Correct Answer: C. -12xy

Explanation

(x - 3y)² - (x + 3y)² = [(x - 3y) - (x + 3y)]

[(x + 3y + x + 3y)] = [-6y] [2x]

= -12xy

Explanation

(a) \((\sqrt{3} - 5\sqrt{2})(\sqrt{3} + \sqrt{2}) = \sqrt{3 \times 3} + \sqrt{3 \times 2} - 5\sqrt{2 \times 3} - 5\sqrt{2 \times 2}\)

= \(3 + \sqrt{6} - 5\sqrt{6} - 5(2)\)

= \(-7 - 4\sqrt{6}\)

\(\therefore\) a = -7 and b = -4.

(b) \(\frac{2^{1 - y} \times 2^{y - 1}}{2^{y + 2}} = 8^{2 - 3y}\)

\(\frac{2^{1 - y + y - 1}}{2^{y + 2}} = (2^{3})^{2 - 3y}\)

\(2^{0 - (y + 2)} = 2^{6 - 9y} \implies 2^{- y - 2} = 2^{6 - 9y}\)

\(- y - 2 = 6 - 9y \implies - y + 9y = 6 + 2\)

\(8y = 8 \implies y = 1\).

Correct Answer: C. 2x² - 5x - 3 = 0

Explanation

x = \(\frac{-1}{2}\)

x = 3

2x = -1, x = 3; 2x + 1 = 0

x - 3 = 0

(2x + 1)(x - 3) = 0

2x² - 6x + x - 3 = 0

2x² - 5x - 3 = 0

Correct Answer: B. \(f(x) = x^{3} + \frac{1}{x^{4}} + 2\)

Explanation

\(\frac{\mathrm d y}{\mathrm d x} = 3x^{2} - \frac{4}{x^5} = 3x^{2} - 4x^{-5}\)

\(y = \int (3x^{2} - 4x^{-5}) \mathrm {d} x \)

\(y = x^{3} + \frac{1}{x^{4}} + c\)

f(1) = 4; \(4 = 1^{3} + \frac{1}{1^{4}} + c \implies 4 = 2 + c\)

\(c = 2\)

\(f(x) = x^{3} + \frac{1}{x^{4}} + 2\)

Correct Answer: B. 90°

Explanation

Tan \(\theta = \frac{6}{6} = 1\)

\(\theta\) = tab - 1(1) = 45^o

< top = 20

= 2 x 45^o = 90^o

Correct Answer: C. their intersection is an empty set

Correct Answer: A. (0, 4)

Correct Answer: D. \(\frac{1}{2}(\text{Upper quartile - Lower quartile})\)

Correct Answer: C. \(\frac{7}{15}\)

Explanation

\(P(\text{both bulbs are good}) = \frac{7}{10} \times \frac{6}{9}\)

= \(\frac{7}{15}\)

Explanation

(a)(i) \(T_{3} = 2 ; ar^{2} = 2 ... (i)\)

\(T_{6} = 54 ; ar^{5} = 54 ... (ii)\)

\(\frac{T_{6}}{T_{3}} = \frac{ar^{5}}{ar^{2}} = \frac{54}{2}\)

\(r^{3} = 27 \implies r = 3 \)

(ii) First term = a

\(T_{3} = ar^{2} = 2\)

\(a = \frac{2}{3^{2}} = \frac{2}{9}\)

(iii) \(S_{10} = \frac{a(r^{n} - 1)}{r - 1}\)

= \(\frac{2(3^{10} - 1)}{9(3 - 1)}\)

= \(\frac{3^{10} - 1}{9}\)

= \(\frac{59049 - 1}{9} \approxeq 6561\) (nearest whole number).

(b) \((1 + 2x)^{n} = 1 + ^{n}C_{1} (1)^{n - 1} (2x) + ^{n}C_{2} (1)^{n - 2} (2x)^{2} + ^{n}C_{3} (1)^{n - 3} (2x)^{3} + ^{n}C_{4} (1)^{n - 4} (2x)^{4}+ ...\)

\(\frac{\text{coefficient of} x^{4}}{\text{coefficient of } x^{3}} = \frac{3}{1}\)

\(\frac{n!}{4! (n - 4)!} \times 2^{4} \times \frac{3! (n - 3)!}{n!} = \frac{3}{1}\)

\(\frac{16 \times 3! (n - 3)(n - 4)!}{4 \times 3! (n - 4)! \times 8} = \frac{3}{1}\)

\(\frac{2(n - 3)}{4} = \frac{3}{1} \implies 2n - 6 = 12\)

\(2n = 18 \implies n = 9\)