Mixed Practice

We know that,

Discounted price = (100 - Discount)% × MRP

Cost (in ₹) in Calcutta	Discount %	GST %	Discounted price
1050		18	1050
1750		18	1750
1100	25	18	75% of 1100 = 825
1425	30	18	70% of 1425 = 997.50
1725	40	18	60% of 1725 = 1035
Total			5657.50

As this is a inter-state transaction, hence IGST will apply.

IGST = 18% of 5657.50 = $\frac{18}{100} \times 5657.50$ = 1018.35

Total amount = ₹ 5657.50 + ₹ 1018.35 = ₹ 6675.85

Hence, amount of bill = ₹ 6675.85

S.P. in Mathura = ₹ 20,000

GST = IGST = 12% of ₹ 20,000 = $\frac{12}{100} \times 20000$ = ₹ 2400 (Inter state transaction)

GST charged by the dealer in Mathura = Input GST for the dealer in Ratlam = ₹ 2400.

S.P. in Ratlam = ₹ 20000 + ₹ 3750 = ₹ 23750.

GST charged by the dealer in Ratlam = CGST + SGST (Intra-state transaction)

CGST = 6% of ₹ 23750 = $\frac{6}{100} \times 23750$ = ₹ 1425

SGST = CGST = ₹ 1425

GST = CGST + SGST = ₹ 1425 + ₹ 1425 = ₹ 2850

Cost of the article in Indore = S.P. for dealer in Ratlam + GST

= ₹ 23750 + ₹ 2850

= ₹ 26600.

Hence, cost of article in Indore = ₹ 26600.

(i) Let monthly deposit be ₹ P.

Given,

Time = 4 years = 4 × 12 = 48 months.

By formula,

Interest (I) = P × $\frac{n(n + 1)}{2 \times 12} \times \frac{r}{100}$

Substituting values we get :

$$\begin{aligned}\Rightarrow 21560 = P \times \frac{48 \times (48 + 1)}{2 \times 12} \times \frac{10}{100} \\[1em] \Rightarrow 21560 = P \times \frac{48 \times 49}{24} \times \frac{1}{10} \\[1em] \Rightarrow P = \frac{21560 \times 24 \times 10}{48 \times 49} \\[1em] \Rightarrow P = \frac{107800}{49} = ₹ 2,200.\end{aligned}$$

Hence, monthly installment = ₹ 2,200.

(ii) By formula,

Maturity value = P × n + Interest

= ₹ 2200 × 48 + ₹ 21560

= ₹ 105600 + ₹ 21560

= ₹ 1,27,160.

Hence, amount of maturity = ₹ 1,27,160.

Let time be n months.

Given,

M.V. = ₹ 1,20,400

By formula,

M.V. = P × n + P × $\frac{n(n + 1)}{2 \times 12} \times \frac{r}{100}$

Substituting values we get :

$$\begin{aligned}\Rightarrow 120400 = 1600 × n + 1600 \times \frac{n(n + 1)}{2 \times 12} \times \frac{10}{100} \\[1em] \Rightarrow 120400 = 1600\Big(n + \frac{n(n + 1)}{24} \times \frac{10}{100}\Big) \\[1em] \Rightarrow \frac{120400}{1600} = n + \frac{n(n + 1)}{240} \\[1em] \Rightarrow \frac{301}{4} = \frac{240n + n^2 + n}{240} \\[1em] \Rightarrow \frac{301 \times 240}{4} = 240n + n^2 + n \\[1em] \Rightarrow 301 \times 60 = n^2 + 241n \\[1em] \Rightarrow n^2 + 241n - 18060 = 0 \\[1em] \Rightarrow n^2 + 301n - 60n - 18060 = 0 \\[1em] \Rightarrow n(n + 301) - 60(n + 301) = 0 \\[1em] \Rightarrow (n - 60)(n + 301) = 0 \\[1em] \Rightarrow n = 60 \text{ or } n = -301.\end{aligned}$$

Since, time cannot be negative.

∴ n = 60 months or $\frac{60}{12}$ = 5 years.

Hence, time = 5 years.

Given, investment = ₹ 24000, div % = 7%, N.V. = ₹ 100,

⇒ M.V. = N.V. - discount

⇒ M.V. = ₹ 100 - $\frac{20}{100} \times 100$ = ₹ 100 - ₹ 20 = ₹ 80.

(i) No. of shares = $\frac{\text{Investment}}{\text{M.V. of each share}} = \frac{₹ 24000}{₹ 80}$ = 300.

∴ Dividend on 1 share = 7% of ₹ 100 = ₹ 7

∴ Rajat's dividend for the first year = ₹ 7 × 300 = ₹ 2100.

Hence, Rajat gained ₹ 2100 in first year.

(ii) Since, each share is sold for ₹ 75

∴ Proceeds (including dividend) = 300 × ₹ 75 + ₹ 2100 = ₹ 22500 + ₹ 2100 = ₹ 24600.

N.V. of each share = ₹ 25

M.V. of each share = N.V. + 64% of N.V.

= ₹ 25 + $\frac{64}{100} \times 25$

= ₹ 25 + ₹ 16

= ₹ 41

Dividend = 18%

∴ No. of shares bought = $\frac{\text{Investment}}{\text{M.V. of each share}} = \frac{₹ 24600}{₹ 41}$ = 600.

Dividend on 1 share = 18% of ₹ 25 = $\frac{18 \times 25}{100}$ = ₹ 4.50

Annual dividend (income) from second investment = 600 × ₹ 4.50 = ₹ 2700.

Hence, annual dividend (income) from second investment = ₹ 2700.

(iii) Increase in return = ₹ 2700 - ₹ 2100 = ₹ 600.

Percentage increase in return (on original investment) = $\frac{600}{24000} \times 100 = \frac{100}{40}$ = 2.5 %.

Hence, percentage increase in return = 2.5 %.

N.V. of share = ₹ 20

M.V. of share = N.V. - Discount

= ₹ 20 - 10% of 20

= ₹ 20 - $\frac{10}{100} \times 20$

= ₹ 20 - ₹ 2 = ₹ 18.

∴ Dividend on 1 share = 8% of ₹ 20 = $\frac{8}{100} \times 20$ = ₹ 1.60

Let no. of shares purchased be x.

Dividend on x number of shares = 1.6x

S.P. of each share = ₹ 18

S.P. of x shares (Sum invested) = ₹ 18x

In 2nd investment

N.V. of share = ₹ 10

M.V. of share = N.V. + Premium

= ₹ 10 + 50% of 10

= ₹ 10 + ₹ 5

= ₹ 15

No. of shares purchased = $\frac{\text{Investment}}{\text{M.V. of share}} = \frac{18x}{15}$

∴ Dividend on 1 share = 12% of ₹ 10 = $\frac{12}{100} \times 10$ = ₹ 1.2.

Total income = 1.2 × $\frac{18x}{15} = \frac{21.6x}{15}$ = 1.44x

Given,

Change in income = ₹ 600

⇒ 1.6x - 1.44x = 600

⇒ 0.16x = 600

⇒ x = $\frac{600}{0.16}$ = 3750.

Hence, no. of shares sold = 3750.

Solving L.H.S. of the above inequation :

$$\begin{aligned}\Rightarrow -2 \le \frac{1}{2} - \frac{2x}{3} \\[1em] \Rightarrow \frac{2x}{3} \le \frac{1}{2} + 2 \\[1em] \Rightarrow \frac{2x}{3} \le \frac{5}{2} \\[1em] \Rightarrow x \le \frac{5}{2} \times \frac{3}{2} \\[1em] \Rightarrow x \le \frac{15}{4} \text{ .........(1) }\end{aligned}$$

Solving R.H.S. of the above inequation :

$$\begin{aligned}\Rightarrow \frac{1}{2} - \frac{2x}{3} \le 1\frac{5}{6} \\[1em] \Rightarrow \frac{1}{2} - \frac{2x}{3} \le \frac{11}{6} \\[1em] \Rightarrow \frac{2x}{3} \ge \frac{1}{2} - \frac{11}{6}\\[1em] \Rightarrow \frac{2x}{3} \ge \frac{3 - 11}{6} \\[1em] \Rightarrow x \ge -\frac{8}{6} \times \frac{3}{2} \\[1em] \Rightarrow x \ge -2 \text{ .......(2)}\end{aligned}$$

From (1) and (2), we get :

-2 ≤ x ≤ $\frac{15}{4}$

Since, x ∈ W.

x = {0, 1, 2, 3}

Hence, x = {0, 1, 2, 3}.

Given, equation :

$$\begin{aligned}\Rightarrow \frac{x}{x + 1} + \frac{x + 1}{x} = 2\frac{4}{15} \\[1em] \Rightarrow \frac{x^2 + (x + 1)^2}{x(x + 1)} = \frac{34}{15} \\[1em] \Rightarrow 15[x^2 + (x + 1)^2] = 34x(x + 1) \\[1em] \Rightarrow 15x^2 + 15(x + 1)^2 = 34x^2 + 34x \\[1em] \Rightarrow 15x^2 + 15(x^2 + 1 + 2x) = 34x^2 + 34x \\[1em] \Rightarrow 15x^2 + 15x^2 + 15 + 30x = 34x^2 + 34x \\[1em] \Rightarrow 34x^2 - 15x^2 - 15x^2 + 34x - 30x - 15 = 0 \\[1em] \Rightarrow 4x^2 + 4x - 15 = 0.\end{aligned}$$

Comparing above equation, with ax² + bx + c = 0, we get :

a = 4, b = 4 and c = -15.

By formula,

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Substituting values we get :

$$\begin{aligned}x = \frac{-4 \pm \sqrt{4^2 - 4 \times 4 \times (-15)}}{2 \times 4} \\[1em] = \frac{-4 \pm \sqrt{16 + 240}}{8} \\[1em] = \frac{-4 \pm \sqrt{256}}{8} \\[1em] = \frac{-4 \pm 16}{8} \\[1em] = \frac{-4 + 16}{8}, \frac{-4 - 16}{8} \\[1em] = \frac{12}{8}, -\frac{20}{8} \\[1em] = \frac{3}{2}, -\frac{5}{2}\end{aligned}$$

Hence, x = $-\frac{5}{2}$, $\frac{3}{2}$

Let cost price be ₹ x.

Profit = ₹ (24 - x)

According to question,

Profit percent = x%.

By formula,

$\Rightarrow \text{Profit}$ % = $\frac{\text{Profit}}{\text{C.P.}} \times 100$

$$\begin{aligned}\Rightarrow x = \frac{24 - x}{x} \times 100 \\[1em] \Rightarrow x^2 = 100(24 - x) \\[1em] \Rightarrow x^2 = 2400 - 100x \\[1em] \Rightarrow x^2 + 100x - 2400 = 0 \\[1em] \Rightarrow x^2 + 120x - 20x - 2400 = 0 \\[1em] \Rightarrow x(x + 120) - 20(x + 120) = 0 \\[1em] \Rightarrow (x - 20)(x + 120) = 0 \\[1em] \Rightarrow x - 20 = 0 \text{ or } x + 120 = 0 \\[1em] \Rightarrow x = 20 \text{ or } x = -120.\end{aligned}$$

Since, C.P. cannot be negative.

∴ x = ₹ 20.

Hence, cost price = ₹ 20.

Let A take x days to complete the work so B will take (x - 16) days.

Work done by A alone in 1 day = $\frac{1}{x}$

Work done by B alone in 1 day = $\frac{1}{x - 16}$

Work done by A and B together in 1 day is $\frac{1}{x} + \frac{1}{x - 16}$

Given,

If both work together, they finish the work in 15 days.

$$\begin{aligned}\therefore 15\Big(\frac{1}{x} + \frac{1}{x - 16}\Big) = 1 \\[1em] \Rightarrow \frac{x - 16 + x}{x(x - 16)} = \frac{1}{15} \\[1em] \Rightarrow \frac{2x - 16}{x^2 - 16x} = \frac{1}{15} \\[1em] \Rightarrow 15(2x - 16) = x^2 - 16x \\[1em] \Rightarrow 30x - 240 = x^2 - 16x \\[1em] \Rightarrow x^2 - 16x - 30x + 240 = 0 \\[1em] \Rightarrow x^2 - 46x + 240 = 0 \\[1em] \Rightarrow x^2 - 40x - 6x + 240 = 0 \\[1em] \Rightarrow x(x - 40) - 6(x - 40) = 0 \\[1em] \Rightarrow (x - 6)(x - 40) = 0 \\[1em] \Rightarrow x - 6 = 0 \text{ or } x - 40 = 0 \\[1em] \Rightarrow x = 6 \text{ or } x = 40.\end{aligned}$$

Value of x will be greater than 16.

∴ x = 40 and x - 16 = 40 - 16 = 24.

Hence, B alone can finish the work in 24 days.

Let half of the first part, one-third of the second part and one-sixth of the third part be equal to x.

⇒ $\frac{1}{2}$ First part = x

⇒ First part = 2x

⇒ $\frac{1}{3}$ Second part = x

⇒ Second part = 3x

⇒ $\frac{1}{6}$ Third part = x

⇒ Third part = 6x.

We know that,

⇒ Total = ₹ 1870

⇒ 2x + 3x + 6x = ₹ 1870

⇒ 11x = ₹ 1870

⇒ x = $\frac{1870}{11}$ = ₹ 170.

First part = 2x = 2 × 170 = ₹ 340

Second part = 3x = 3 × 170 = ₹ 510

Third part = 6x = 6 × 170 = ₹ 1020

Hence, first part = ₹ 340, second part = ₹ 510 and third part = ₹ 1020.

Given,

a + c = be and $\frac{1}{b} + \frac{1}{d} = \frac{e}{c}$.

