(a) Finding fog, its Domain and Range

Given data

$$f(x) = \sqrt{x}, \qquad g(x) = \sqrt{3-x}$$

Computing fog

$$fog(x) = f(g(x)) = f\left(\sqrt{3-x}\right) = \sqrt{\sqrt{3-x}} = (3-x)^{1/4}$$

$$\boxed{fog(x) = (3-x)^{1/4}}$$

Domain

Domain analysis via composition:

• $g(x) = \sqrt{3-x}$ requires $3 - x \geq 0 \Rightarrow x \leq 3$, so $D_g = (-\infty, 3]$.
• $f(x) = \sqrt{x}$ requires $x \geq 0$, so $D_f = [0,\infty)$.
• For $fog$ we need $x \in D_g$ and $g(x) \in D_f$. Since $g(x) = \sqrt{3-x} \geq 0$ wherever it is defined, the second condition is automatically satisfied.

$$\therefore\ D_{fog} = (-\infty, 3]$$

Range

On $(-\infty, 3]$:

• At $x = 3$: $(3-3)^{1/4} = 0$.
• As $x \to -\infty$: $(3-x)^{1/4} \to +\infty$.

The function is continuous and decreasing, so it attains all values in $[0, \infty)$.

$$\therefore\ R_{fog} = [0, +\infty)$$

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(b) Cost as a Function of Width

Given data

• Volume $V = 20\ \text{m}^3$
• Length $= 2 \times$ width
• Base cost $= \text{Rs } 10/\text{m}^2$
• Side cost $= \text{Rs } 4/\text{m}^2$
• Open top

Variables

Let width $= w$, length $= 2w$, height $= h$.

Volume constraint

$$V = (2w)(w)(h) = 2w^2 h = 20 \Rightarrow h = \frac{10}{w^2}$$

Areas

Base area: $$A_{\text{base}} = 2w \cdot w = 2w^2$$

Sides (4 walls, open top):

• Two of size $2w \times h$: $2(2wh) = 4wh$
• Two of size $w \times h$: $2(wh) = 2wh$

$$A_{\text{sides}} = 4wh + 2wh = 6wh$$

Cost

$$C = 10(2w^2) + 4(6wh) = 20w^2 + 24wh$$

Substitute $h = \dfrac{10}{w^2}$:

$$C(w) = 20w^2 + 24w\cdot\frac{10}{w^2} = 20w^2 + \frac{240}{w}$$

$$\boxed{C(w) = 20w^2 + \frac{240}{w}, \quad w > 0}$$

Volume of Solid Obtained by Rotating About the Y-Axis

Step 1 - Extract (Given Data)

• Region bounded by the curves:
  • $y = x$
  • $y = x^2$
• Axis of rotation: the y-axis

Step 2 - Solve

Find Intersection Points

Set the two curves equal:

$$x = x^2 \implies x^2 - x = 0 \implies x(x-1) = 0$$

$$\therefore x = 0 \quad \text{and} \quad x = 1$$

On the interval $0 < x < 1$, we have $x > x^2$, so $y = x$ is the upper curve and $y = x^2$ is the lower curve.

Method: Cylindrical Shells

Rotating about the y-axis, a typical shell at position $x$ has:

• radius $= x$
• height $= (\text{top}) - (\text{bottom}) = x - x^2$
• thickness $= dx$

$$V = 2\pi \int_0^1 (\text{radius})(\text{height}), dx = 2\pi \int_0^1 x,(x - x^2), dx$$

Evaluate the Integral

$$V = 2\pi \int_0^1 (x^2 - x^3), dx$$

$$V = 2\pi \left[ \frac{x^3}{3} - \frac{x^4}{4} \right]_0^1$$

$$V = 2\pi \left[ \left(\frac{1}{3} - \frac{1}{4}\right) - 0 \right] = 2\pi \left[ \frac{4 - 3}{12} \right] = 2\pi \cdot \frac{1}{12}$$

$$\boxed{V = \dfrac{\pi}{6}}$$

Verification by Washer (Disk) Method

Express curves as $x$ in terms of $y$: from $y = x$, $x = y$; from $y = x^2$, $x = \sqrt{y}$. For $0 < y < 1$, $\sqrt{y} > y$, so outer radius $= \sqrt{y}$, inner radius $= y$.

$$V = \pi \int_0^1 \left[ (\sqrt{y})^2 - (y)^2 \right] dy = \pi \int_0^1 (y - y^2), dy$$

$$V = \pi \left[ \frac{y^2}{2} - \frac{y^3}{3} \right]_0^1 = \pi \left( \frac{1}{2} - \frac{1}{3} \right) = \pi \cdot \frac{1}{6} = \frac{\pi}{6}$$

Both methods agree.

Final Answer

$$V = \dfrac{\pi}{6} \approx 0.5236 \text{ cubic units}$$

Sequence: Definition and Convergence of $a_n = \dfrac{n}{\sqrt{5+n}}$

Given Data

• General term: $a_n = \dfrac{n}{\sqrt{5+n}}$, $n \in \mathbb{N}$
• Required: definition of a sequence; determine convergence of ${a_n}$

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Part 1: What is a Sequence?

A sequence is a function whose domain is the set of natural numbers $\mathbb{N}$. It is an ordered list of numbers arranged in a definite order:

$$a_1, a_2, a_3, \ldots, a_n, \ldots$$

where $a_n$ is the $n$-th term (general term), and the sequence is denoted ${a_n}$.

