Part (a): Domain of \(f(x) = \frac{x^{2} + 2x + 1}{x^{2} – 8x + 12}\)

The function is defined for all real x except where denominator = 0.

Denominator: \(x^{2} – 8x + 12 = 0\)

Factorizing: \(x^{2} – 8x + 12\)\( = (x – 2)(x – 6) \)\(= 0\)

Thus, \(x = 2\) or \(x = 6\) make denominator zero.

Therefore, domain = \(\mathbb{R} \setminus \{2, 6\}\)

In interval notation: \((-\infty, 2) \cup (2, 6) \cup (6, \infty)\)

Part (b): Range of \(f(x) = \frac{x^{2}}{1 + x^{2}}\)

Let \(y = \frac{x^{2}}{1 + x^{2}}\)

Since \(x^{2} \geq 0\) for all real x, we have \(y \geq 0\).

Also, \(x^{2} < 1 + x^{2}\) for all x, so \(\frac{x^{2}}{1 + x^{2}} < 1\).

To find if y can be 1: \(y = 1 \Rightarrow \frac{x^{2}}{1 + x^{2}} = 1 \)\(\Rightarrow x^{2} = 1 + x^{2} \)\(\Rightarrow 0 = 1\), impossible.

Thus, y < 1 for all x.

As \(x \to \infty\), \(y \to 1^{-}\). As \(x = 0\), \(y = 0\).

Therefore, range = \([0, 1)\)

Main: \(\frac{(\sin 7x + \sin 5x) + (\sin 9x + \sin 3x)}{(\cos 7x + \cos 5x) + (\cos 9x + \cos 3x)} = \tan 6x\)

Step 1: Use sum-to-product formulas:

\(\sin C + \sin D\)\( = 2 \sin\frac{C+D}{2} \cos\frac{C-D}{2}\)

\(\cos C + \cos D \)\(= 2 \cos\frac{C+D}{2} \cos\frac{C-D}{2}\)

Step 2: Group numerator terms:

\(\sin 7x + \sin 5x \)\(= 2 \sin\frac{7x+5x}{2} \cos\frac{7x-5x}{2}\)\( = 2 \sin 6x \cos x\)

\(\sin 9x + \sin 3x \)\(= 2 \sin\frac{9x+3x}{2} \cos\frac{9x-3x}{2} \)\(= 2 \sin 6x \cos 3x\)

Numerator = \(2 \sin 6x (\cos x + \cos 3x)\)

Step 3: Group denominator terms:

\(\cos 7x + \cos 5x\)\( = 2 \cos\frac{7x+5x}{2} \cos\frac{7x-5x}{2}\)\( = 2 \cos 6x \cos x\)

\(\cos 9x + \cos 3x \)\(= 2 \cos\frac{9x+3x}{2} \cos\frac{9x-3x}{2}\)\( = 2 \cos 6x \cos 3x\)

Denominator = \(2 \cos 6x (\cos x + \cos 3x)\)

Step 4: Cancel common factor \(2(\cos x + \cos 3x)\):

LHS = \(\frac{2 \sin 6x (\cos x + \cos 3x)}{2 \cos 6x (\cos x + \cos 3x)} = \frac{\sin 6x}{\cos 6x}\)\( = \tan 6x\) = RHS

Hence proved.

OR: \(\tan 4x = \frac{4\tan x(1 – \tan^{2} x)}{1 – 6\tan^{2} x + \tan^{4} x}\)

Using \(\tan 2\theta = \frac{2\tan\theta}{1 – \tan^{2}\theta}\)

\(\tan 4x = \tan 2(2x) = \frac{2\tan 2x}{1 – \tan^{2} 2x}\)

Now, \(\tan 2x = \frac{2\tan x}{1 – \tan^{2} x}\)

Let \(t = \tan x\). Then \(\tan 2x = \frac{2t}{1 – t^{2}}\)

\(\tan 4x = \frac{2 \cdot \frac{2t}{1-t^{2}}}{1 – \left(\frac{2t}{1-t^{2}}\right)^{2}} = \frac{\frac{4t}{1-t^{2}}}{1 – \frac{4t^{2}}{(1-t^{2})^{2}}}\)

Simplify denominator: \(1 – \frac{4t^{2}}{(1-t^{2})^{2}} = \frac{(1-t^{2})^{2} – 4t^{2}}{(1-t^{2})^{2}} \)\(= \frac{1 – 2t^{2} + t^{4} – 4t^{2}}{(1-t^{2})^{2}} = \frac{1 – 6t^{2} + t^{4}}{(1-t^{2})^{2}}\)

Therefore, \(\tan 4x = \frac{\frac{4t}{1-t^{2}}}{\frac{1 – 6t^{2} + t^{4}}{(1-t^{2})^{2}}} = \frac{4t}{1-t^{2}} \cdot \frac{(1-t^{2})^{2}}{1 – 6t^{2} + t^{4}}\)

\(= \frac{4t(1-t^{2})}{1 – 6t^{2} + t^{4}}\)

Substituting back \(t = \tan x\):

\(\tan 4x = \frac{4\tan x(1 – \tan^{2} x)}{1 – 6\tan^{2} x + \tan^{4} x}\)

Hence proved.

Step 1: Let the first AP have first term A and common difference D.

Let the second AP have first term a and common difference d.