Solving,

a + c = be

Divide the equation by b,

$\Rightarrow \frac{a}{b} + \frac{c}{b}$ = e ............(1)

Now solving,

$\frac{1}{b} + \frac{1}{d} = \frac{e}{c}$

Multiplying the equation by c,

$\Rightarrow \frac{c}{b} + \frac{c}{d}$ = e

Putting value of e from equation (1) in above equation, we get :

$$\begin{aligned}\Rightarrow \frac{c}{b} + \frac{c}{d} = \frac{a}{b} + \frac{c}{b} \\[1em] \Rightarrow \frac{c}{d} = \frac{a}{b}.\end{aligned}$$

Since, $\frac{a}{b} = \frac{c}{d}$, hence, a, b, c and d are in proportion.

Given,

Equation : $\frac{4x + 3y}{4x - 3y} = \frac{7}{4}$

Applying componendo and dividendo we get :

$$\begin{aligned}\Rightarrow \frac{4x + 3y + (4x - 3y)}{4x + 3y - (4x - 3y)} = \frac{7 + 4}{7 - 4} \\[1em] \Rightarrow \frac{8x}{6y} = \frac{11}{3} \\[1em] \Rightarrow \frac{x}{y} = \frac{6 \times 11}{3 \times 8} \\[1em] \Rightarrow \frac{x}{y} = \frac{11}{4}\end{aligned}$$

Squaring both sides,

$$\begin{aligned}\Rightarrow \Big(\frac{x}{y}\Big)^2 = \frac{121}{16} \\[1em] \Rightarrow \frac{x^2}{y^2} = \frac{121}{16}\end{aligned}$$

Multiplying both sides by $\frac{2}{11}$, we get :

$$\begin{aligned}\Rightarrow \frac{2x^2}{11y^2} = \frac{121}{16} \times \frac{2}{11} \\[1em] \Rightarrow \frac{2x^2}{11y^2} = \frac{11}{8}\end{aligned}$$

Applying componendo and dividendo we get :

$$\begin{aligned}\Rightarrow \frac{2x^2 + 11y^2}{2x^2 - 11y^2} = \frac{11 + 8}{11 - 8} \\[1em] \Rightarrow \frac{2x^2 + 11y^2}{2x^2 - 11y^2} = \frac{19}{3} \\[1em] \Rightarrow \frac{2x^2 - 11y^2}{2x^2 + 11y^2} = \frac{3}{19}.\end{aligned}$$

Hence, $\frac{2x^2 - 11y^2}{2x^2 + 11y^2} = \frac{3}{19}$.

Given,

Equation : $\frac{\sqrt{12x + 1} + \sqrt{2x - 3}}{\sqrt{12x + 1} - \sqrt{2x - 3}} = \frac{3}{2}$.

Applying componendo and dividendo we get :

$$\begin{aligned}\Rightarrow \frac{\sqrt{12x + 1} + \sqrt{2x - 3} + (\sqrt{12x + 1} - \sqrt{2x - 3})}{\sqrt{12x + 1} + \sqrt{2x - 3} - (\sqrt{12x + 1} - \sqrt{2x - 3})} = \frac{3 + 2}{3 - 2} \\[1em] \Rightarrow \frac{2\sqrt{12x + 1}}{2\sqrt{2x - 3}} = 5 \\[1em] \Rightarrow \frac{\sqrt{12x + 1}}{\sqrt{2x - 3}} = 5\end{aligned}$$

Squaring both sides we get :

$$\begin{aligned}\Rightarrow \Big(\frac{\sqrt{12x + 1}}{\sqrt{2x - 3}}\Big)^2 = 5^2 \\[1em] \Rightarrow \frac{12x + 1}{2x - 3} = 25 \\[1em] \Rightarrow 12x + 1 = 25(2x - 3) \\[1em] \Rightarrow 12x + 1 = 50x - 75 \\[1em] \Rightarrow 50x - 12x = 75 + 1 \\[1em] \Rightarrow 38x = 76 \\[1em] \Rightarrow x = \frac{76}{38} = 2.\end{aligned}$$

Hence, x = 2.

Given,

⇒ 2x + 1 = 0

⇒ x = -$\frac{1}{2}$.

Let number added be a.

Polynomial = 2x³ - 3x² - 8x + 3 + a

By remainder theorem,

When a polynomial p(x) is divided by (x - a), then the remainder = f(a).

$$\begin{aligned}\Rightarrow 2\Big(-\frac{1}{2}\Big)^3 - 3\Big(-\frac{1}{2}\Big)^2 - 8\Big(-\frac{1}{2}\Big) + 3 + a = 10 \\[1em] \Rightarrow 2 \times -\frac{1}{8} - 3 \times \frac{1}{4} + 4 + 3 + a = 10 \\[1em] \Rightarrow -\frac{1}{4} - \frac{3}{4} + a = 10 - 4 - 3 \\[1em] \Rightarrow \frac{-4}{4} + a = 3 \\[1em] \Rightarrow -1 + a = 3 \\[1em] \Rightarrow a = 3 + 1 = 4.\end{aligned}$$

Hence, no. to be added = 4.

Given,

A² + B² = (A + B)²

$$\Rightarrow \begin{bmatrix*}[r] 1 & -1 \\ 2 & -1 \end{bmatrix*}^2 + \begin{bmatrix*}[r] x & 1 \\ 4 & -1 \end{bmatrix*}^2 = \Big(\begin{bmatrix*}[r] 1 & -1 \\ 2 & -1 \end{bmatrix*} + \begin{bmatrix*}[r] x & 1 \\ 4 & -1 \end{bmatrix*}\Big)^2 \\[1em] \Rightarrow \begin{bmatrix*}[r] 1 & -1 \\ 2 & -1 \end{bmatrix*}\begin{bmatrix*}[r] 1 & -1 \\ 2 & -1 \end{bmatrix*} + \begin{bmatrix*}[r] x & 1 \\ 4 & -1 \end{bmatrix*}\begin{bmatrix*}[r] x & 1 \\ 4 & -1 \end{bmatrix*} \\[1em] = \Big(\begin{bmatrix*}[r] 1 + x & 0 \\ 6 & -2 \end{bmatrix*}\Big)^2 \\[1em] \Rightarrow \begin{bmatrix*}[r] 1 \times 1 + (-1) \times 2 & 1 \times -1 + (-1) \times(-1) \\ 2 \times 1 + (-1) \times 2 & 2 \times -1 + (-1) \times (-1) \end{bmatrix*} + \begin{bmatrix*}[r] x \times x + 1 \times 4 & x \times 1 + 1 \times(-1) \\ 4 \times x + (-1) \times 4 & 4 \times 1 + (-1) \times (-1) \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] (1 + x)^2 + 0 \times 6 & (1 + x) \times 0 + 0 \times -2 \\ 6(1 + x) + (-2) \times 6 & 6 \times 0 + (-2) \times(-2) \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 1 - 2 & -1 + 1 \\ 2 - 2 & -2 + 1 \end{bmatrix*} + \begin{bmatrix*}[r] x^2 + 4 & x - 1 \\ 4x - 4 & 4 + 1 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] (1 + x)^2 & 0 \\ 6 + 6x - 12 & 4 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] -1 & 0 \\ 0 & -1 \end{bmatrix*} + \begin{bmatrix*}[r] x^2 + 4 & x - 1 \\ 4x - 4 & 5 \end{bmatrix*} = \begin{bmatrix*}[r] (1 + x)^2 & 0 \\ 6x - 6 & 4 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] x^2 + 4 - 1 & x - 1 \\ 4x - 4 & 5 - 1 \end{bmatrix*} = \begin{bmatrix*}[r] (1 + x)^2 & 0 \\ 6x - 6 & 4 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] x^2 + 3 & x - 1 \\ 4x - 4 & 4 \end{bmatrix*} = \begin{bmatrix*}[r] (1 + x)^2 & 0 \\ 6x - 6 & 4 \end{bmatrix*}$$

⇒ x - 1 = 0

⇒ x = 1.

A² + B² and (A + B)² are not always equal. As,

By formula,

(A + B)² = A² + B² + 2AB.

Hence, x = 1.

Given, sequence : 10, 8, 6, ........

The above sequence is an A.P. with first term (a) = 10 and common difference (d) = (8 - 10) = -2.

By formula,

⇒ a_n = a + (n - 1)d

⇒ a₁₀ = 10 + (10 - 1) × -2

⇒ a₁₀ = 10 + 9 × -2 = 10 - 18 = -8.

Hence, 10th term of the sequence = -8.

Let first term be a and common difference be d of A.P.

By formula,

⇒ a_n = a + (n - 1)d

5^th term = 16

⇒ a₅ = 16

⇒ a + (5 - 1)d = 16

⇒ a + 4d = 16 .........(1)

11^th term = 34

⇒ a₁₁ = 34

⇒ a + (11 - 1)d = 34

⇒ a + 10d = 34 ........(2)

Subtracting equation (1) from (2), we get :

⇒ a + 10d - (a + 4d) = 34 - 16

⇒ a - a + 10d - 4d = 18

⇒ 6d = 18

⇒ d = 3.

Substituting value of d in equation (1), we get :

⇒ a + 4(3) = 16

⇒ a + 12 = 16

⇒ a = 4.

A.P. = a, (a + d), (a + 2d), ........

= 4, (4 + 3), (4 + 2 × 3), ........

= 4, 7, 10, .......

Hence, A.P. = 4, 7, 10, .......

Let first term of A.P. be a and common term be d.

By formula,

⇒ a_n = a + (n - 1)d

Given,

⇒ p^th term = q

⇒ a + (p - 1)d = q ..........(1)

⇒ q^th term = p

⇒ a + (q - 1)d = p ..........(2)

Subtracting (2) from (1), we get :

⇒ a + (p - 1)d - [a + (q - 1)d] = q - p

⇒ a - a + (p - 1)d - (q - 1)d = (q - p)

⇒ d[p - 1 - (q - 1)] = (q - p)

⇒ d[p - 1 - q + 1] = (q - p)

⇒ d[p - q] = (q - p)

⇒ d[p - q] = -(p - q)

⇒ d = -1.

Substituting value of d in equation (1),

⇒ a + (p - 1)(-1) = q

⇒ a - p + 1 = q

⇒ a = p + q - 1.

r^th term = a_r

= a + (r - 1)d

= p + q - 1 + (r - 1)(-1)

⇒ p + q - 1 - r + 1

⇒ p + q - r.

Hence, proved that r^th term = p + q - r.

Given,

⇒ a_n = (2n - 1)

⇒ a₇ = (2 × 7 - 1) = 13.

Hence, 7^th term = 13.

By formula,

S_n = $\frac{n}{2}[2a + (n - 1)d]$

Given,

S_n = 3n² + 2n

Equating values we get :

$$\begin{aligned}\Rightarrow 3n^2 + 2n = \frac{n}{2}[2a + (n - 1)d] \\[1em] \Rightarrow 3n^2 + 2n = \frac{n}{2}[a + a + (n - 1)d] \\[1em] \Rightarrow n(3n + 2) = \frac{n}{2}[a + a_n] \\[1em] \Rightarrow 3n + 2 = \frac{a + a_n}{2} \\[1em] \Rightarrow 2(3n + 2) = a + a_n \\[1em] \Rightarrow 6n + 4 = a + a_n \text{ .........(1)}\end{aligned}$$

As,

S_n = 3n² + 2n

S₁ = 3(1)² + 2(1) = 3 + 2 = 5.

∴ a = 5.

Substituting value in (1), we get :

⇒ 6n + 4 = 5 + a_n

⇒ a_n = 6n - 1

⇒ a_r = 6r - 1

Hence, r_th term = 6r - 1.

Given,

Common difference (d) = 2

Last term (l) = 18

Let no. of terms be n.

⇒ l = a + (n - 1)d

⇒ 18 = a + 2(n - 1)

⇒ 18 = a + 2n - 2

⇒ a + 2n = 20

⇒ a = 20 - 2n .........(1)

By formula,

S_n = $\frac{n}{2}(a + l)$

Substituting values we get :

$$\begin{aligned}\Rightarrow 60 = \frac{n}{2}(a + 18) \\[1em] \Rightarrow 120 = n(a + 18)\end{aligned}$$

Substituting value of a in above equation, we get :

$$\begin{aligned}\Rightarrow 120 = n(20 - 2n + 18) \\[1em] \Rightarrow 120 = n(38 - 2n) \\[1em] \Rightarrow 120 = 38n - 2n^2 \\[1em] \Rightarrow 2n^2 - 38n + 120 = 0 \\[1em] \Rightarrow 2(n^2 - 19n + 60) = 0 \\[1em] \Rightarrow n^2 - 19n + 60 = 0 \\[1em] \Rightarrow n^2 - 15n - 4n + 60 = 0 \\[1em] \Rightarrow n(n - 15) - 4(n - 15) = 0 \\[1em] \Rightarrow (n - 4)(n - 15) = 0 \\[1em] \Rightarrow n - 4 = 0 \text{ or } n - 15 = 0 \\[1em] \Rightarrow n = 4 \text{ or } n = 15.\end{aligned}$$

Hence, no. of terms = 4 or 15.

Let first term of G.P. be a and common ratio be r.

By formula,

⇒ a_n = ar^{n - 1}

⇒ a₅ = ar^{5 - 1}

⇒ 48 = ar⁴ ........(1)

⇒ a₈ = 384

⇒ ar^{8 - 1} = 384

⇒ ar⁷ = 384 ........(2)

Dividing equation (2) by (1), we get :

$$\begin{aligned}\Rightarrow \frac{ar^7}{ar^4} = \frac{384}{48} \\[1em] \Rightarrow r^3 = 8 \\[1em] \Rightarrow r^3 = (2)^3 \\[1em] \Rightarrow r = 2.\end{aligned}$$

Substituting value of r in equation (1), we get :

⇒ a(2)⁴ = 48

⇒ 16a = 48

⇒ a = $\frac{48}{16}$ = 3.

