Explanation

(a) \(T_{n} = 4T_{n - 1} - 3\)

\(T_{3} = 4T_{2} - 3 ; T_{2} = 4T_{1} - 3\)

\(2T_{3} = 5T_{2} \implies T_{3} = \frac{5}{2} T_{2}\)

\(\frac{5}{2} T_{2} = 4T_{2} - 3 \implies 3 = \frac{3}{2} T_{2}\)

\(T_{2} = \frac{3}{\frac{3}{2}} = 2\)

\(2 = 4T_{1} - 3 \implies 5 = 4T_{1}\)

\(T_{1} = \frac{5}{4}\)

\(2T_{3} = 5(2) = 10\)

\(T_{3} = 5\)

The first three terms of the sequence are \(\frac{5}{4}, 2, 5\).

(b) \(\begin{pmatrix} 2 & 0 & 1 \\ 5 & -3 & 1 \\ 0 & 4 & 6 \end{pmatrix} \begin{pmatrix} 1 \\ m \\ r \end{pmatrix} = \begin{pmatrix} k \\ 2 \\ 26 \end{pmatrix}\)

\(\begin{pmatrix} 2(1) + r \\ 5(1) - 3m + r \\ 4m + 6r \end{pmatrix} = \begin{pmatrix} k \\ 2 \\ 26 \end{pmatrix}\)

\(2 + r = k ... (1)\)

\(5 - 3m + r = 2 \implies 3m - r = 3 ... (2)\)

\(4m + 6r = 26 ... (3)\)

\((2) \times 6 : 18m - 6r = 18\)

\((2) + (3) : 22m = 44 \implies m = 2\)

\(4(2) + 6r = 26 \implies 6r = 18\)

\(r = 3\)

\(2 + r = k \implies 2 + 3 = 5\)

\((k, m, r) = (5, 2, 3)\).

Correct Answer: D. \(R = \frac{P^2-3Qt}{Q}\)

Correct Answer: A. 19,650

Correct Answer: C. -6 ≤ x ≤ 3

Correct Answer: B. x + 1 ≤ 0, y + x ≥ 0, y - x ≤ 0

Correct Answer: B. 25°

Explanation

MPQ = PQM = 3x; ?MPQ

MPQ + PQM + PMQ = 180^o; 3x + 3x + 30^o = 180^o

6x \times 30^o = 180^o; 6x + 180^o - 30^o; 6x + 150^o

\(x = \frac{150}{6} = 25^{\circ}\)

Correct Answer: C. 42.4%

Correct Answer: D. nx\(^2\) - mx - 1 = 0

Explanation

x\(^2\) + mx - n = 0

a = 1, b = m, c = -n

α + β = \(\frac{-b}{a}\) = \(\frac{-m}{1}\) = -m

αβ = \(\frac{c}{a}\) = \(\frac{-n}{1}\) = -n

the roots are = \(\frac{1}{α}\) and \(\frac{1}{β}\)

sum of the roots = \(\frac{1}{α}\) + \(\frac{1}{β}\)

\(\frac{1}{α}\) + \(\frac{1}{β}\) = \(\frac{α+β}{αβ}\)

α + β = -m

αβ = -n

\(\frac{α+β}{αβ}\) = \(\frac{-m}{-n}\) → \(\frac{m}{n}\)

product of the roots = \(\frac{1}{α}\) * \(\frac{1}{β}\)

\(\frac{1}{α}\) + \(\frac{1}{β}\) = \(\frac{1}{αβ}\) → \(\frac{1}{-n}\)

x\(^2\) - (sum of roots)x + (product of roots)

x\(^2\) - ( m/n )x + ( 1/-n ) = 0

multiply through by n

nx\(^2\) - mx - 1 = 0

Correct Answer: D. \(\frac{3}{7}\) + \(\frac{?2}{7}\)

Explanation

Rationalization with conjugate \({3+√2}\)

\(\frac{1}{3-√2}\)

→ \(\frac{1 * [3+√2]}{[3-√2][3+√2]}\)

= \(\frac{3+√2}{9 -3√2 + 3√2 + √4}\)

= \(\frac{3+√2}{9 - 2}\) → \(\frac{3+√2}{7}\)

= \(\frac{3}{7}\) + \(\frac{√2}{7}\)

Explanation

(a) \((2x - 1) - p(x^{2} + 2) = 0\)

\(2x - 1 - px^{2} - 2p = 0\)

\(px^{2} - 2x + (2p + 1) = 0\)

For real roots, \(b^{2} - 4ac \geq 0\)

\(2^{2} - 4(p)(2p + 1) \geq 0\)

\(4 - 8p^{2} - 4p \geq 0 \implies 2p^{2} + p - 1 \leq 0\)

(b) \(2p^{2} - p + 2p - 1 \leq 0 \implies p(2p - 1) + 1(2p - 1) \leq 0\)

\((p + 1)(2p - 1) \leq 0 \)

\(2p - 1 \leq 0 \implies p \leq \frac{1}{2}\)

\(p + 1 \leq 0 \implies p \leq -1\)

Check : For \(p \leq 1 , \text{let p = -2}\)

\(2(-2)^{2} + (-2) - 1 = 8 - 2 - 1 = 5 \)

For \(p \leq \frac{1}{2}, \text{let p = 0}\)

\(2(0^{2}) + 0 - 1 = -1 \leq 0\)

\(\therefore -1 \leq p \leq \frac{1}{2}\).