Class 11 Mathematics

Solution:

Let $U$ be the set of total surveyed students, $T$ be the set of students taking tea, and $C$ be the set of students taking coffee.

• Total students $n(U) = 600$
• $n(T) = 150$
• $n(C) = 225$
• $n(T \cap C) = 100$

Number of students taking either tea or coffee $n(T \cup C)$:

$n(T \cup C) = n(T) + n(C) - n(T \cap C) = 150 + 225 - 100 = 275$

Number of students taking neither tea nor coffee $n(T' \cap C')$:

$n(T' \cap C') = n(U) - n(T \cup C) = 600 - 275 = 325 \text{ students.}$

Solution:

1. Domain: For $f(x)$ to be a real function, the quantity inside the square root must be non-negative:

$9 - x^2 \ge 0 \implies x^2 \le 9 \implies -3 \le x \le 3$

Hence, $\text{Domain} = [-3, 3]$.

2. Range: Let $y = \sqrt{9 - x^2}$. Since square root values are always non-negative, $y \ge 0$.

Squaring both sides: $y^2 = 9 - x^2 \implies x^2 = 9 - y^2$.

Since $-3 \le x \le 3$, $0 \le x^2 \le 9$, which gives $0 \le 9 - y^2 \le 9 \implies 0 \le y \le 3$.

Hence, $\text{Range} = [0, 3]$.

Solution:

By definition of operations on real functions:

• $(f + g)(x) = f(x) + g(x) = x^2 + 2x + 1 = (x + 1)^2$
• $(f - g)(x) = f(x) - g(x) = x^2 - (2x + 1) = x^2 - 2x - 1$
• $\left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)} = \frac{x^2}{2x + 1}, \quad \text{where } x \neq -\frac{1}{2}$

Solution:

Using transformation formulas:

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

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

Apply to LHS:

$$\text{LHS} = \frac{2 \sin\left(\frac{5x+3x}{2}\right) \cos\left(\frac{5x-3x}{2}\right)}{2 \cos\left(\frac{5x+3x}{2}\right) \cos\left(\frac{5x-3x}{2}\right)} = \frac{2 \sin 4x \cos x}{2 \cos 4x \cos x}$$

Cancelling common factors $2\cos x$:

$$\text{LHS} = \frac{\sin 4x}{\cos 4x} = \tan 4x = \text{RHS. (Hence Proved)}$$

Solution:

Multiply LHS by 2 and divide by 2, using $2\cos A \cos B = \cos(A+B) + \cos(A-B)$:

$$\text{LHS} = \frac{1}{2} \left[ 2\cos 2x \cos \frac{x}{2} - 2\cos \frac{9x}{2} \cos 3x \right]$$

$$= \frac{1}{2} \left[ \left(\cos \frac{5x}{2} + \cos \frac{3x}{2}\right) - \left(\cos \frac{15x}{2} + \cos \frac{3x}{2}\right) \right]$$

$$= \frac{1}{2} \left[ \cos \frac{5x}{2} - \cos \frac{15x}{2} \right]$$

Apply $\cos C - \cos D = -2\sin\left(\frac{C+D}{2}\right)\sin\left(\frac{C-D}{2}\right)$:

$$= \frac{1}{2} \left[ -2 \sin\left(\frac{10x}{2}\right) \sin\left(\frac{-5x}{2}\right) \right] = \sin 5x \sin \frac{5x}{2} = \text{RHS.}$$

Solution:

Rearranging the given equation:

$$\cos 4x - \cos 2x = 0$$

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

$$-2 \sin 3x \sin x = 0 \implies \sin 3x = 0 \quad \text{or} \quad \sin x = 0$$

• If $\sin 3x = 0 \implies 3x = n\pi \implies x = \frac{n\pi}{3}, \quad n \in \mathbb{Z}$
• If $\sin x = 0 \implies x = n\pi, \quad n \in \mathbb{Z}$

General Solution: $x = \frac{n\pi}{3}, \quad n \in \mathbb{Z}$ (since $n\pi$ is a subset of $\frac{n\pi}{3}$).

Solution:

Numerator: $(3)^2 - (i\sqrt{5})^2 = 9 - i^2(5) = 9 - (-1)(5) = 14$

Denominator: $\sqrt{3} + \sqrt{2}i - \sqrt{3} + \sqrt{2}i = 2\sqrt{2}i$

Expression becomes:

$$\frac{14}{2\sqrt{2}i} = \frac{7}{\sqrt{2}i} \times \frac{-i}{-i} = \frac{-7i}{\sqrt{2}} = 0 - \frac{7\sqrt{2}}{2}i$$

In $a + ib$ form: $0 + \left(-\frac{7\sqrt{2}}{2}\right)i$

Solution:

Simplify $z$ by rationalizing:

$$z = \frac{(1+i)(1+i)}{(1-i)(1+i)} = \frac{1 + i^2 + 2i}{1 - i^2} = \frac{1 - 1 + 2i}{1 + 1} = \frac{2i}{2} = i$$

Here $z = 0 + 1i$, so $x = 0, y = 1$.