G.P. = a, ar, ar², .........

= 3, 3 × 2, 3 × 2², ........

= 3, 6, 12, ......

Hence, G.P. = 3, 6, 12, ......

Common ratio (r) = $\frac{-\frac{1}{3}}{\frac{2}{9}} = -\frac{9}{2 \times 3} = -\frac{3}{2}$.

Let no. of terms be n.

By formula,

Sum of G.P. (when |r| > 1 ) = $\frac{a(r^n - 1)}{(r - 1)}$

Substituting value we get :

$$\begin{aligned}\Rightarrow \frac{\frac{2}{9}\Big[\Big(-\frac{3}{2}\Big)^n - 1\Big]}{\Big(-\frac{3}{2} - 1\Big)} = \frac{55}{72} \\[1em] \Rightarrow \frac{\frac{2}{9}\Big[\Big(-\frac{3}{2}\Big)^n - 1\Big]}{\Big(\frac{-3 - 2}{2}\Big)} = \frac{55}{72} \\[1em] \Rightarrow \frac{\frac{2}{9}\Big[\Big(-\frac{3}{2}\Big)^n - 1\Big]}{\Big(-\frac{5}{2}\Big)} = \frac{55}{72} \\[1em] \Rightarrow \frac{2}{9}\Big[\Big(-\frac{3}{2}\Big)^n - 1\Big] = \frac{55}{72} \times -\frac{5}{2} \\[1em] \Rightarrow \Big[\Big(-\frac{3}{2}\Big)^n - 1\Big] = \frac{55}{72} \times -\frac{5}{2} \times \frac{9}{2} \\[1em] \Rightarrow \Big[\Big(-\frac{3}{2}\Big)^n - 1\Big] = -\frac{275}{32} \\[1em] \Rightarrow \Big(-\frac{3}{2}\Big)^n = -\frac{275}{32} + 1 \\[1em] \Rightarrow \Big(-\frac{3}{2}\Big)^n = \frac{-275 + 32}{32}\\[1em] \Rightarrow \Big(-\frac{3}{2}\Big)^n = -\frac{243}{32} \\[1em] \Rightarrow \Big(-\frac{3}{2}\Big)^n = \Big(-\frac{3}{2}\Big)^5 \\[1em] \Rightarrow n = 5.\end{aligned}$$

Hence, sum of 5 terms = $\frac{55}{72}$.

Given,

Sequence : 5 + 55 + 555 + .......

⇒ 5(1 + 11 + 111 + ........)

⇒ $\frac{5}{9} \times 9(1 + 11 + 111 + .........$ upto n terms)

⇒ $\frac{5}{9}(9 + 99 + 999 + .........$ upto n terms)

⇒ $\frac{5}{9}[(10 - 1) + (10^2 - 1) + (10^3 - 1) + .......$ upto n terms]

⇒ $\frac{5}{9}[(10 + 10^2 + 10^3 + \text{ upto n terms}) - n]$

⇒ $\frac{5}{9}\Big[\Big(\frac{10(10^n - 1)}{10 - 1} - n\Big)\Big]$

⇒ $\frac{5}{9} \times \frac{10(10^n - 1)}{9} - \frac{5}{9}n$

⇒ $\frac{50(10^n - 1)}{81} - \frac{5}{9}n$

Hence, sum = $\frac{50(10^n - 1)}{81} - \frac{5}{9}n$.

We know that,

y-coordinate on x-axis = 0.

Let the point on x-axis be (x, 0).

Distance formula = $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Since, (x, 0) is equidistant from (6, 7) and (4, -3)

$$\begin{aligned}\therefore \sqrt{(x - 6)^2 + (0 - 7)^2} = \sqrt{(x - 4)^2 + [0 - (-3)]^2} \\[1em] \Rightarrow \sqrt{x^2 + 36 - 12x + (-7)^2} = \sqrt{x^2 + 16 - 8x + 3^2} \\[1em] \Rightarrow \sqrt{x^2 + 36 - 12x + 49} = \sqrt{x^2 + 16 - 8x + 9}\end{aligned}$$

Squaring both sides we get :

$$\begin{aligned}\Rightarrow x^2 - 12x + 85 = x^2 - 8x + 25 \\[1em] \Rightarrow x^2 - x^2 - 8x + 12x = 85 - 25 \\[1em] \Rightarrow 4x = 60 \\[1em] \Rightarrow x = 15.\end{aligned}$$

(x, 0) = (15, 0).

Hence, (15, 0) is equidistant from (6, 7) and (4, -3).

We know that, y-coordinate on any point of the line y = 2 is 2. Let coordinate be x.

Point = (x, 2)

Let ratio in which point divides line segment AB is k : 1.

By section-formula,

(x, y) = $\Big(\frac{m_1x_2 + m_2x_1}{m_1 + m_2}, \frac{m_1y_2 + m_2y_1}{m_1 + m_2}\Big)$

Substituting values for y-coordinate we get,

$$\begin{aligned}\Rightarrow 2 = \frac{k \times (-3) + 1 \times 5}{k + 1} \\[1em] \Rightarrow 2(k + 1) = -3k + 5 \\[1em] \Rightarrow 2k + 2 = -3k + 5 \\[1em] \Rightarrow 2k + 3k = 5 - 2 \\[1em] \Rightarrow 5k = 3 \\[1em] \Rightarrow k = \frac{3}{5}.\end{aligned}$$

k : 1 = $\frac{3}{5} : 1$ = 3 : 5.

Hence, ratio in which the line segment A(6, 5) and B(4, -3) is divided by the line y = 2 is 3 : 5.

From graph,

The lines intersect at point I, J, K and L.

By two point formula,

Equation of line : y - y₁ = $\frac{y_2- y_1}{x_2 - x_1}(x - x_1)$

Equation of diagonal IK is

$$\begin{aligned}\Rightarrow y - 2 = \frac{-1 - 2}{4 - (-1)}[x - (-1)] \\[1em] \Rightarrow y - 2 = -\frac{3}{5}[x + 1] \\[1em] \Rightarrow 5(y - 2) = -3[x + 1] \\[1em] \Rightarrow 5y - 10 = -3x - 3 \\[1em] \Rightarrow 5y + 3x - 10 + 3 = 0 \\[1em] \Rightarrow 5y + 3x - 7 = 0.\end{aligned}$$

Equation of diagonal LJ is

$$\begin{aligned}\Rightarrow y - (-1) = \frac{2 - (-1)}{4 - (-1)}[x - (-1)] \\[1em] \Rightarrow y + 1 = \frac{3}{5}[x + 1] \\[1em] \Rightarrow 5(y + 1) = 3[x + 1] \\[1em] \Rightarrow 5y + 5 = 3x + 3 \\[1em] \Rightarrow 5y - 3x + 5 - 3 = 0 \\[1em] \Rightarrow 5y - 3x + 2 = 0.\end{aligned}$$

Hence, equation of diagonals are 5y + 3x - 7 = 0 and 5y - 3x + 2 = 0

(i) We know that,

y-coordinate at x-axis = 0.

Substituting y = 0 in 4x + 5y + 20 = 0, we get :

⇒ 4x + 5(0) + 20 = 0

⇒ 4x = -20

⇒ x = -5.

A = (-5, 0).

x-coordinate at y-axis = 0.

Substituting x = 0 in 4x + 5y + 20 = 0, we get :

⇒ 4(0) + 5y + 20 = 0

⇒ 5y = -20

⇒ y = $\frac{-20}{5}$

⇒ y = -4.

B = (0, -4).

Hence, A = (-5, 0) and B = (0, -4).

(ii) Given,

AB : BP = 5 : 3

Let AB = 5a and BP = 3a.

From figure,

⇒ AB = AP + PB

⇒ 5a = AP + 3a

⇒ AP = 5a - 3a = 2a.

AP : BP = 2a : 3a = 2 : 3.

Let coordinates of P be (x, y).

By section-formula,

(x, y) = $\Big(\frac{m_1x_2 + m_2x_1}{m_1 + m_2}, \frac{m_1y_2 + m_2y_1}{m_1 + m_2}\Big)$

Substituting values for x-coordinate we get :

$$\begin{aligned}\Rightarrow x = \frac{2 \times 0 + 3 \times -5}{2 + 3} \\[1em] \Rightarrow x = \frac{0 - 15}{5} \\[1em] \Rightarrow x = \frac{-15}{5} \\[1em] \Rightarrow x = -3.\end{aligned}$$

Substituting values for y-coordinate we get :

$$\begin{aligned}\Rightarrow y = \frac{2 \times -4 + 3 \times 0}{2 + 3} \\[1em] \Rightarrow y = \frac{-8 + 0}{5} \\[1em] \Rightarrow y = \frac{-8}{5} = -1\frac{3}{5}.\end{aligned}$$

P = (x, y) = $\Big(-3, -1\frac{3}{5}\Big)$.

(iii) By formula,

Slope = $\frac{y_2 - y_1}{x_2 - x_1}$

Substituting values we get :

Slope of AB = $\frac{-4 - 0}{0 - (-5)} = -\frac{4}{5}$

Let slope of line perpendicular to AB be m.

We know that,

Product of slope of perpendicular lines = -1.

$$\begin{aligned}\therefore -\frac{4}{5} \times m = -1 \\[1em] \Rightarrow m = \frac{5}{4}.\end{aligned}$$

By point-slope form,

Equation of line : y - y₁ = m(x - x₁)

Substituting values we get :

$$\begin{aligned}\Rightarrow y - \Big(-1\frac{3}{5}\Big) = \frac{5}{4}[x - (-3)] \\[1em] \Rightarrow y + \Big(\frac{8}{5}\Big) = \frac{5}{4}[x + 3] \\[1em] \Rightarrow \frac{5y + 8}{5} = \frac{5x + 15}{4} \\[1em] \Rightarrow 4(5y + 8) = 5(5x + 15) \\[1em] \Rightarrow 20y + 32 = 25x + 75 \\[1em] \Rightarrow 25x - 20y + 75 - 32 = 0 \\[1em] \Rightarrow 25x - 20y + 43 = 0\end{aligned}$$

Hence, equation of line through P and perpendicular to AB is 25x - 20y + 43 = 0.

(i) Given,

DE || BC

In △ADE and △ABC,

∠ADE = ∠ABC [Corresponding angles are equal]

∠DAE = ∠BAC [Common angle]

△ADE ~ △ABC [By AA axiom]

Given,

AE : EC = 5 : 4

AE = 5x and EC = 4x

AC = AE + EC = 5x + 4x = 9x.

We know that,

Ratio of corresponding sides of two triangles are proportional to each other.

$$\begin{aligned}\Rightarrow \frac{DE}{BC} = \frac{AE}{AC} \\[1em] \Rightarrow \frac{DE}{BC} = \frac{5x}{9x} = \frac{5}{9}.\end{aligned}$$

Hence, DE : BC = 5 : 9.

(ii) In △DOE and △BOC,

∠DOE = ∠BOC [Vertically opposite angles are equal]

∠ODE = ∠OCB [Alternate angles are equal]

△DOE ~ △BOC [By AA axiom]

Given,

DE : BC = 5 : 9

We know that,

Ratio of corresponding sides of two triangles are proportional to each other.

$$\begin{aligned}\Rightarrow \frac{DO}{OC} = \frac{DE}{BC} \\[1em] \Rightarrow \frac{DO}{OC} = \frac{5}{9}.\end{aligned}$$

As,

DO : OC = 5 : 9

Let,

DO = 5y or OC = 9y

From figure,

DC = DO + OC = 5y + 9y = 14y.

DO : DC = 5y : 14y = 5 : 14.

Hence, DO : DC = 5 : 14.

(iii) We know that,

Ratio of areas of two triangles having equal heights is equal to the ratio of the corresponding bases.

$\Rightarrow \frac{\text{Area of △DOE}}{\text{Area of △DCE}} = \frac{DO}{DC} = \frac{5}{14}$

Hence, area of △DOE : area of △ DCE = 5 : 14.

In △PAC and △PDB

∠APC = ∠DPB (Vertically opposite angles are equal)

∠CAP = ∠BDP (Angles in same segment are equal)

∴ △PAC ~ △PDB

We know that,

When two triangles are similar, the ratio of the lengths of their corresponding sides are proportional.

∴ $\frac{PA}{PD} = \frac{PC}{PB}$

⇒ PA × PB = PC × PD.

Hence, proved that PA × PB = PC × PD.

From figure,

In △PQR,

By pythagoras theorem,

⇒ PQ² = PR² + QR²

⇒ 17² = (x + 9)² + (x + 2)²

⇒ 289 = x² + 81 + 18x + x² + 4 + 4x

⇒ 289 = 2x² + 22x + 85

⇒ 2x² + 22x + 85 - 289 = 0

⇒ 2x² + 22x - 204 = 0

⇒ 2(x² + 11x - 102) = 0

⇒ x² + 11x - 102 = 0

⇒ x² + 17x - 6x - 102 = 0

⇒ x(x + 17) - 6(x + 17) = 0

⇒ (x - 6)(x + 17) = 0

⇒ x - 6 = 0 or x + 17 = 0

⇒ x = 6 or x = -17.

Since, radius cannot be negative.

⇒ x = 6 cm.

Hence, equation is x² + 11x - 102 = 0 and x = 6 cm.

(i) Steps of construction :

1. Draw a line segment AB = 6.4 cm

2. Draw perpendicular bisector of AB touching AB at point O.

3. With O as center and radius OA or OB draw a circle.

4. Draw a line LM parallel to AB at a distance of 2.5 cm.

5. With centre O and radius 3.2 + 2.5 = 5.7 cm, draw an arc intersecting the line LM at D.

6. With centre D and radius 2.5 cm, draw a circle which touches the given circle at E and AB produced at C.

Hence, above is the required circle.