Convergence: A sequence ${a_n}$ converges to a finite limit $L$ if for every $\varepsilon > 0$ there exists $N$ such that

$$|a_n - L| < \varepsilon \quad \text{for all } n > N.$$

If no finite $L$ exists, the sequence diverges.

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Part 2: Convergence of $a_n = \dfrac{n}{\sqrt{5+n}}$

We test the limit:

$$\lim_{n \to \infty} \frac{n}{\sqrt{5+n}}$$

Step 1: Rewrite. Write $n = \sqrt{n}\cdot\sqrt{n}$ and factor $n$ under the root:

$$\sqrt{5+n} = \sqrt{n},\sqrt{\tfrac{5}{n}+1}$$

So

$$\frac{n}{\sqrt{5+n}} = \frac{\sqrt{n}\cdot\sqrt{n}}{\sqrt{n},\sqrt{\tfrac{5}{n}+1}} = \frac{\sqrt{n}}{\sqrt{\tfrac{5}{n}+1}}$$

Step 2: Take the limit. As $n \to \infty$:

• $\sqrt{n} \to \infty$
• $\dfrac{5}{n} \to 0 \implies \sqrt{\tfrac{5}{n}+1} \to 1$

Hence

$$\lim_{n \to \infty} \frac{\sqrt{n}}{\sqrt{\tfrac{5}{n}+1}} = \frac{\infty}{1} = \infty$$

Quick check with small values: $a_1 = \frac{1}{\sqrt6} \approx 0.408$, $a_{100} = \frac{100}{\sqrt{105}} \approx 9.76$, $a_{10000} = \frac{10000}{\sqrt{10005}} \approx 99.98$. The terms grow without bound, confirming divergence.

Step 3: Conclusion. Since the limit is not a finite number,

$$\boxed{\therefore \ a_n = \frac{n}{\sqrt{5+n}} \text{ is divergent.}}$$

Vector Perpendicular to a Plane Through Three Points

Given Data

• $P(1, 4, 6)$
• $Q(-2, 5, -1)$
• $R(1, -1, 1)$

Concept

A normal vector to the plane is the cross product of two vectors lying in the plane: $$\vec{n} = \overrightarrow{PQ} \times \overrightarrow{PR}$$

Step 1: Vectors in the Plane

$$\overrightarrow{PQ} = Q - P = (-2-1,\ 5-4,\ -1-6) = (-3,\ 1,\ -7)$$ $$\overrightarrow{PR} = R - P = (1-1,\ -1-4,\ 1-6) = (0,\ -5,\ -5)$$

Step 2: Cross Product

$$\vec{n} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \ -3 & 1 & -7 \ 0 & -5 & -5 \end{vmatrix}$$

i-component: $$(1)(-5) - (-7)(-5) = -5 - 35 = -40$$

j-component: $$-\left[(-3)(-5) - (-7)(0)\right] = -[15 - 0] = -15$$

k-component: $$(-3)(-5) - (1)(0) = 15 - 0 = 15$$

Result

$$\boxed{\vec{n} = -40,\vec{i} - 15,\vec{j} + 15,\vec{k} = (-40,\ -15,\ 15)}$$

Verification (perpendicularity check):

• $\vec{n}\cdot\overrightarrow{PQ} = (-40)(-3) + (-15)(1) + (15)(-7) = 120 - 15 - 105 = 0$ ✓
• $\vec{n}\cdot\overrightarrow{PR} = (-40)(0) + (-15)(-5) + (15)(-5) = 0 + 75 - 75 = 0$ ✓

Both dot products are zero, confirming $\vec{n}$ is perpendicular to the plane.

Any scalar multiple (e.g. dividing by 5: $(-8, -3, 3)$) is also a valid normal vector.

Local Maxima, Minima, and Saddle Points of $f(x,y) = x^4 + y^4 - 4xy + 1$

Step 1: Given Data

Function: $f(x,y) = x^4 + y^4 - 4xy + 1$

Step 2: Find Critical Points

$$ f_x = 4x^3 - 4y = 0 \Rightarrow y = x^3 \tag{1} $$

$$ f_y = 4y^3 - 4x = 0 \Rightarrow x = y^3 \tag{2} $$

Substitute (1) into (2): $$x = (x^3)^3 = x^9 \Rightarrow x^9 - x = 0 \Rightarrow x(x^8 - 1) = 0$$

So $x = 0$ or $x = \pm 1$.

• $x = 0 \Rightarrow y = 0$: point $(0,0)$
• $x = 1 \Rightarrow y = 1$: point $(1,1)$
• $x = -1 \Rightarrow y = -1$: point $(-1,-1)$

Step 3: Second Derivative Test

$$f_{xx} = 12x^2, \quad f_{yy} = 12y^2, \quad f_{xy} = -4$$ $$D = f_{xx}f_{yy} - (f_{xy})^2 = 144x^2y^2 - 16$$

At $(0,0)$: $$D = 144(0) - 16 = -16 < 0 \Rightarrow \text{Saddle point}$$

At $(1,1)$: $$D = 144(1)(1) - 16 = 128 > 0, \quad f_{xx} = 12 > 0 \Rightarrow \text{Local minimum}$$ $$f(1,1) = 1 + 1 - 4 + 1 = -1$$

At $(-1,-1)$: $$D = 144(1)(1) - 16 = 128 > 0, \quad f_{xx} = 12 > 0 \Rightarrow \text{Local minimum}$$ $$f(-1,-1) = 1 + 1 - 4(-1)(-1) + 1 = 1 + 1 - 4 + 1 = -1$$

Summary

Local minimum value: $-1$ at $(1,1)$ and $(-1,-1)$ Saddle point: $(0,0)$ No local maximum exists.