• Modulus $|z|$: $\sqrt{x^2 + y^2} = \sqrt{0^2 + 1^2} = 1$
• Argument $\theta$: Since $x=0$ and $y > 0$, $\theta = \frac{\pi}{2}$ (or $90^\circ$)

Solution:

Comparing with $ax^2 + bx + c = 0$: $a = 1, b = 3, c = 9$.

Discriminant $D = b^2 - 4ac = 3^2 - 4(1)(9) = 9 - 36 = -27 < 0$.

Using quadratic formula $x = \frac{-b \pm \sqrt{D}}{2a}$:

$$x = \frac{-3 \pm \sqrt{-27}}{2} = \frac{-3 \pm 3\sqrt{3}i}{2}$$

Solution:

Multiply both sides by LCM $(3, 4, 5) = 60$:

$$20(2x - 1) \ge 15(3x - 2) - 12(2 - x)$$

$$40x - 20 \ge 45x - 30 - 24 + 12x$$

$$40x - 20 \ge 57x - 54$$

$$54 - 20 \ge 57x - 40x \implies 34 \ge 17x \implies x \le 2$$

Solution set: $x \in (-\infty, 2]$

Solution Steps:

1. Draw line $3x + 4y = 12$ passing through $(4,0)$ and $(0,3)$. Test $(0,0) \implies 0 \le 12$ (true), so shade region towards origin.
2. Draw vertical line $x = 1$. Shade the region to the right ($x \ge 1$).
3. Draw horizontal line $y = 2$. Shade the region above ($y \ge 2$).

The feasible region is the triangular region bounded by the intersection of these three shaded half-planes.

Solution:

We need to select and arrange 4 distinct digits out of 9 digits (1 to 9):

$$\text{Total 4-digit numbers} = {}^9P_4 = \frac{9!}{(9-4)!} = 9 \times 8 \times 7 \times 6 = 3024$$

Solution:

Using permutation formula ${}^nP_r = \frac{n!}{(n-r)!}$:

$$\frac{\frac{(n-1)!}{(n-1-3)!}}{\frac{n!}{(n-4)!}} = \frac{1}{9} \implies \frac{(n-1)!}{(n-4)!} \times \frac{(n-4)!}{n!} = \frac{1}{9}$$

$$\frac{(n-1)!}{n(n-1)!} = \frac{1}{9} \implies \frac{1}{n} = \frac{1}{9} \implies n = 9$$

Solution:

At least 3 girls means either 3 girls or 4 girls:

• Case 1 (3 Girls + 4 Boys): ${}^4C_3 \times {}^9C_4 = 4 \times 126 = 504$
• Case 2 (4 Girls + 3 Boys): ${}^4C_4 \times {}^9C_3 = 1 \times 84 = 84$

Total Ways $= 504 + 84 = 588$

Solution:

1. General Term $T_{r+1}$:

$$T_{r+1} = {}^{12}C_r (x^2)^{12-r} \left(-\frac{1}{x}\right)^r = {}^{12}C_r (-1)^r x^{24 - 2r - r} = {}^{12}C_r (-1)^r x^{24 - 3r}$$

2. Term Independent of $x$: Set exponent of $x$ to $0$:

$24 - 3r = 0 \implies 3r = 24 \implies r = 8$

$$\text{Independent Term } T_9 = {}^{12}C_8 (-1)^8 = {}^{12}C_4 = \frac{12 \times 11 \times 10 \times 9}{4 \times 3 \times 2 \times 1} = 495$$

Solution:

Write $9^{n+1} = (1 + 8)^{n+1}$ and expand using Binomial Theorem:

$$9^{n+1} = {}^{n+1}C_0 + {}^{n+1}C_1 (8) + {}^{n+1}C_2 (8^2) + \dots + {}^{n+1}C_{n+1} (8^{n+1})$$

$$9^{n+1} = 1 + (n+1)8 + 64 \left[ {}^{n+1}C_2 + {}^{n+1}C_3 (8) + \dots \right]$$

$$9^{n+1} = 1 + 8n + 8 + 64k \quad (\text{where } k \text{ is an integer})$$

$$9^{n+1} - 8n - 9 = 64k$$

Hence, $9^{n+1} - 8n - 9$ is divisible by 64.

Solution:

Let $S_n = 8 + 88 + 888 + \dots + \text{n terms}$.

$$S_n = 8 (1 + 11 + 111 + \dots) = \frac{8}{9} (9 + 99 + 999 + \dots)$$

$$S_n = \frac{8}{9} \left[ (10 - 1) + (10^2 - 1) + (10^3 - 1) + \dots + (10^n - 1) \right]$$

Summing the GP $10 + 10^2 + \dots + 10^n$:

$$S_n = \frac{8}{9} \left[ \frac{10(10^n - 1)}{10 - 1} - n \right] = \frac{80}{81}(10^n - 1) - \frac{8n}{9}$$

Solution:

Given $\text{A.M.} = \frac{a+b}{2} = 10 \implies a + b = 20$.

Given $\text{G.M.} = \sqrt{ab} = 8 \implies ab = 64$.