Given,

External radius (R) = $\frac{21}{2}$ = 10.5 cm

Internal radius (r) = $\frac{14}{2}$ = 7 cm

By formula,

Total surface area of hollow hemisphere (T)

= Curved surface area of inner hemispherical surface + Curved surface area of outer hemispherical surface + Area of shaded annular disc.

= 2πr² + 2πR² + πR² - πr²

= 3πR² + πr²

Substituting values we get :

$$\begin{aligned}T = 3 \times \frac{22}{7} \times (10.5)^2 - \frac{22}{7} \times 7^2 \\[1em] = \frac{66}{7} \times 110.25 + 154 \\[1em] = 66 \times 15.75 + 154 \\[1em] = 1039.5 + 154 \\[1em] = 1193.5 \text{ cm}^2\end{aligned}$$

Given,

Cost of silver plating of 1 cm² of its surface is ₹ 32.

Total cost = 1193.5 × ₹ 32 = ₹ 38192

Hence, cost of silver plating the vessel = ₹ 38192.

To prove:

$\frac{\text{cosec A + 1}}{\text{cosec A - 1}}$ = (sec A + tan A)²

Solving L.H.S. of the equation :

$$\begin{aligned}\Rightarrow \frac{\frac{1}{\text{sin A}} + 1}{\frac{1}{\text{sin A}} - 1} \\[1em] \Rightarrow \frac{\frac{\text{1 + sin A}}{\text{sin A}}}{\frac{\text{1 - sin A}}{\text{sin A}}} \\[1em] \Rightarrow \frac{\text{1 + sin A}}{\text{1 - sin A}}\end{aligned}$$

Multiplying numerator and denominator by (1 + sin A), we get :

$$\begin{aligned}\Rightarrow \frac{\text{1 + sin A}}{\text{1 - sin A}} \times \frac{\text{1 + sin A}}{\text{1 + sin A}} \\[1em] \Rightarrow \frac{(\text{1 + sin A})^2}{\text{1 - sin}^2 A}\end{aligned}$$

By formula,

1 - sin² A = cos² A

$$\begin{aligned}\Rightarrow \frac{\text{(1 + sin A)}^2}{\text{cos}^2 A} \\[1em] \Rightarrow \Big(\frac{\text{1 + sin A}}{\text{cos A}}\Big)^2 \\[1em] \Rightarrow \Big(\frac{1}{\text{cos A}} + \frac{\text{sin A}}{\text{cos A}}\Big)^2 \\[1em] \Rightarrow (\text{sec A + tan A})^2.\end{aligned}$$

Since, L.H.S. = R.H.S.

Hence, proved that $\frac{\text{cosec A + 1}}{\text{cosec A - 1}}$ = (sec A + tan A)².

Solving R.H.S. of the equation :

$$\begin{aligned}\Rightarrow \frac{\text{2 cos}^2 A - 1}{\text{sin A cos A}} \\[1em] \Rightarrow \frac{\text{cos}^2 A - 1 + \text{cos}^2 A}{\text{sin A cos A}} \\[1em] \Rightarrow \frac{\text{cos}^2 A - \text{(1 - cos}^2 A)}{\text{sin A cos A}}\end{aligned}$$

By formula,

1 - cos² A = sin² A

$$\begin{aligned}\Rightarrow \frac{\text{cos}^2 A - \text{sin}^2 A}{\text{sin A cos A}} \\[1em] \Rightarrow \frac{\text{cos}^2 A}{\text{sin A cos A}} - \frac{\text{sin}^2 A}{\text{sin A cos A}} \\[1em] \Rightarrow \frac{\text{cos A}}{\text{sin A}} - \frac{\text{sin A}}{\text{cos A}} \\[1em] \Rightarrow \text{cot A - tan A}.\end{aligned}$$

Since, L.H.S. = R.H.S.

Hence, proved that cot A - tan A = $\frac{\text{2 cos}^2 A - 1}{\text{sin A cos A}}$.

To prove:

$\sqrt{\frac{\text{1 + cos A}}{\text{1 - cos A}}} + \sqrt{\frac{\text{1 - cos A}}{\text{1 + cos A}}}$ = 2 cosec A

Solving L.H.S. of the equation :

$$\begin{aligned}\Rightarrow \sqrt{\frac{\text{1 + cos A}}{\text{1 - cos A}}} + \sqrt{\frac{\text{1 - cos A}}{\text{1 + cos A}}} \\[1em] \Rightarrow \sqrt{\frac{\text{1 + cos A}}{\text{1 - cos A}} \times \frac{\text{1 + cos A}}{\text{1 + cos A}}} + \sqrt{\frac{\text{1 - cos A}}{\text{1 + cos A}} \times \frac{\text{1 - cos A}}{\text{1 - cos A}}} \\[1em] \Rightarrow \sqrt{\frac{(\text{1 + cos A})^2}{\text{1 - cos}^2 A}} + \sqrt{\frac{(\text{1 - cos A})^2}{\text{1 - cos}^2 A}} \\[1em]\end{aligned}$$

By formula,

1 - cos² A = sin² A

$$\begin{aligned}\Rightarrow \sqrt{\frac{(\text{1 + cos A})^2}{\text{sin}^2 A}} + \sqrt{\frac{(\text{1 - cos A})^2}{\text{sin}^2 A}} \\[1em] \Rightarrow \frac{\text{1 + cos A}}{\text{sin A}} + \frac{\text{1 - cos A}}{\text{sin A}} \\[1em] \Rightarrow \frac{1}{\text{sin A}} + \frac{\text{cos A}}{\text{sin A}} + \frac{1}{\text{sin A}} - \frac{\text{cos A}}{\text{sin A}} \\[1em] \Rightarrow \text{cosec A + cot A + cosec A - cot A} \\[1em] \Rightarrow \text{2 cosec A}.\end{aligned}$$

Since, L.H.S. = R.H.S.

Hence, proved that $\sqrt{\frac{\text{1 + cos A}}{\text{1 - cos A}}} + \sqrt{\frac{\text{1 - cos A}}{\text{1 + cos A}}}$ = 2 cosec A.

To prove:

$\frac{\text{cos A}}{\text{cosec A + 1}} + \frac{\text{cos A}}{\text{cosec A - 1}}$ = 2 tan A

Solving L.H.S. of the equation

$$\begin{aligned}\Rightarrow \frac{\text{cos A}}{\text{cosec A + 1}} + \frac{\text{cos A}}{\text{cosec A - 1}} \\[1em] \Rightarrow \frac{\text{cos A(cosec A - 1) + cos A(cosec A + 1)}}{\text{(cosec A + 1)(cosec A - 1)}} \\[1em] \Rightarrow \frac{\text{cos A.cosec A - cos A + cos A.cosec A+ cos A}}{\text{cosec}^2 A - 1}\end{aligned}$$

By formula,

cosec² A - 1 = cot² A

$$\begin{aligned}\Rightarrow \frac{\text{2 cos A cosec A}}{\text{cot}^2 A} \\[1em] \Rightarrow \frac{\text{2 cos A} \times \frac{1}{\text{sin A}}}{\text{cot}^2 A} \\[1em] \Rightarrow \frac{\text{2 cot A}}{\text{cot}^2 A} \\[1em] \Rightarrow \frac{2}{\text{cot A}} \\[1em] \Rightarrow \text{2 tan A}.\end{aligned}$$

Since, L.H.S. = R.H.S.

Hence, proved that $\frac{\text{cos A}}{\text{cosec A + 1}} + \frac{\text{cos A}}{\text{cosec A - 1}}$ = 2 tan A.

(i) Given,

⇒ 4 cos² 2x - 3 = 0

⇒ 4 cos² 2x = 3

⇒ cos² 2x = $\frac{3}{4}$

⇒ cos 2x = $\sqrt{\frac{3}{4}}$

⇒ cos 2x = $\frac{\sqrt{3}}{2}$

⇒ cos 2x = cos 30°

⇒ 2x = 30°

⇒ x = $\frac{30°}{2}$ = 15°.

Hence, x = 15°.

(ii) Given,

⇒ 2 sin² x - sin x = 0

⇒ sin x(2 sin x - 1) = 0

⇒ sin x = 0 or 2 sin x - 1 = 0

⇒ sin x = 0 or 2 sin x = 1

⇒ sin x = 0 or sin x = $\frac{1}{2}$

⇒ sin x = sin 0° or sin x = sin 30°

⇒ x = 0° or x = 30°.

Hence, x = 0° or 30°.

In the figure AB and CD represent the same ladder. So, their length must be equal.

Let length of the ladder be h.

∴ AB = CD = h

In △AEB :

sin α = $\frac{AE}{AB}$

AE = AB sin α = h sin α.

In △AEB :

cos α = $\frac{BE}{AB}$

BE = AB cos α = h cos α.

In △DEC :

sin β = $\frac{DE}{CD}$

DE = CD sin β = h sin β

cos β = $\frac{CE}{CD}$

CE = CD cos β = h cos β

From figure,

$$\begin{aligned}\Rightarrow \frac{a}{b} = \frac{BC}{AD} \\[1em] \Rightarrow \frac{a}{b} = \frac{CE - BE}{AE - DE} \\[1em] \Rightarrow \frac{a}{b} = \frac{\text{h cos β} - \text{h cos α}}{\text{h sin α} - \text{h sin β}} \\[1em] \Rightarrow \frac{a}{b} = \frac{\text{cos β} - \text{cos α}}{\text{sin α} - \text{sin β}}\end{aligned}$$

Hence, proved that $\frac{a}{b} = \frac{\text{cos β} - \text{cos α}}{\text{sin α} - \text{sin β}}$.

Cumulative frequency distribution table is as follows :

C.I.	f	Cumulative frequency
450 - 550	40	40
550 - 650	68	108 (40 + 68)
650 - 750	86	194 (108 + 86)
750 - 850	120	314 (194 + 120)
850 - 950	90	404 (314 + 90)
950 - 1050	40	444 (404 + 40)
1050 - 1150	6	450 (444 + 6)

Here no. of terms = 450, which is even.

By formula,

Median = $\frac{n}{2}$ th term = 225.

Steps of construction of ogive :

1. Take 2 cm = 100 units (C.I.) on x-axis.

2. Take 1 cm = 50 units (frequency) on y-axis.

3. A kink is shown near x-axis as it starts from 450. Plot the point (450, 0) as ogive starts on x-axis representing lower limit of first class.

4. Plot the points (550, 40), (650, 108), (750, 194), (850, 314), (950, 404), (1050, 444) and (1150, 450).

5. Join the points by a free-hand curve.

6. Draw a line parallel to x-axis from point I (frequency) = 225, touching the graph at point J. From point J draw a line parallel to y-axis touching x-axis at point K.

From graph, K = 780

Hence, median = 780.

Steps :

1. Draw a histogram of the given distribution.

2. Inside the highest rectangle, which represents the maximum frequency (or modal class), draw two lines AC and BD diagonally from the upper corners A and B of adjacent rectangles.

3. Through point K (the point of intersection of diagonals AC and BD), draw KL perpendicular to the horizontal axis.

4. The value of point L on the horizontal axis represents the value of mode.

∴ Mode = 803.

Hence, mode = 803.

Total cards = 52

∴ No. of possible outcomes = 52

There are 4 ace and 4 jacks in a pack.

∴ No. of favourable outcomes = 8

P(drawing an ace or jack) = $\frac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \frac{8}{52} = \frac{2}{13}$.

Hence, probability of drawing an ace or a jack = $\frac{2}{13}$.

Let no. of blue balls be x.

Total balls = (x + 5).

∴ No. of possible outcomes = x + 5

There are 5 red balls

∴ No. of favourable outcomes for drawing a red ball = 5

P(drawing a red ball) = $\frac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \frac{5}{5 + x}$.

There are x blue balls

∴ No. of favourable outcomes for drawing a blue ball = x

P(drawing a blue ball) = $\frac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \frac{x}{5 + x}$.

Given,

Probability of drawing a blue ball is double that of red ball.

$$\begin{aligned}\therefore \frac{x}{5 + x} = 2 \times \frac{5}{5 + x} \\[1em] \Rightarrow \frac{x}{5 + x} = \frac{10}{5 + x} \\[1em] \Rightarrow x = 10.\end{aligned}$$

Hence, no. of blue balls = 10.

In a single throw of two dice.

∴ No. of possible outcomes = 6 × 6 = 36.

(i) Favourable outcomes for getting a total of 9 = {(3, 6), (4, 5), (5, 4), (6, 3)}.

∴ No. of favourable outcomes = 4.

P(of getting a total of 9) = $\frac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \frac{4}{36} = \frac{1}{9}$.

Hence, probability of getting a total of 9 = $\frac{1}{9}$.

(ii) Favourable outcomes for getting two aces = {(1, 1)}.

∴ No. of favourable outcomes = 1.

P(of getting two aces) = $\frac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \frac{1}{36}$.

Hence, probability of getting two aces = $\frac{1}{36}$.

(iii) Favourable outcomes for getting at least one ace = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (3, 1), (4, 1), (5, 1), (6, 1)}.

∴ No. of favourable outcomes = 11.

P(of getting at least one ace) = $\frac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \frac{11}{36}$.

Hence, probability of getting at least one ace = $\frac{11}{36}$.

(iv) Favourable outcomes for getting a doublet = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}.

∴ No. of favourable outcomes = 6.

P(of getting a doublet) = $\frac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \frac{6}{36} = \frac{1}{6}$.

Hence, probability of getting a doublet = $\frac{1}{6}$.

(v) Favourable outcomes for getting 5 on one die and 6 on other = {(5, 6), (6, 5)}.

∴ No. of favourable outcomes = 2.

P(of getting 5 on one die and 6 on other)

= $\frac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \frac{2}{36} = \frac{1}{18}$.

Hence, probability of getting 5 on one die and 6 on other = $\frac{1}{18}$.

