}{2}(y^2 - 1)$

Step 1: Compute the Left Hand Side, $y'$

Using the quotient rule with $u = 1 + ce^t$, $v = 1 - ce^t$:

$$u' = ce^t, \qquad v' = -ce^t$$

$$y' = \frac{u'v - uv'}{v^2} = \frac{ce^t(1 - ce^t) - (1 + ce^t)(-ce^t)}{(1 - ce^t)^2}$$

Simplify the numerator:

$$ce^t(1 - ce^t) + ce^t(1 + ce^t) = ce^t\left[(1 - ce^t) + (1 + ce^t)\right] = ce^t(2) = 2ce^t$$

Therefore:

$$y' = \frac{2ce^t}{(1 - ce^t)^2}$$

Step 2: Compute the Right Hand Side, $\tfrac{1}{2}(y^2 - 1)$

$$y^2 = \frac{(1 + ce^t)^2}{(1 - ce^t)^2}$$

$$y^2 - 1 = \frac{(1 + ce^t)^2 - (1 - ce^t)^2}{(1 - ce^t)^2}$$

Using $A^2 - B^2 = (A+B)(A-B)$ with $A = 1 + ce^t$, $B = 1 - ce^t$:

$$A + B = 2, \qquad A - B = 2ce^t$$

$$(1 + ce^t)^2 - (1 - ce^t)^2 = (2)(2ce^t) = 4ce^t$$

So:

$$y^2 - 1 = \frac{4ce^t}{(1 - ce^t)^2}$$

Multiply by $\tfrac{1}{2}$:

$$\frac{1}{2}(y^2 - 1) = \frac{1}{2}\cdot\frac{4ce^t}{(1 - ce^t)^2} = \frac{2ce^t}{(1 - ce^t)^2}$$

Step 3: Compare

$$y' = \frac{2ce^t}{(1 - ce^t)^2} = \frac{1}{2}(y^2 - 1)$$

The two sides are identical for every value of the arbitrary constant $c$.

Conclusion

Since $y'$ equals $\tfrac{1}{2}(y^2 - 1)$ for all $c$, every member of the family $y = \dfrac{1 + ce^t}{1 - ce^t}$ is a solution of the differential equation $y' = \tfrac{1}{2}(y^2 - 1)$. $\blacksquare$

115 marksNumericalPartial derivativesAnswer

If $f(x,y) = 2x^3 + x^2y^2 - y^4$ , find $f_x(1,-2)$ , $f_y(1,-1)$ and $f_{yx}(1,-1)$ . [5]

If $f(x,y) = 2x^3 + x^2y^2 - y^4$, find $f_x(1,-2)$, $f_y(1,-1)$ and $f_{yx}(1,-1)$. [5]

Given Data

$f(x, y) = 2x^3 + x^2y^2 - y^4$

Required:

$f_x(1, -2)$
$f_y(1, -1)$
$f_{yx}(1, -1)$

Step 1: Compute $f_x$

Differentiate with respect to $x$ (treat $y$ constant):

$f_x = 6x^2 + 2xy^2 - 0 = 6x^2 + 2xy^2$

Evaluate at $(1, -2)$ :

$f_x(1,-2) = 6(1)^2 + 2(1)(-2)^2 = 6 + 2(4) = 6 + 8 = 14$

Step 2: Compute $f_y$

Differentiate with respect to $y$ (treat $x$ constant):

$f_y = 0 + 2x^2y - 4y^3 = 2x^2y - 4y^3$

Evaluate at $(1, -1)$ :

$f_y(1,-1) = 2(1)^2(-1) - 4(-1)^3 = -2 - 4(-1) = -2 + 4 = 2$

Step 3: Compute $f_{yx}$

Differentiate $f_y = 2x^2y - 4y^3$ with respect to $x$ :

$f_{yx} = 4xy - 0 = 4xy$

Evaluate at $(1, -1)$ :

$f_{yx}(1,-1) = 4(1)(-1) = -4$

Summary

Expression	Value
$f_x(1,-2)$	$14$
$f_y(1,-1)$	$2$
$f_{yx}(1,-1)$	$-4$

All computations verified.

Given Data

$$f(x, y) = 2x^3 + x^2y^2 - y^4$$

Required:

$f_x(1, -2)$
$f_y(1, -1)$
$f_{yx}(1, -1)$

Step 1: Compute $f_x$

Differentiate with respect to $x$ (treat $y$ constant):

$$f_x = 6x^2 + 2xy^2 - 0 = 6x^2 + 2xy^2$$

Evaluate at $(1, -2)$:

$$f_x(1,-2) = 6(1)^2 + 2(1)(-2)^2 = 6 + 2(4) = 6 + 8 = 14$$

Step 2: Compute $f_y$

Differentiate with respect to $y$ (treat $x$ constant):

$$f_y = 0 + 2x^2y - 4y^3 = 2x^2y - 4y^3$$

Evaluate at $(1, -1)$:

$$f_y(1,-1) = 2(1)^2(-1) - 4(-1)^3 = -2 - 4(-1) = -2 + 4 = 2$$

Step 3: Compute $f_{yx}$

Differentiate $f_y = 2x^2y - 4y^3$ with respect to $x$:

$$f_{yx} = 4xy - 0 = 4xy$$

Evaluate at $(1, -1)$:

$$f_{yx}(1,-1) = 4(1)(-1) = -4$$

Summary

Expression	Value
$f_x(1,-2)$	$14$
$f_y(1,-1)$	$2$
$f_{yx}(1,-1)$	$-4$

All computations verified.