Using identity $(a - b)^2 = (a + b)^2 - 4ab$:

$(a - b)^2 = 20^2 - 4(64) = 400 - 256 = 144 \implies a - b = 12$.

Solving $a+b=20$ and $a-b=12$ gives:

$a = 16, b = 4 \quad \text{or} \quad a = 4, b = 16$.

Solution:

Solving $4x + 7y = 3$ and $2x - 3y = -1$ simultaneously:

Multiply second equation by $2$: $4x - 6y = -2$.

Subtract: $13y = 5 \implies y = \frac{5}{13}$.

Substitute back: $2x = 3\left(\frac{5}{13}\right) - 1 = \frac{2}{13} \implies x = \frac{1}{13}$. Point is $\left(\frac{1}{13}, \frac{5}{13}\right)$.

Line with equal intercepts $a$ has equation $\frac{x}{a} + \frac{y}{a} = 1 \implies x + y = a$.

Substitute point: $a = \frac{1}{13} + \frac{5}{13} = \frac{6}{13}$.

Equation of line: $x + y = \frac{6}{13} \implies 13x + 13y = 6$.

Solution:

Perpendicular distance formula $d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$:

$$d = \frac{|3(1) - 4(-2) - 26|}{\sqrt{3^2 + (-4)^2}} = \frac{|3 + 8 - 26|}{\sqrt{9 + 16}} = \frac{|-15|}{5} = 3 \text{ units}$$

Solution:

Comparing with $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$: $a^2 = 36 \implies a = 6$, $b^2 = 16 \implies b = 4$.

• $c = \sqrt{a^2 - b^2} = \sqrt{36 - 16} = \sqrt{20} = 2\sqrt{5}$
• Vertices: $(\pm a, 0) = (\pm 6, 0)$
• Foci: $(\pm c, 0) = (\pm 2\sqrt{5}, 0)$
• Major Axis Length: $2a = 2(6) = 12$
• Eccentricity $e$: $\frac{c}{a} = \frac{2\sqrt{5}}{6} = \frac{\sqrt{5}}{3}$

Solution:

Using identity $1 - \cos \theta = 2 \sin^2\left(\frac{\theta}{2}\right)$:

$$\lim_{x \to 0} \frac{-(1 - \cos 2x)}{-(1 - \cos x)} = \lim_{x \to 0} \frac{2 \sin^2 x}{2 \sin^2(x/2)} = \lim_{x \to 0} \frac{\left(\frac{\sin x}{x}\right)^2 \cdot x^2}{\left(\frac{\sin(x/2)}{x/2}\right)^2 \cdot \frac{x^2}{4}}$$

Since $\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$:

$$\text{Limit} = \frac{1 \cdot x^2}{1 \cdot \frac{x^2}{4}} = 4$$

Solution:

By First Principle: $f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$

$$f'(x) = \lim_{h \to 0} \frac{\sin(x+h) - \sin x}{h}$$

Apply $\sin C - \sin D = 2 \cos\left(\frac{C+D}{2}\right) \sin\left(\frac{C-D}{2}\right)$:

$$f'(x) = \lim_{h \to 0} \frac{2 \cos\left(x + \frac{h}{2}\right) \sin\left(\frac{h}{2}\right)}{h} = \lim_{h \to 0} \cos\left(x + \frac{h}{2}\right) \cdot \lim_{h \to 0} \frac{\sin(h/2)}{h/2}$$

$$f'(x) = \cos(x + 0) \cdot 1 = \cos x$$

Solution:

Using Quotient Rule $\frac{d}{dx}\left[\frac{u}{v}\right] = \frac{v u' - u v'}{v^2}$ where $u = x + \cos x$ and $v = \tan x$:

• $u' = 1 - \sin x$
• $v' = \sec^2 x$

$$\frac{dy}{dx} = \frac{\tan x (1 - \sin x) - (x + \cos x)\sec^2 x}{\tan^2 x}$$

Solution:

1. Mean ($\bar{x}$):

$$\bar{x} = \frac{\sum x_i}{n} = \frac{1 + 2 + \dots + n}{n} = \frac{\frac{n(n+1)}{2}}{n} = \frac{n+1}{2}$$

2. Variance ($\sigma^2$):

$$\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2 = \frac{\frac{n(n+1)(2n+1)}{6}}{n} - \left(\frac{n+1}{2}\right)^2$$

$$= \frac{(n+1)(2n+1)}{6} - \frac{(n+1)^2}{4} = \frac{n+1}{12} \left[ 2(2n+1) - 3(n+1) \right]$$

$$\sigma^2 = \frac{n^2 - 1}{12}$$

Solution:

Total ways to draw 2 cards from 52 cards $= {}^{52}C_2$.

Total ways to draw 2 aces from 4 aces in deck $= {}^4C_2$.

$$P(\text{Both Aces}) = \frac{{}^4C_2}{{}^{52}C_2} = \frac{\frac{4 \times 3}{2}}{\frac{52 \times 51}{2}} = \frac{6}{1326} = \frac{1}{221}$$