(vi) Favourable outcomes for getting a multiple of 2 on one die and a multiple of 3 on other = {(2, 3), (2, 6), (3, 2), (3, 4), (3, 6), (4, 3), (4, 6), (6, 2), (6, 3), (6, 4), (6, 6)}.

∴ No. of favourable outcomes = 11.

P(of getting a multiple of 2 on one die and a multiple of 3 on other)

= $\frac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \frac{11}{36}$.

Hence, probability of getting a multiple of 2 on one die and a multiple of 3 on other = $\frac{11}{36}$.

Give,

M.R.P. = ₹ 30,000

Discount = 25%

Amount after deducting discount = M.R.P. - Discount

⇒ ₹ 30000 - ₹ $\frac{25}{100} \times 30000$

⇒ ₹ 30000 - ₹ 7500

⇒ ₹ 22500

Amount (including G.S.T.) = Amount after deducting discount + G.S.T.

= ₹ 22500 + ₹ $\frac{12}{100} \times 22500$

= ₹ 22500 + ₹ 12 × 225

= ₹ 22500 + ₹ 2700

= ₹ 25200.

Hence, amount of bill = ₹ 25200.

Given,

Time (n) = 1 year 6 months = 12 + 6 = 18 months.

Monthly installment (P) = ₹ 2400.

Maturity value = ₹ 47304.

Let rate of interest be r%.

By formula,

Maturity value = P × n + P × $\frac{n(n + 1)}{2 \times 12} \times \frac{r}{100}$

Substituting values we get :

$$\begin{aligned}\Rightarrow 47304 = 2400 \times 18 + 2400 \times \frac{18 \times 19}{2 \times 12} \times \frac{r}{100} \\[1em] \Rightarrow 47304 = 43200 + 2400 \times \frac{3 \times 19}{2 \times 2} \times \frac{r}{100} \\[1em] \Rightarrow 47304 - 43200 = 2400 \times \frac{57}{4} \times \frac{r}{100} \\[1em] \Rightarrow 4104 = 6 \times 57r \\[1em] \Rightarrow r = \frac{4104}{6 \times 57} \\[1em] \Rightarrow r = \frac{4104}{342} = 12 %.\end{aligned}$$

Hence, rate of interest = 12%.

Let monthly installment be ₹ P.

Given,

Time (n) = 2 years = 24 months.

Rate (r) = 10%.

Maturity value = ₹ 42400.

By formula,

Maturity value = P × n + P × $\frac{n(n + 1)}{2 \times 12} \times \frac{r}{100}$

Substituting values we get :

$$\begin{aligned}\Rightarrow 42400 = P \times 24 + P \times \frac{24 \times 25}{2 \times 12} \times \frac{10}{100} \\[1em] \Rightarrow 42400 = 24P + P \times 25 \times \frac{1}{10} \\[1em] \Rightarrow 42400 = 24P + 2.5P \\[1em] \Rightarrow 26.5P = 42400 \\[1em] \Rightarrow P = \frac{42400}{26.5} \\[1em] \Rightarrow P = 1600.\end{aligned}$$

Hence, monthly installment = ₹ 1600.

For company A,

Let invested money be ₹ x

Nominal value of each share (N.V.) = ₹ 100

Discount = 20%

M.V. of share = F.V. - Discount

= ₹ 100 - $\frac{20}{100} \times 100$

= ₹ 100 - ₹ 20 = ₹ 80.

No. of shares = $\frac{\text{Invested money}}{\text{M.V.}} = \frac{x}{80}$

Rate of dividend = 15%

Total dividend for company A = No. of shares × Dividend × 100

= $\frac{x}{80} \times \frac{15}{100} \times 100$

= $\frac{3x}{16}$

For company B,

Let invested money be ₹ x

Nominal value of each share (N.V.) = ₹ 25

Premium = 20%

M.V. of share = N.V. + Premium

= ₹ 25 + $\frac{20}{100} \times 25$

= ₹ 25 + ₹ 5 = ₹ 30.

No. of shares = $\frac{\text{Invested money}}{\text{M.V.}} = \frac{x}{30}$

Let rate of dividend = d%

Total dividend for company B = No. of shares × Dividend × 100

= $\frac{x}{30} \times \frac{d}{100} \times 25$

= $\frac{xd}{120}$

Since, man gets equal dividend from both companies so,

$$\begin{aligned}\Rightarrow \frac{3x}{16} = \frac{xd}{120} \\[1em] \Rightarrow d = \frac{3x \times 120}{x \times 16} \\[1em] \Rightarrow d = \frac{360}{16} = 22.5 %.\end{aligned}$$

Hence, dividend paid by company B = 22.5%.

For first company,

Let money invested be ₹ x.

N.V. = ₹ 100

Premium = 40%

M.V. = N.V. + Premium

= 100 + $\frac{40}{100} \times 100$

= ₹ 100 + ₹ 40

= ₹ 140.

No. of shares = $\frac{\text{Invested money}}{\text{M.V.}} = \frac{x}{140}$

Rate of dividend = 6%

Total dividend for first company = No. of shares × Dividend × 100

= $\frac{x}{140} \times \frac{6}{100} \times 100$

= $\frac{3x}{70}$

For second company,

Let money invested be ₹ (54000 - x).

N.V. = ₹ 100

Premium = 25%

M.V. = N.V. + Premium

= 100 + $\frac{25}{100} \times 100$

= ₹ 100 + ₹ 25

= ₹ 125.

No. of shares = $\frac{\text{Invested money}}{\text{M.V.}} = \frac{54000 - x}{125}$

Rate of dividend = 5%

Total dividend for second company = No. of shares × Dividend × 100

= $\frac{54000 - x}{125} \times \frac{5}{100} \times 100$

= $\frac{54000 - x}{25}$

Since, total income = ₹ 2240

$$\begin{aligned}\therefore \frac{54000 - x}{25} + \frac{3x}{70} = 2240 \\[1em] \Rightarrow \frac{14(54000 - x) + 3x \times 5}{350} = 2240 \\[1em] \Rightarrow \frac{756000 - 14x + 15x}{350} = 2240 \\[1em] \Rightarrow 756000 + x = 2240 \times 350 \\[1em] \Rightarrow 756000 + x = 784000 \\[1em] \Rightarrow x = 784000 - 756000 = 28000.\end{aligned}$$

₹ (54000 - x) = ₹ 54000 - ₹ 28000 = ₹ 26000.

Hence, money invested in second company = ₹ 26000.

(i) Given,

⇒ 2x - 9 < 7

⇒ 2x < 7 + 9

⇒ 2x < 16

⇒ x < $\frac{16}{2}$

⇒ x < 8 ...........(1)

Also,

⇒ 3x + 9 ≤ 25

⇒ 3x ≤ 25 - 9

⇒ 3x ≤ 16

⇒ x ≤ $\frac{16}{3}$

⇒ x ≤ $5\frac{1}{3}$..........(2)

From (1) and (2), we get :

⇒ x ≤ $5\frac{1}{3}$

Hence, solution = {x : x ≤ $5\frac{1}{3}$, x ∈ R}.

(ii) Given,

⇒ 3x - 2 > 19

⇒ 3x > 19 + 2

⇒ 3x > 21

⇒ x > $\frac{21}{3}$

⇒ x > 7 ........(1)

Also,

⇒ 3 - 2x ≥ 7

⇒ 2x ≤ 3 - 7

⇒ 2x ≤ -4

⇒ x ≤ $-\frac{4}{2}$

⇒ x ≤ -2 ........(2)

From (1) and (2), we get :

x > 7 or x ≤ -2

Hence, solution = {x : x > 7 or x ≤ -2, x ∈ R}.

Given,

x² + x - (a + 1)(a + 2) = 0

Comparing above equation with ax² + bx + c = 0, we get :

a = 1, b = 1 and c = -(a + 1)(a + 2)

By formula,

x = $\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Substituting values we get :

$$\begin{aligned}\Rightarrow x = \frac{-1 \pm \sqrt{1^2 - 4 \times 1 \times -(a + 1)(a + 2)}}{2 \times 1} \\[1em] \Rightarrow x = \frac{-1 \pm \sqrt{1 + 4(a + 1)(a + 2)}}{2} \\[1em] \Rightarrow x = \frac{-1 \pm \sqrt{1 + 4(a^2 + 2a + a + 2)}}{2} \\[1em] \Rightarrow x = \frac{-1 \pm \sqrt{1 + 4(a^2 + 3a + 2)}}{2} \\[1em] \Rightarrow x = \frac{-1 \pm \sqrt{1 + 4a^2 + 12a + 8}}{2} \\[1em] \Rightarrow x = \frac{-1 \pm \sqrt{4a^2 + 12a + 9}}{2} \\[1em] \Rightarrow x = \frac{-1 \pm \sqrt{(2a + 3)^2}}{2} \\[1em] \Rightarrow x = \frac{-1 \pm (2a + 3)}{2} \\[1em] \Rightarrow x = \frac{-1 + (2a + 3)}{2}, \frac{-1 - (2a + 3)}{2} \\[1em] \Rightarrow x = \frac{2a + 3 - 1}{2}, \frac{-1 - 2a - 3}{2} \\[1em] \Rightarrow x = \frac{2a + 2}{2}, \frac{-2a - 4}{2} \\[1em] \Rightarrow x = \frac{2(a + 1)}{2}, \frac{-2(a + 2)}{2} \\[1em] \Rightarrow x = (a + 1), -(a + 2).\end{aligned}$$

Hence, x = (a + 1) or -(a + 2).

Let cost price be ₹ x.

S.P. = ₹ 96

Profit = S.P. - C.P. = ₹ (96 - x)

Given,

Man gains as much percent as its cost price.

Profit percent = x %.

Substituting values we get :

$$\begin{aligned}\Rightarrow \frac{\text{Profit}}{\text{C.P.}} \times 100 = x \\[1em] \Rightarrow \frac{96 - x}{x} \times 100 = x \\[1em] \Rightarrow 100(96 - x) = x^2 \\[1em] \Rightarrow x^2 = 9600 - 100x \\[1em] \Rightarrow x^2 + 100x - 9600 = 0 \\[1em] \Rightarrow x^2 + 160x - 60x - 9600 = 0 \\[1em] \Rightarrow x(x + 160) - 60(x + 160) = 0 \\[1em] \Rightarrow (x - 60)(x + 160) = 0 \\[1em] \Rightarrow x - 60 = 0 \text{ or } x + 160 = 0 \\[1em] \Rightarrow x = 60 \text{ or } x = -160.\end{aligned}$$

Since, cost price cannot be negative.

∴ x = 60.

Hence, C.P. = ₹ 60.

Let no. of articles brought be x.

Cost of each article = ₹ $\frac{900}{x}$

5 articles were damaged.

No. of articles left = (x - 5).

Given,

No. of articles left were sold at ₹ 2 more than what he paid for it.

S.P. of each article = ₹ $\frac{900}{x} + 2$

Given,

⇒ Gain = ₹ 80.

⇒ Total S.P. - Total C.P. = ₹ 80

$$\begin{aligned}\Rightarrow (x - 5)\Big(\frac{900}{x} + 2\Big) - 900 = 80 \\[1em] \Rightarrow x \times \frac{900}{x} + 2x - 5 \times \frac{900}{x} - 5 \times 2 = 80 + 900 \\[1em] \Rightarrow 900 + 2x - \frac{4500}{x} - 10 = 980 \\[1em] \Rightarrow \frac{900x + 2x^2 - 4500 - 10x}{x} = 980 \\[1em] \Rightarrow 900x + 2x^2 - 4500 - 10x = 980x \\[1em] \Rightarrow 2x^2 + 900x - 980x - 10x - 4500 = 0 \\[1em] \Rightarrow 2x^2 - 90x - 4500 = 0 \\[1em] \Rightarrow 2(x^2 - 45x - 2250) = 0 \\[1em] \Rightarrow x^2 - 45x - 2250 = 0 \\[1em] \Rightarrow x^2 - 75x + 30x - 2250 = 0 \\[1em] \Rightarrow x(x - 75) + 30(x - 75) = 0 \\[1em] \Rightarrow (x + 30)(x - 75) = 0 \\[1em] \Rightarrow x = -30 \text{ or } x = 75.\end{aligned}$$

Since, no. of articles cannot be negative.

Hence, no. of articles = 75.

Given,

Number of remaining boxes in the three trucks are in the ratio 4 : 6 : 5.

So, let remaining boxes be

In first truck = 4x, second truck = 6x and third truck = 5x

Total boxes including rotten boxes in each truck are

In first truck = 4x + 7, second truck = 6x + 12 and third truck = 5x + 8

Total boxes = 1077

⇒ 4x + 7 + 6x + 12 + 5x + 8 = 1077

⇒ 15x + 27 = 1077

⇒ 15x = 1077 - 27

⇒ 15x = 1050

⇒ x = $\frac{1050}{15}$ = 70.

So, no. of boxes :

In first truck = 4x + 7 = 4(70) + 7 = 280 + 7 = 287,

second truck = 6x + 12 = 6(70) + 12 = 420 + 12 = 432,

third truck = 5x + 8 = 5(70) + 8 = 358.

Hence, no. of boxes originally in each truck were 287, 432 and 358.

Given,

a : b is the duplicate ratio of (a + c) and (b + c).

$$\begin{aligned}\therefore \frac{a}{b} = \frac{a + c}{b + c} \times \frac{a + c}{b + c} \\[1em] \Rightarrow \frac{a}{b} = \frac{a^2 + ac + ac + c^2}{b^2 + bc + bc + c^2} \\[1em] \Rightarrow \frac{a}{b} = \frac{a^2 + 2ac + c^2}{b^2 + 2bc + c^2} \\[1em] \Rightarrow a(b^2 + 2bc + c^2) = b(a^2 + 2ac + c^2) \\[1em] \Rightarrow ab^2 + 2abc + ac^2 = ba^2 + 2abc + bc^2 \\[1em] \Rightarrow ac^2 - bc^2 = ba^2 - ab^2 + 2abc - 2abc \\[1em] \Rightarrow c^2(a - b) = ab(a - b) \\[1em] \Rightarrow c^2 = ab.\end{aligned}$$

Since, c² = ab.