Step 2: Sum of n terms of first AP: \(S_n = \frac{n}{2}[2A + (n-1)D]\)

Sum of n terms of second AP: \(s_n = \frac{n}{2}[2a + (n-1)d]\)

Step 3: Given ratio: \(\frac{S_n}{s_n} = \frac{5n + 4}{9n + 6}\)

\(\frac{\frac{n}{2}[2A + (n-1)D]}{\frac{n}{2}[2a + (n-1)d]} = \frac{5n + 4}{9n + 6}\)

\(\frac{2A + (n-1)D}{2a + (n-1)d} = \frac{5n + 4}{9n + 6}\)

Step 4: We need ratio of 18th terms: \(\frac{A + 17D}{a + 17d}\)

To get this, we put \(n-1 = 17 \Rightarrow n = 18\) in the ratio formula.

Step 5: Substitute n = 18 in the right-hand side:

\(\frac{5(18) + 4}{9(18) + 6} = \frac{90 + 4}{162 + 6} = \frac{94}{168} = \frac{47}{84}\)

Step 6: But \(\frac{2A + (18-1)D}{2a + (18-1)d} = \frac{2A + 17D}{2a + 17d} = \frac{47}{84}\)

This is not directly the ratio of 18th terms. We need \(\frac{A + 17D}{a + 17d}\).

Step 7: Let \(x = \frac{A + 17D}{a + 17d}\) be the required ratio.

From step 6: \(\frac{2A + 17D}{2a + 17d} = \frac{47}{84}\)

Divide numerator and denominator by 2: \(\frac{A + \frac{17D}{2}}{a + \frac{17d}{2}} = \frac{47}{84}\)

This is not matching x. There’s a standard result: For ratio of nth terms, we put n = 2n-1 in the sum ratio formula.

Step 8: To get ratio of mth terms, replace n by 2m-1 in the sum ratio.

For 18th term, m = 18 ⇒ 2m-1 = 35

Step 9: Ratio of 18th terms = \(\frac{5(35) + 4}{9(35) + 6} = \frac{175 + 4}{315 + 6} = \frac{179}{321}\)

Step 10: Simplify: 179 is prime, 321 = 3 × 107, no common factors.

Thus, ratio = \(\frac{179}{321}\)

Part (a): \(y = \frac{x^{n} – a^{n}}{x – a}\)

This is of the form \(\frac{x^n – a^n}{x – a}\). We know that:

\(\frac{x^n – a^n}{x – a} = x^{n-1} + x^{n-2}a + x^{n-3}a^2 \)\(+ \ldots + a^{n-1}\) (sum of n terms)

Therefore, \(y = x^{n-1} + x^{n-2}a + x^{n-3}a^2 \)\(+ \ldots + a^{n-1}\)

Differentiating term by term:

\(\frac{dy}{dx} = (n-1)x^{n-2} + (n-2)x^{n-3}a \)\(+ (n-3)x^{n-4}a^2 + \ldots + 0\)

The last term \(a^{n-1}\) is constant, so its derivative is 0.

Alternatively, using quotient rule directly would also give the same result.

Part (b): \(y = (5x^{3} + 3x – 1)(x – 1)\)

Method 1 – Product Rule:

Let \(u = 5x^3 + 3x – 1\), \(v = x – 1\)

\(\frac{du}{dx} = 15x^2 + 3\), \(\frac{dv}{dx} = 1\)

\(\frac{dy}{dx} = u\frac{dv}{dx} + v\frac{du}{dx} \)\(= (5x^3 + 3x – 1)(1) + (x – 1)(15x^2 + 3)\)

\(= 5x^3 + 3x – 1 + (x – 1)(15x^2 + 3)\)

Expand second term: \((x – 1)(15x^2 + 3) = 15x^3 + 3x – 15x^2 – 3\)

Add: \(5x^3 + 3x – 1 + 15x^3 + 3x – 15x^2 – 3\)

\(= 20x^3 – 15x^2 + 6x – 4\)

Method 2 – Expand first:

\(y = (5x^3 + 3x – 1)(x – 1) \)\(= 5x^4 – 5x^3 + 3x^2 – 3x – x + 1\)

\(= 5x^4 – 5x^3 + 3x^2 – 4x + 1\)

\(\frac{dy}{dx} = 20x^3 – 15x^2 + 6x – 4\) ✓

OR: First principle for \(f(x) = \sin x \cos x\)

\(f(x) = \sin x \cos x = \frac{1}{2}\sin 2x\) (using \(\sin 2\theta = 2\sin\theta\cos\theta\))

By first principle: \(f'(x) = \lim_{h \to 0} \frac{f(x+h) – f(x)}{h}\)

\(f(x+h) – f(x) \)\(= \frac{1}{2}[\sin 2(x+h) – \sin 2x]\)

Using \(\sin C – \sin D = 2 \cos\frac{C+D}{2} \sin\frac{C-D}{2}\):

\(\sin 2(x+h) – \sin 2x \)\(= 2 \cos\frac{2(x+h)+2x}{2} \sin\frac{2(x+h)-2x}{2}\)

\(= 2 \cos(2x + h) \sin h\)

Therefore, \(f'(x) = \lim_{h \to 0} \frac{\frac{1}{2} \cdot 2 \cos(2x + h) \sin h}{h} \)\(= \lim_{h \to 0} \cos(2x + h) \cdot \frac{\sin h}{h}\)

As \(h \to 0\), \(\cos(2x + h) \to \cos 2x\) and \(\frac{\sin h}{h} \to 1\)

Thus, \(f'(x) = \cos 2x\)

We can also write \(\cos 2x = \cos^2 x – \sin^2 x\)