125 marksNumericalVolumes of cylindrical cellsAnswer

Use cylindrical shells to find the volume of the solid obtained by rotating about the x-axis the region under the curve $y = \sqrt{x}$ for $0$ to $1$ . [5]

Use cylindrical shells to find the volume of the solid obtained by rotating about the x-axis the region under the curve $y = \sqrt{x}$ for $0$ to $1$. [5]

Curve: $y = \sqrt{x}$ - Region: under the curve from $x = 0$ to $x = 1$ - Axis of rotation: the x-axis - Method required: cylindrical shells The method of cylindrical shells uses strips parallel to the axis of rotation. Since we rotate...

Curve: $y = \sqrt{x}$ - Region: under the curve from $x = 0$ to $x = 1$ - Axis of rotation: the x-axis - Method required: cylindrical shells The method of cylindrical shells uses strips parallel to the axis of rotation. Since we rotate...

Mathematics I 2081 question paper

Model Answer

Given Data

Part 1: Graph, Domain and Range of f(x)=x2f(x) = x^2f(x)=x2

Domain

Range

Sketching Features

Sketch

Part 2: Evaluate the Limit

Step 1: Simplify the inner expression

Step 2: Limit of the inner expression

Step 3: Continuity of sin⁡−1\sin^{-1}sin−1

Step 4: Evaluate

Model Answer

Given Data

Part 1: Graph, Domain and Range of $f(x) = x^2$

Domain

Range

Sketching Features

Sketch

Part 2: Evaluate the Limit

Step 1: Simplify the inner expression

Step 2: Limit of the inner expression

Step 3: Continuity of $\sin^{-1}$

Step 4: Evaluate

Evaluate ∫031+x2⋅x3 dx\int_0^3 \sqrt{1+x^2}\cdot x^3\, dx∫03​1+x2​⋅x3dx

Step 1 - Given Data

Step 2 - Substitution

Step 3 - Transform

Step 4 - Integrate

Step 5 - Apply Limits

Step 6 - Combine

Final Answer

Evaluate $\int_0^3 \sqrt{1+x^2}\cdot x^3, dx$

Step 1 - Given Data

Step 2 - Substitution

Step 3 - Transform

Step 4 - Integrate

Step 5 - Apply Limits

Step 6 - Combine

Final Answer

Solution

Given Data

Step 1: Compute the Left Hand Side, y′y'y′

Step 2: Compute the Right Hand Side, 12(y2−1)\tfrac{1}{2}(y^2 - 1)21​(y2−1)

Step 3: Compare

Conclusion

Solution

Given Data

Step 1: Compute the Left Hand Side, $y'$

Step 2: Compute the Right Hand Side, $\tfrac{1}{2}(y^2 - 1)$

Step 3: Compare

Conclusion

Given Data

Step 1: Compute fxf_xfx​

Step 2: Compute fyf_yfy​

Step 3: Compute fyxf_{yx}fyx​

Summary

Given Data

Step 1: Compute $f_x$

Step 2: Compute $f_y$

Step 3: Compute $f_{yx}$

Summary

Model Answer

Given Data

Part 1: Graph, Domain and Range of f(x)=x2f(x) = x^2f(x)=x2

Domain

Range

Sketching Features

Sketch

Part 2: Evaluate the Limit

Step 1: Simplify the inner expression

Step 2: Limit of the inner expression

Step 3: Continuity of sin⁡−1\sin^{-1}sin−1

Step 4: Evaluate

Evaluate ∫031+x2⋅x3 dx\int_0^3 \sqrt{1+x^2}\cdot x^3\, dx∫03​1+x2​⋅x3dx

Step 1 - Given Data

Step 2 - Substitution

Step 3 - Transform

Step 4 - Integrate

Part 1: Graph, Domain and Range of $f(x) = x^2$

Step 3: Continuity of $\sin^{-1}$

Evaluate $\int_0^3 \sqrt{1+x^2}\cdot x^3\, dx$

Step 1: Compute the Left Hand Side, $y'$

Step 2: Compute the Right Hand Side, $\tfrac{1}{2}(y^2 - 1)$

Step 1: Compute $f_x$

Step 2: Compute $f_y$

Step 3: Compute $f_{yx}$

Part 1: Graph, Domain and Range of $f(x) = x^2$

Step 3: Continuity of $\sin^{-1}$

Evaluate $\int_0^3 \sqrt{1+x^2}\cdot x^3\, dx$

Step 1: Compute the Left Hand Side, $y'$

Step 2: Compute the Right Hand Side, $\tfrac{1}{2}(y^2 - 1)$

Step 1: Compute $f_x$

Step 2: Compute $f_y$

Step 3: Compute $f_{yx}$