Hence, proved that a, c and b are in continued proportion.

Given,

$$\begin{aligned}16\Big(\frac{a - x}{a + x}\Big)^3 = \Big(\frac{a + x}{a - x}\Big) \\[1em] \Rightarrow 16\Big(\frac{a - x}{a + x}\Big)^3 = \frac{1}{\frac{a - x}{a + x}}\end{aligned}$$

Let $\Big(\frac{a - x}{a + x}\Big)$ = t.

Substituting $\Big(\frac{a - x}{a + x}\Big)$ = t in given equation, we get :

$$\begin{aligned}\Rightarrow 16t^3 = \frac{1}{t} \\[1em] \Rightarrow 16t^4 = 1 \\[1em] \Rightarrow t^4 = \frac{1}{16} \\[1em] \Rightarrow t^4 = \Big(\frac{1}{2}\Big)^4 \\[1em] \Rightarrow t = \frac{1}{2} \\[1em] \Rightarrow \frac{a - x}{a + x} = \frac{1}{2} \\[1em] \Rightarrow 2(a - x) = a + x \\[1em] \Rightarrow 2a - 2x = a + x \\[1em] \Rightarrow 2a - a = x + 2x \\[1em] \Rightarrow a = 3x.\end{aligned}$$

Hence, proved that a = 3x.

Applying componendo and dividendo, we get :

$$\begin{aligned}\Rightarrow \frac{1 + x + x^2 + 1 - x + x^2}{1 + x + x^2 - (1 - x + x^2)} = \frac{62(1 + x) + 63(1 - x)}{62(1 + x) - 63(1 - x)} \\[1em] \Rightarrow \frac{2 + 2x^2}{1 - 1 + x + x + x^2- x^2} = \frac{62 + 62x + 63 - 63x}{62 + 62x - 63 + 63x} \\[1em] \Rightarrow \frac{2(1 + x^2)}{2x} = \frac{125 - x}{125x - 1} \\[1em] \Rightarrow \frac{1 + x^2}{x} = \frac{125 - x}{125x - 1} \\[1em] \Rightarrow (125x - 1)(1 + x^2) = x(125 - x)\\[1em] \Rightarrow 125x + 125x^3 - 1 - x^2 = 125x - x^2 \\[1em] \Rightarrow 125x - 125x - x^2 + x^2 + 125x^3 - 1 = 0 \\[1em] \Rightarrow 125x^3 - 1 = 0 \\[1em] \Rightarrow 125x^3 = 1 \\[1em] \Rightarrow x^3 = \frac{1}{125} \\[1em] \Rightarrow x^3 = \Big(\frac{1}{5}\Big)^3 \\[1em] \Rightarrow x = \frac{1}{5}.\end{aligned}$$

Hence, x = $\frac{1}{5}$.

Given,

⇒ 2x + 7 = 0

⇒ 2x = -7

⇒ x = -$\frac{7}{2}$

Substituting x = -$\frac{7}{2}$ in L.H.S. of 2x³ + 7x² - 4x - 14 = 0, we get :

$$\begin{aligned}\phantom{=} 2 \times \Big(-\frac{7}{2}\Big)^3 + 7 \times \Big(-\frac{7}{2}\Big)^2 - 4 \times -\frac{7}{2} - 14 \\[1em] = 2 \times -\frac{343}{8} + 7 \times \frac{49}{4} + 14 - 14 \\[1em] = -\frac{343}{4} + \frac{343}{4} + 14 - 14 \\[1em] = 0.\end{aligned}$$

Since, L.H.S. = R.H.S. = 0.

∴ 2x + 7 is a factor of 2x³ + 7x² - 4x - 14 = 0.

Dividing 2x³ + 7x² - 4x - 14 = 0 by 2x + 7:

$$\begin{array}{l} \phantom{2x + 7)}{x^2 - 2} \\ 2x + 7\overline{\smash{\big)}2x^3 + 7x^2 - 4x - 14} \\ \phantom{2x + 7}\underline{\underset{-}{}2x^3 \underset{-}{+}7x^2} \\ \phantom{{2x + 7}2x^3} \times \phantom{+25^2} - 4x - 14 \\ \phantom{{2x + 7}x^3+2+25^2}\underline{\underset{+}{-}4x \underset{+}{-} 14} \\ \phantom{{2x + 7}x^3+2-5x^23-3}\times \end{array}$$

∴ 2x³ + 7x² - 4x - 14 = (2x + 7)(x² - 2)

= (2x + 7)(x + $\sqrt{2})(x - \sqrt{2}$)

Solving,

⇒ 2x + 7 = 0

⇒ 2x = -7

⇒ x = -$\frac{7}{2}$

Also,

⇒ x + $\sqrt{2}$ = 0

⇒ x = -$\sqrt{2}$ = -1.41

Also,

⇒ x - $\sqrt{2}$ = 0

⇒ x = $\sqrt{2}$ = 1.41

Hence, x = -$\frac{7}{2}$, 1.41, -1.41

Let no. to be subtracted be a.

So, resulting polynomial is 2x³ - 5x² + 5x + 8 - a.

By factor theorem,

A polynomial f(x) has a factor (x - a), if and only if, f(a) = 0.

Given,

Factor : 2x - 3 = 0

x = $\frac{3}{2}$

Substituting value of x = $\frac{3}{2}$ in 2x³ - 5x² + 5x + 8 - a, will give remainder = 0.

$$\begin{aligned}\Rightarrow 2 \times \Big(\frac{3}{2}\Big)^3 - 5 \times \Big(\frac{3}{2}\Big)^2 + 5 \times \frac{3}{2} + 8 - a = 0 \\[1em] \Rightarrow 2 \times \frac{27}{8} - 5 \times \frac{9}{4} + \frac{15}{2} + 8 - a = 0 \\[1em] \Rightarrow \frac{27}{4} - \frac{45}{4} + \frac{15 + 16}{2} - a = 0 \\[1em] \Rightarrow -\frac{18}{4} + \frac{31}{2} - a = 0 \\[1em] \Rightarrow \frac{-18 + 62}{4} - a = 0 \\[1em] \Rightarrow \frac{44}{4} - a = 0 \\[1em] \Rightarrow 11 - a = 0 \\[1em] \Rightarrow a = 11.\end{aligned}$$

Hence, number to be subtracted is 11.

By remainder theorem,

When a polynomial p(x) is divided by a linear polynomial (x - a), then the remainder is equal to p(a).

⇒ x + 1 = 0

⇒ x = -1

Given,

4x³ - bx² + x - c leaves remainder 0 on dividing it by (x + 1).

∴ 4(-1)³ - b(-1)² + (-1) - c = 0

⇒ 4(-1) - b(1) - 1 - c = 0

⇒ -4 - b - 1 - c = 0

⇒ b + c = -5

⇒ b = -5 - c .........(1)

Given,

4x³ - bx² + x - c leaves remainder 30 on dividing it by (2x - 3).

2x - 3 = 0

⇒ 2x = 3

⇒ x = $\frac{3}{2}$

Substituting x = $\frac{3}{2}$ in 4x³ - bx² + x - c should give result as 30,

$$\begin{aligned}\Rightarrow 4 \times \Big(\frac{3}{2}\Big)^3 - b \times \Big(\frac{3}{2}\Big)^2 + \Big(\frac{3}{2}\Big) - c = 30 \\[1em] \Rightarrow 4 \times \Big(\frac{27}{8}\Big) - b \times \Big(\frac{9}{4}\Big) + \Big(\frac{3}{2}\Big) - c = 30 \\[1em] \Rightarrow \frac{27}{2} - \frac{9b}{4} + \frac{3}{2} - c = 30 \\[1em] \Rightarrow \frac{54 - 9b + 6 - 4c}{4} = 30 \\[1em] \Rightarrow 60 - 9b - 4c = 120\end{aligned}$$

Substituting value of b from (1) in above equation :

⇒ 60 - 9(-5 - c) - 4c = 120

⇒ 60 + 45 + 9c - 4c = 120

⇒ 5c + 105 = 120

⇒ 5c = 120 - 105

⇒ 5c = 15

⇒ c = $\frac{15}{5}$ = 3.

Substituting value of c in (1), we get :

⇒ b = -5 - c = -5 - 3 = -8.

Hence, b = -8 and c = 3.

Let order of matrix M be a × b.

$$M_{a \times b} \times \begin{bmatrix*}[r] 3 & 2 \\ 2 & -1 \end{bmatrix*}_{2 \times 2} = \begin{bmatrix*}[r] -1 & 4 \end{bmatrix*}_{1 \times 2}$$

Since, the product of matrices is possible, only when the number of columns in the first matrix is equal to the number of rows in the second.

∴ b = 2

Also, the no. of rows of product (resulting) matrix is equal to no. of rows of first matrix.

∴ a = 1

Order of matrix M = a × b = 1 × 2.

Let M = $$\begin{bmatrix*}[r] x & y \end{bmatrix*}$$.

$$\Rightarrow \begin{bmatrix*}[r] x & y \end{bmatrix*} \times \begin{bmatrix*}[r] 3 & 2 \\ 2 & -1 \end{bmatrix*} =\begin{bmatrix*}[r] -1 & 4 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] x \times 3 + y \times 2 & x \times 2 + y \times -1 \end{bmatrix*} = \begin{bmatrix*}[r] -1 & 4 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 3x + 2y & 2x - y \end{bmatrix*} = \begin{bmatrix*}[r] -1 & 4 \end{bmatrix*}$$

By definition of equality of matrices we get,

⇒ 3x + 2y = -1 and 2x - y = 4

From 2x - y = 4

⇒ y = 2x - 4

Substituting value of y in 3x + 2y = -1

⇒ 3x + 2(2x - 4) = -1

⇒ 3x + 4x - 8 = -1

⇒ 7x = -1 + 8

⇒ 7x = 7

⇒ x = 1.

⇒ y = 2x - 4 = 2(1) - 4 = 2 - 4 = -2.

∴ M = $$\begin{bmatrix*}[r] x & y \end{bmatrix*} = \begin{bmatrix*}[r] 1 & -2 \end{bmatrix*}$$.

Hence, M = $$\begin{bmatrix*}[r] 1 & -2 \end{bmatrix*}.$$

Let first term of A.P. be a and common difference be d.

By formula,

Sum of first n terms of an A.P. = $\frac{n}{2}[2a + (n - 1)d]$

Given,

S₂₀ = S₂₈

$$\begin{aligned}\therefore \frac{20}{2}[2a + (20 - 1)d] = \frac{28}{2}[2a + (28 - 1)d] \\[1em] \Rightarrow 10[2a + 19d] = 14[2a + 27d] \\[1em] \Rightarrow 20a + 190d = 28a + 378d \\[1em] \Rightarrow 28a - 20a = 190d - 378d \\[1em] \Rightarrow 8a = -188d \\[1em] \Rightarrow a = -\frac{188}{8}d \text{ .........(1)}\end{aligned}$$

Sum of 48 terms is

$$\begin{aligned}S_{48} = \frac{48}{2}[2a + (48 - 1)d] \\[1em] = 24 \times [2 \times -\frac{188}{8}d + 47d] \\[1em] = 24 \times \Big[-\frac{376}{8}d + 47d\Big] \\[1em] = 24 \times \Big[-47d + 47d\Big] \\[1em] = 24 \times 0 \\[1em] = 0.\end{aligned}$$

Hence, S₄₈ = 0.

Given,

a, b and c are in A.P.

∴ b - a = c - b

⇒ b + b = a + c

⇒ 2b = a + c

If (b + c), (c + a) and (a + b) are in A.P., then

⇒ (a + b) - (c + a) = (c + a) - (b + c)

⇒ a + b - c - a = c + a - b - c

⇒ b - c = a - b

⇒ b + b = a + c

⇒ 2b = a + c

Since, 2b = a + c is proved above.

Hence, proved that (b + c), (c + a) and (a + b) are also in A.P.

Given,

a, x, b are in A.P.

∴ b - x = x - a

⇒ 2x = a + b

⇒ x = $\frac{a + b}{2}$ .........(1)

Given,

b, y, c are in A.P.

∴ c - y = y - b

⇒ 2y = b + c

⇒ y = $\frac{b + c}{2}$ ...........(2)

Given,

a, b and c are in G.P.

⇒ $\frac{b}{a} = \frac{c}{b}$

⇒ b² = ac ..........(3)

To prove :

$\frac{1}{x} + \frac{1}{y} = \frac{2}{b}$

Substituting value of x and y from equation (1) and (2), in L.H.S. of above equation :

$$\begin{aligned}\Rightarrow \frac{1}{\frac{a + b}{2}} + \frac{1}{\frac{b + c}{2}} \\[1em] \Rightarrow \frac{2}{a + b} + \frac{2}{b + c} \\[1em] \Rightarrow \frac{2(b + c) + 2(a + b)}{(a + b)(b + c)} \\[1em] \Rightarrow \frac{2b + 2c + 2a + 2b}{ab + ac + b^2 + bc} \\[1em] \Rightarrow \frac{2a + 2c + 4b}{ab + b^2 + b^2 + bc} \space [\text{From} (3)] \\[1em] \Rightarrow \frac{2(a + c + 2b)}{ab + 2b^2 + bc} \\[1em] \Rightarrow \frac{2(a + c + 2b)}{b(a + 2b + c)} \\[1em] \Rightarrow \frac{2}{b}.\end{aligned}$$

Hence, proved that $\frac{1}{x} + \frac{1}{y} = \frac{2}{b}$.

Given,

⇒ 9 + 99 + 999 + ........ upto n terms.

⇒ (10 - 1) + (100 - 1) + (1000 - 1) + ......... upto n terms

⇒ (10 + 100 + 1000 + ....... upto n terms) + (-1 + -1 + -1 + ......... upto n terms)

⇒ (10 + 10² + 10³ + ....... upto n terms) + (-n)

(10 + 10² + 10³ + ....... upto n terms) is a G.P. with first term (a) = 10 and common ratio (r) = 10.

By formula,

Sum of G.P. = $\frac{a(r^n - 1)}{r - 1}$

$$\begin{aligned}\Rightarrow \frac{10(10^n - 1)}{10 - 1} + (-n) \\[1em] \Rightarrow \frac{10(10^n - 1)}{9} - n.\end{aligned}$$

Hence, 9 + 99 + 999 + ........ upto n terms = $\frac{10(10^n - 1)}{9} - n.$

We know that,

x-coordinate of the point on the y-axis is 0.

Let point on y-axis be (0, y).

By distance formula,

D = $\sqrt{(y_2 - y_1)^2 + (x_2 - x_1)^2}$

Given,

Distances of (0, y) from the points (3, 2) and (-1, 1.5) are in the ratio 2 : 1.

$$\begin{aligned}\Rightarrow \frac{\sqrt{(2 - y)^2 + (3 - 0)^2}}{\sqrt{(1.5 - y)^2 + (-1 - 0)^2}} = \frac{2}{1} \\[1em] \Rightarrow \frac{\sqrt{4 + y^2 - 4y + 9}}{\sqrt{2.25 + y^2 - 3y + 1}} = \frac{2}{1} \\[1em] \Rightarrow \frac{\sqrt{y^2 - 4y + 13}}{\sqrt{y^2 - 3y + 3.25}} = \frac{2}{1}\end{aligned}$$

Squaring both sides we get,

$$\begin{aligned}\Rightarrow \frac{y^2 - 4y + 13}{y^2 - 3y + 3.25} = \frac{4}{1} \\[1em] \Rightarrow y^2 - 4y + 13 = 4(y^2 - 3y + 3.25) \\[1em] \Rightarrow y^2 - 4y + 13 = 4y^2 - 12y + 13 \\[1em] \Rightarrow 4y^2 - y^2 - 12y + 4y + 13 - 13 = 0 \\[1em] \Rightarrow 3y^2 - 8y = 0 \\[1em] \Rightarrow y(3y - 8) = 0 \\[1em] \Rightarrow y = 0 \text{ or } 3y - 8 = 0 \\[1em] \Rightarrow y = 0 \text{ or } 3y = 8 \\[1em] \Rightarrow y = 0 \text{ or } y = \frac{8}{3} = 2\frac{2}{3}. \\[1em]\end{aligned}$$

Since, point lies on y-axis.

So, y ≠ 0

Point = (0, y) = $\Big(0, 2\frac{2}{3}\Big)$.

Hence, required point = $\Big(0, 2\frac{2}{3}\Big)$.

Let ratio be k : 1.

By section formula,

(x, y) = $\Big(\frac{m_1x_2 + m_2x_1}{m_1 + m_2}, \frac{m_1y_2 + m_2y_1}{m_1 + m_2}\Big)$

Substituting values for y-coordinate we get :

$$\begin{aligned}\Rightarrow 2 = \frac{k \times 4 + 1 \times -3}{k + 1} \\[1em] \Rightarrow 2(k + 1) = 4k - 3 \\[1em] \Rightarrow 2k + 2 = 4k - 3 \\[1em] \Rightarrow 4k - 2k = 2 + 3 \\[1em] \Rightarrow 2k = 5 \\[1em] \Rightarrow k = \frac{5}{2}\end{aligned}$$

k : 1 = $\frac{5}{2} : 1$ = 5 : 2.

So, m₁ : m₂ = 5 : 2.

Substituting values for x-coordinate we get :

$$\begin{aligned}\Rightarrow a = \frac{5 \times -9 + 2 \times 5}{5 + 2} \\[1em] \Rightarrow a = \frac{-45 + 10}{7} \\[1em] \Rightarrow a = \frac{-35}{7} = -5.\end{aligned}$$

Hence, a = -5 and ratio = 5 : 2.

Given,

A straight line makes on the coordinate axes positive intercepts whose sum is 5.

Let intercept be a on x-axis and b on y-axis.

a + b = 5

a = 5 - b .........(1)

Intercept form of equation : $\frac{x}{a} + \frac{y}{b} = 1$

Substituting values we get :

$$\begin{aligned}\Rightarrow \frac{-3}{a} + \frac{4}{b} = 1 \\[1em] \Rightarrow \frac{-3b + 4a}{ab} = 1 \\[1em] \Rightarrow \frac{4a - 3b}{ab} = 1 \\[1em] \Rightarrow 4a - 3b = ab\end{aligned}$$

Substituting value of a from equation (1) in above equation :

$$\begin{aligned}\Rightarrow 4(5 - b) - 3b = (5 - b)b \\[1em] \Rightarrow 20 - 4b - 3b = 5b - b^2 \\[1em] \Rightarrow 20 - 7b = 5b - b^2 \\[1em] \Rightarrow b^2 - 7b - 5b + 20 = 0 \\[1em] \Rightarrow b^2 - 12b + 20 = 0 \\[1em] \Rightarrow b^2 - 10b - 2b + 20 = 0 \\[1em] \Rightarrow b(b - 10) - 2(b - 10) = 0 \\[1em] \Rightarrow (b - 2)(b - 10) = 0 \\[1em] \Rightarrow b = 2 \text{ or } b = 10.\end{aligned}$$

Substituting values of b in equation (1), we get :

When b = 2,

a = 5 - b = 5 - 2 = 3

When b = 10,

a = 5 - b = 5 - 10 = -5.

Since, both intercepts are positive,

So, a ≠ -5.

∴ a = 3 and b = 2

Substituting value of a and b in $\frac{x}{a} + \frac{y}{b} = 1$, we get :

$$\begin{aligned}\Rightarrow \frac{x}{3} + \frac{y}{2} = 1 \\[1em] \Rightarrow \frac{2x + 3y}{6} = 1 \\[1em] \Rightarrow 2x + 3y = 6.\end{aligned}$$

Hence, required equation is 2x + 3y = 6.

(i) We know that,

y-coordinate of any point on x-axis = 0.

So, let point A be (a, 0)

⇒ 3a - 4(0) + 12 = 0

⇒ 3a + 12 = 0

⇒ 3a = -12

⇒ a = $-\frac{12}{3}$ = -4.

A = (-4, 0).

x-coordinate of any point on y-axis = 0.

So, let point B be (0, b).

⇒ 3(0) - 4b + 12 = 0

⇒ 4b = 12

⇒ b = $\frac{12}{4}$ = 3.

B = (0, 3).

Hence, point A = (-4, 0) and B = (0, 3).

(ii) By formula,

Mid-point = $\Big(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\Big)$

Mid-point of AB = $\Big(\frac{-4 + 0}{2}, \frac{0 + 3}{2}\Big) = \Big(\frac{-4}{2}, \frac{3}{2}\Big)$ = (-2, 1.5).

By formula,

Slope = $\frac{y_2 - y_1}{x_2 - x_1}$

Slope of AB = $\frac{3 - 0}{0 - (-4)} = \frac{3}{4}$.

We know that,

Product of slope of perpendicular lines = -1.

⇒ Slope of AB × Slope of perpendicular line = -1

⇒ $\frac{3}{4} \times$ Slope of perpendicular line = -1

⇒ Slope of perpendicular line (m) = $-\frac{4}{3}$.

By point-slope form, equation of line :

⇒ y - y₁ = m(x - x₁)

⇒ y - 1.5 = $-\frac{4}{3}$[x - (-2)]

⇒ 3(y - 1.5) = -4[x + 2]

⇒ 3y - 4.5 = -4x - 8

⇒ 4x + 3y - 4.5 + 8 = 0

⇒ 4x + 3y + 3.5 = 0

⇒ 4x + 3y + $\frac{35}{10}$ = 0

⇒ 4x + 3y + $\frac{7}{2}$ = 0

⇒ $\frac{8x + 6y + 7}{2}$ = 0

⇒ 8x + 6y + 7 = 0.

Hence, equation of perpendicular bisector of AB is 8x + 6y + 7 = 0.

(i) From figure,

∠QRP = ∠MRP

and

∠LPQ = ∠MRP

Also,

∠RQP = ∠LQP

and

∠LQP = ∠RPM

In △PQL and △RPM,

∠LPQ = ∠MRP [Proved above]

∠LQP = ∠RPM [Proved above]

∴ △PQL ~ △RPM [By A.A. axiom]

Hence, proved that △PQL ~ △RPM.

(ii) We know that,

Ratio of corresponding sides of similar triangle are proportional.

$\therefore \frac{QL}{PM} = \frac{PL}{RM}$

⇒ QL × RM = PL × PM

Hence, proved that QL × RM = PL × PM.

(iii) In △LPQ and △PQR,

∠Q = ∠Q [Common]

∠QPL = ∠PRQ [Given]

∴ △LPQ ~ △PQR [By A.A. axiom]

We know that,

Ratio of corresponding sides of similar triangle are proportional.

$\therefore \frac{PQ}{QR} = \frac{QL}{PQ}$

⇒ PQ² = QL × QR

Hence, proved that PQ² = QL × QR.

Given,

cot α = $\frac{3}{2}$

By formula,

⇒ cosec² α = 1 + cot² α

⇒ cosec² α = 1 + $\Big(\frac{3}{2}\Big)^2$

⇒ cosec² α = 1 + $\frac{9}{4}$

⇒ cosec² α = $\frac{4 + 9}{4}$

⇒ cosec² α = $\frac{13}{4}$

⇒ cosec α = $\sqrt{\frac{13}{4}} = \frac{\sqrt{13}}{2}$.

By formula,

⇒ cosec α = $\frac{\text{Hypotenuse}}{\text{Perpendicular}}$

⇒ cosec α = $\frac{AC}{AD}$

$\Rightarrow \frac{\sqrt{13}}{2} = \frac{15}{AD}$

$\Rightarrow AD = \frac{15 \times 2}{\sqrt{13}} = \frac{30}{\sqrt{13}}$

By formula,

⇒ cot α = $\frac{\text{Base}}{\text{Perpendicular}}$

⇒ cot α = $\frac{CD}{AD}$

$\Rightarrow \frac{3}{2} = \frac{CD}{\frac{30}{\sqrt{13}}}$

$\Rightarrow CD = \frac{3}{2} \times \frac{30}{\sqrt{13}} = \frac{45}{\sqrt{13}}$

Perimeter = 2(length + breadth) = 2(CD + AD)

= $2\Big(\frac{45}{\sqrt{13}} + \frac{30}{\sqrt{13}}\Big)$

= $2\Big(\frac{75}{\sqrt{13}}\Big)$

= $\Big(\frac{150}{\sqrt{13}}\Big)$

= $\Big(\frac{150}{\sqrt{13}} \times \frac{\sqrt{13}}{\sqrt{13}}\Big)$      [Rationalising]

= $\frac{150\sqrt{13}}{13}$ cm.

Area = length × breadth

= CD × AD

= $\frac{45}{\sqrt{13}} \times \frac{30}{\sqrt{13}} = \frac{1350}{13} = 103\frac{11}{13}$ cm².

Hence, perimeter = $\frac{150\sqrt{13}}{13}$ cm and area = $103\frac{11}{13}$ cm².

Join BC.

We know that,

The diameter of a circle subtends an angle of 90° at any point on circle.

∴ ∠ACB = 90°

We know that,

A tangent line is perpendicular to the radius line from the center to the point of contact

∴ ∠ABP = 90°

In △ACB and △ABP,

∠ACB = ∠ABP = 90°

∠A = ∠A [Common]

∴ △ACB ~ △ABP [By A.A. axiom]

We know that :

Ratio of corresponding sides of similar triangle are proportional.

$\therefore \frac{AB}{AP} = \frac{AC}{AB}$

⇒ AB² = AC × AP

Hence, proved that AB² = AC × AP.

Steps of construction :

1. Draw a line BC = 6 cm.

2. Draw two arc intersecting each other at A having radius 7.5 cm and centre B and C.

3. Join AC and AB.

4. Draw perpendicular bisector of BC as AD.

5. With centre A and radius 2.8 cm, draw a circle which intersects AD at E.

6. Join EB and EC and draw perpendicular bisector of EB intersecting AD at O.

7. With centre O and radius OE or OB or OC draw a circle which passes through B and C and touches the first circle at E externally.

Hence, above is the required circle.

△ABC is shown in the figure below:

In △ABC,

By pythagoras theorem,

⇒ BC² = AB² + AC²

⇒ 10² = 6² + AC²

⇒ AC² = 100 - 36

⇒ AC² = 64

⇒ AC = $\sqrt{64}$ = 8 cm.

A tangent line is perpendicular to the radius line from the center to the point of contact

∴ DF ⊥ BC, DG ⊥ AC and DH ⊥ AB.

Let radius of circle be r.

∴ DF = DG = DH = r

By formula,

Area of triangle = $\frac{1}{2} \times$ base × height

From figure,

Area of △ABC = Area of △ADC + Area of △CDB + Area of △ADB

$$\begin{aligned}\Rightarrow \frac{1}{2} \times AC \times AB = \frac{1}{2} \times DG \times AC + \frac{1}{2} \times DF \times BC + \frac{1}{2} \times DH \times AB \\[1em] \Rightarrow \frac{1}{2} \times 8 \times 6 = \frac{1}{2} \times r \times 8 + \frac{1}{2} \times r \times 10 + \frac{1}{2} \times r \times 6 \\[1em] \Rightarrow \frac{1}{2} \times 48 = \frac{1}{2}\Big(8r + 10r + 6r\Big) \\[1em] \Rightarrow 48 = 8r + 10r + 6r \\[1em] \Rightarrow 24r = 48 \\[1em] \Rightarrow r = \frac{48}{24} = 2\text{ cm}.\end{aligned}$$

Area of triangle (excluding circle) = Area of triangle - Area of circle

= $\frac{1}{2} \times AC \times AB$ - πr²

= $\frac{1}{2} \times 8 \times 6 - \frac{22}{7}$ × (2)²

= 24 - $\frac{88}{7}$

= $\frac{168 - 88}{7}$

= $\frac{80}{7}$

= $11\frac{3}{7}$ cm².

Hence, area of triangle excluding circle = $11\frac{3}{7}$ cm².

Given,

Radius of conical vessel (R) = AC = 6 cm

Height of conical vessel (H) = OC = 8 cm

Radius of sphere (r)

∴ PC = PD = r cm.

We know that,

Since, lengths of two tangents from an external point to a circle are equal.

∴ AC = AD = 6 cm

As radius from center to the point of tangent are perpendicular to each other.

△OCA and △OPD are right angle triangles.

In △OCA,

By pythagoras theorem,

$$\begin{aligned}\Rightarrow OA^2 = OC^2 + AC^2 \\[1em] \Rightarrow OA = \sqrt{OC^2 + AC^2} \\[1em] \Rightarrow OA = \sqrt{8^2 + 6^2} \\[1em] \Rightarrow OA = \sqrt{64+ 36} \\[1em] \Rightarrow OA = \sqrt{100} = 10 \text{ cm}.\end{aligned}$$

In △OPD,

By pythagoras theorem,

$\Rightarrow OP^2 = OD^2 + PD^2 \text{ ........(1)}$

From figure,

OD = OA - AD = 10 - 6 = 4 cm.

OP = OC - PC = 8 - r

Substituting values of OP, OD and PD in equation (1), we get :

$$\begin{aligned}\Rightarrow (8 - r)^2 = 4^2 + r^2 \\[1em] \Rightarrow 64 + r^2 - 16r = 16 + r^2 \\[1em] \Rightarrow r^2 - r^2 + 64 - 16 - 16r = 0 \\[1em] \Rightarrow 16r = 48 \\[1em] \Rightarrow r = \frac{48}{16} = 3 \text{ cm}.\end{aligned}$$

Volume of water overflown = Volume of sphere

= $\frac{4}{3}πr^3 = \frac{4}{3}π(3)^3$

= $\frac{4}{3}π \times 27= 36π$ cm³.

Original volume of water = Volume of cone

= $\frac{1}{3}πR^2H = \frac{1}{3}π \times 6^2 \times 8$ = 96π cm³.

Fraction of water overflown = $\frac{\text{Volume of water overflown}}{\text{Original volume of water}} = \frac{36π}{96π} = \frac{3}{8}$.

Hence, fraction of water overflown = $\frac{3}{8}$.

To prove:

$\frac{\text{1 + cot A}}{\text{cos A}} + \frac{\text{1 + tan A}}{\text{sin A}}$ = 2(sec A + cosec A)

Solving L.H.S. of the above equation :

$$\begin{aligned}\Rightarrow \frac{\text{1 + cot A}}{\text{cos A}} + \frac{\text{1 + tan A}}{\text{sin A}} \\[1em] \Rightarrow \frac{\text{sin A(1 + cot A) + cos A(1 + tan A)}}{\text{cos A sin A}} \\[1em] \Rightarrow \frac{\text{sin A + sin A cot A + cos A + cos A tan A}}{\text{cos A sin A}} \\[1em] \Rightarrow \frac{\text{sin A + sin A} \times \frac{\text{cos A}}{\text{sin A}} + \text{cos A + cos A} \times \frac{\text{sin A}}{\text{cos A}}}{\text{cos A sin A}} \\[1em] \Rightarrow \frac{\text{sin A + cos A + cos A + sin A}}{\text{cos A sin A}} \\[1em] \Rightarrow \frac{\text{2(sin A + cos A)}}{\text{sin A cos A}}.\end{aligned}$$

Solving R.H.S. of the equation :

$$\begin{aligned}\Rightarrow \text{2(sec A + cosec A)} \\[1em] \Rightarrow 2\Big(\frac{1}{\text{cos A}} + \frac{1}{\text{sin A}}\Big) \\[1em] \Rightarrow 2\Big(\frac{\text{sin A + cos A}}{\text{sin A cos A}}\Big) \\[1em] \Rightarrow \frac{2\text{(sin A + cos A)}}{\text{sin A cos A}}.\end{aligned}$$

Since, L.H.S. = R.H.S. = $\frac{\text{2(sin A + cos A)}}{\text{sin A cos A}}$

Hence, proved that $\frac{\text{1 + cot A}}{\text{cos A}} + \frac{\text{1 + tan A}}{\text{sin A}}$ = 2(sec A + cosec A).

To prove:

$\sqrt{\frac{\text{1+ sin A}}{\text{1 - sin A}}} - \sqrt{\frac{\text{1 - sin A}}{\text{1 + sin A}}}$ = 2 tan A

Solving L.H.S. of the equation :

$$\begin{aligned}\Rightarrow \sqrt{\frac{\text{1 + sin A}}{\text{1 - sin A}} \times \frac{\text{1 + sin A}}{\text{1 + sin A}}} - \sqrt{\frac{\text{1 - sin A}}{\text{1 + sin A}} \times \frac{\text{1 - sin A}}{\text{1 - sin A}}} \\[1em] \Rightarrow \sqrt{\frac{(\text{1 + sin A})^2}{\text{1 - sin}^2 A}} - \sqrt{\frac{(\text{1 - sin A})^2}{\text{1 - sin}^2 A}}\end{aligned}$$

By formula,

1 - sin² A = cos² A

$$\begin{aligned}\Rightarrow \sqrt{\frac{(\text{1 + sin A})^2}{\text{cos}^2 A}} - \sqrt{\frac{(\text{1 - sin A})^2}{\text{cos}^2 A}} \\[1em] \Rightarrow \frac{\text{1 + sin A}}{\text{cos A}} - \frac{\text{1 - sin A}}{\text{cos A}} \\[1em] \Rightarrow \frac{\text{1 + sin A - (1 - sin A)}}{\text{cos A}} \\[1em] \Rightarrow \frac{1 - 1 + \text{sin A + sin A}}{\text{cos A}} \\[1em] \Rightarrow \frac{\text{2 sin A}}{\text{cos A}} \\[1em] \Rightarrow \text{2 tan A}.\end{aligned}$$

Since, L.H.S. = R.H.S.

Hence, proved that $\sqrt{\frac{\text{1+ sin A}}{\text{1 - sin A}}} - \sqrt{\frac{\text{1 - sin A}}{\text{1 + sin A}}}$ = 2 tan A.

(i) Given,

⇒ 3 tan² 2x = 1

⇒ tan² 2x = $\frac{1}{3}$

⇒ tan 2x = $\sqrt{\frac{1}{3}}$

⇒ tan 2x = $\frac{1}{\sqrt{3}}$

⇒ tan 2x = tan 30°

⇒ 2x = 30°

⇒ x = $\frac{30°}{2}$ = 15°.

Hence, x = 15°.

(ii) Given,

⇒ tan² x = 3(sec x - 1)

⇒ sec² x - 1 = 3sec x - 3

⇒ sec² x - 3 sec x - 1 + 3 = 0

⇒ sec² x - 3 sec x + 2 = 0

⇒ sec² x - 2 sec x - sec x + 2 = 0

⇒ sec x(sec x - 2) - 1(sec x - 2) = 0

⇒ (sec x - 1)(sec x - 2) = 0

⇒ sec x - 1 = 0 or sec x - 2 = 0

⇒ sec x = 1 or sec x = 2

⇒ sec x = sec 0° or sec x = sec 60°

⇒ x = 0° or x = 60°

Hence, x = 0° or 60°.

Let CO be the tower of height h meters.

In △AOC,

⇒ tan α = $\frac{\text{Perpendicular}}{\text{Base}}$

⇒ tan α = $\frac{OC}{AO}$

⇒ tan α = $\frac{h}{a + x}$

⇒ (a + x)tan α = h

⇒ a tan α + x tan α = h

⇒ x tan α = h - a tan α

⇒ x = $\frac{\text{h - a tan α}}{\text{tan α}}$ ........(1)

In △BOC,

⇒ tan β = $\frac{\text{Perpendicular}}{\text{Base}}$

⇒ tan β = $\frac{OC}{BO}$

⇒ tan β = $\frac{h}{x}$

⇒ x = $\frac{h}{\text{tan β}}$ ............(2)

From (1) and (2)

$$\begin{aligned}\Rightarrow \frac{h}{\text{tan β}} = \frac{\text{h - a tan α}}{\text{tan α}} \\[1em] \Rightarrow \frac{h}{\text{tan β}} =\frac{h}{\text{tan α}} - \frac{\text{a tan α}}{\text{tan α}} \\[1em] \Rightarrow \frac{h}{\text{tan β}} - \frac{h}{\text{tan α}} = -a \\[1em] \Rightarrow \frac{\text{h tan α - h tan β}}{\text{tan α tan β}} = -a \\[1em] \Rightarrow \frac{\text{h (tan α - tan β)}}{\text{tan α tan β}} = -a \\[1em] \Rightarrow h = \frac{-\text{a tan α tan β}}{\text{(tan α - tan β)}} \\[1em] \Rightarrow h = \frac{\text{a tan α tan β}}{\text{(tan β - tan α)}}\end{aligned}$$

Hence, proved that $h = \frac{\text{a tan α tan β}}{\text{(tan β - tan α)}}$.

Class	Frequencies (f)	Class marks (x)	fx
0-20	17	10	170
20-40	f1	30	30f1
40-60	32	50	1600
60-80	f2	70	70f2
80-100	19	90	1710
Total	Σf = f1 + f2 + 68		Σfx = 3480 + 30f1 + 70f2

Given,

Σf = 120

⇒ f₁ + f₂ + 68 = 120

⇒ f₁ + f₂ = 120 - 68

⇒ f₁ + f₂ = 52

⇒ f₁ = 52 - f₂ ..........(1)

By formula,

Mean = $\frac{Σfx}{Σf}$

Substituting values we get :

$$\begin{aligned}\Rightarrow 50 = \frac{3480 + 30f_1 + 70f_2}{120} \\[1em] \Rightarrow 6000 = 3480 + 30f_1 + 70f_2\end{aligned}$$

Substituting value of f₁ from equation (1) in above equation, we get :

⇒ 6000 = 3480 + 30(52 - f₂) + 70f₂

⇒ 6000 = 3480 + 1560 - 30f₂ + 70f₂

⇒ 6000 = 5040 + 40f₂

⇒ 40f₂ = 6000 - 5040

⇒ 40f₂ = 960

⇒ f₂ = $\frac{960}{40}$ = 24.

⇒ f₁ = 52 - f₂ = 52 - 24 = 28.

Hence, f₁ = 28 and f₂ = 24.

Total cards = 52

∴ No. of possible outcomes = 52

(i) There are 4 aces in a deck (1 of each suit).

∴ No. of favourable outcomes = 4

P(drawing an ace) = $\frac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \frac{4}{52} = \frac{1}{13}$.

Hence, probability of drawing an ace = $\frac{1}{13}$.

(ii) There is only one jack of hearts.

∴ No. of favourable outcomes = 1

P(drawing a jack of hearts) = $\frac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \frac{1}{52}$.

Hence, probability of drawing a jack of hearts = $\frac{1}{52}$.

(iii) There is one three of clubs and one six of diamonds.

∴ No. of favourable outcomes = 2

P(drawing a three of clubs or a six of diamonds)

= $\frac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \frac{2}{52} = \frac{1}{26}$.

Hence, probability of drawing a three of clubs or a six of diamonds = $\frac{1}{26}$.

(iv) There are 13 hearts in a deck.

∴ No. of favourable outcomes = 13

P(drawing a heart) = $\frac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \frac{13}{52} = \frac{1}{4}$.

Hence, probability of drawing a heart = $\frac{1}{4}$.

(v) There are 39 other cards except heart.

∴ No. of favourable outcomes = 39

P(drawing any suit except heart) = $\frac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \frac{39}{52} = \frac{3}{4}$.

Hence, probability of drawing any suit except heart = $\frac{3}{4}$.

(vi) There are 13 spade cards and 3 other 10's (1 of each suit except spade)

∴ No. of favourable outcomes = 16

P(drawing a ten or a spade)

= $\frac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \frac{16}{52} = \frac{4}{13}$.

Hence, probability of drawing a ten or a spade = $\frac{4}{13}$.

(vii) There are 13 club cards and 3 other four cards (1 of each suit apart from club)

Total cards = 16

Left cards = 52 - 16 = 36.

∴ No. of favourable outcomes = 36

P(drawing neither a club nor a four)

= $\frac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \frac{36}{52} = \frac{9}{13}$.

Hence, probability of drawing neither a club nor 4 = $\frac{9}{13}$.

(viii) Jack, King and Queen are considered as picture cards.

There are 4 jacks, 4 queens and 4 kings

∴ No. of favourable outcomes = 12

P(drawing a picture card)

= $\frac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \frac{12}{52} = \frac{3}{13}$.

Hence, probability of drawing a picture card = $\frac{3}{13}$.

(ix) No. of spade cards = 13

Jack, king and queen except that of spade are 3 each.

Total cards = 13 + 3 + 3 + 3 = 22.

∴ No. of favourable outcomes = 22

P(drawing a spade or a picture card)

= $\frac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \frac{22}{52} = \frac{11}{26}$.

Hence, probability of drawing a spade or a picture card = $\frac{11}{26}$.