Correct Answer: D. 0 and -1

Explanation

\(\sqrt{x} + \sqrt{x + 1} = \sqrt{2x + 1}\)

Squaring both sides, we have

\((\sqrt{x} + \sqrt{x + 1})^{2} = (\sqrt{2x + 1})^{2}\)

\(x + 2\sqrt{x(x + 1)} + x + 1 = 2x + 1\)

\(2x + 1 + 2\sqrt{x(x+1)} - (2x + 1) = 0\)

\((2\sqrt{x(x + 1)})^{2}= 0^{2} \implies 4(x(x + 1)) = 0\)

\(\therefore x(x + 1) = 0\)

\(x = \text{0 or -1}\)

Explanation

\(\begin{vmatrix} x - 3 & -4 & 3 \\ 5 & 2 & 2 \\ 2 & -4 & 6 - x \end{vmatrix} = -24\)

\((x - 3)[2(6 - x) + 8] + 4[5(6 - x) - 4] + 3[- 20 - 4] = -24\)

\((x - 3)[12 - 2x + 8] + 4[30 - 5x - 4] + 3[-24] = -24\)

\((x - 3)(20 - 2x) + 4(26 - 5x) - 72 = -24\)

\((x - 3)(20 - 2x) + 4(26 - 5x) = - 24 + 72 = 48\)

\(20x - 2x^{2} - 60 + 6x + 104 - 20x = 48\)

\(-2x^{2} + 6x - 4 = 0\)

\(2x^{2} - 6x + 4 = 0\)

Solving the equation, we get x = 1 or 2.

Explanation

p*q = p\(^2\) + 2pq - q\(^2\)

when p = -2 and q = -2

p*q = -2\(^2\) + 2(-2)(-2) - (-2)\(^2\)

p*q = 4 + 8 - 4 = 8

when p = -2 and q = -1

p*q = -2\(^2\) + 2(-2)(-1) - (-1)\(^2\)

p*q = 4 + 4 - 1 = 7

when p = -2 and q = 1

p*q = -2\(^2\) + 2(-2)(1) - (1)\(^2\)

p*q = 4 - 4 -1 = -1

when p = -1 and q = -2

p*q = -1\(^2\) + 2(-1)(-2) - (-2)\(^2\)

p*q = 1 + 4 - 4 = 1

when p = -1 and q = -1

p*q = -1\(^2\) + 2(-1)(-1) - (-1)\(^2\)

p*q = 1 + 2 -1 = 2

when p = -1 and q = 2

p*q = -1\(^2\) + 2(-1)(2) - (2)\(^2\)

p*q = 1 - 4 - 4 = -7

when p = 1 and q = -2

p*q = 1\(^2\) + 2(1)(-2) - (-2)\(^2\)

p*q = 1 - 4 - 4 = -7

when p = 1 and q = -1

p*q = 1\(^2\) + 2(1)(-1) - (-1)\(^2\)

p*q = 1 - 2 - 1 = -2

when p = 1 and q = 1

p*q = 1\(^2\) + 2(1)(1) - (1)\(^2\)

p*q = 1 + 2 - 1 = 2

when p = 1 and q = 2

p*q = 1\(^2\) + 2(1)(2) - (2)\(^2\)

p*q = 1 + 4 - 4 = 1

when p = 2 and q = -2

p*q = 2\(^2\) + 2(2)(-2) - (-2)\(^2\)

p*q = 4 - 8 - 4 = -8

when p = 2 and q = -1

p*q = 2\(^2\) + 2(2)(-1) - (-1)\(^2\)

p*q = 4 - 4 - 1 = -1

when p = 2 and q = 1

p*q = 2\(^2\) + 2(2)(1) - (1)\(^2\)

p*q = 4 + 4 - 1 = 7

when p = 2 and q = 2

p*q = 2\(^2\) + 2(2)(2) - (2)\(^2\)

p*q = 4 + 8 - 4 = 8

Correct Answer: B. onto

Explanation

The function \(y = g(x)\) is one- to- one but \(z = h(y)\) is not one- to- one but onto.

\(\therefore h(g(x)) = onto\)

Correct Answer: B. \(\begin{pmatrix} -5 \\ -1 \end{pmatrix}\)

Explanation

\(4 \begin{pmatrix} 3 \\ 2 \end{pmatrix} + 3 \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} -3 \\ 5 \end{pmatrix}\)

\(\begin{pmatrix} 12 \\ 8 \end{pmatrix} + \begin{pmatrix} 3x \\ 3y \end{pmatrix} = \begin{pmatrix} -3 \\ 5 \end{pmatrix}\)

\(\begin{pmatrix} 3x \\ 3y \end{pmatrix} = \begin{pmatrix} -3 - 12 \\ 5 - 8 \end{pmatrix}\)

\(\begin{pmatrix} 3x \\ 3y \end{pmatrix} = \begin{pmatrix} -15 \\ -3 \end{pmatrix}\)

\(\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} -5 \\ -1 \end{pmatrix}\)

Explanation

\(p(boys) = \frac{8}{14} = \frac{4}{7}\)

\(p(girls) = \frac{6}{14} = \frac{3}{7}\)

(a) p(both are of the same sex) = p(both are boys or both are girls)

= \(\frac{8}{14} \times \frac{7}{13} + \frac{6}{14} \times \frac{5}{13}\)

= \(\frac{86}{182}\)

= \(\frac{43}{91}\)

(b) p(they are of different sex) = p(first is a boy and second is a girl or first is a girl and second is a boy)

= \(\frac{8}{14} \times \frac{6}{13} + \frac{6}{14} \times \frac{8}{13}\)

= \(\frac{96}{182}\)

= \(\frac{48}{91}\)

Explanation

(a) \(f(x) = \frac{4 - 5x}{2} ; g(x) = x + 6, x \in R\)

Let g(x) = y

\(y = x + 6 \implies x = y - 6\)

Let \(x = g^{-1} (x)\) and y = x.

\(g^{-1}(x) = x - 6\)

\(f \cdot g^{-1}(x) = f(g^{-1}(x))\)

= \(\frac{4 - 5(x - 6)}{2}\)

= \(\frac{4 - 5x + 30}{2}\)

= \(\frac{34 - 5x}{2}\)

(b) \(x = \frac{7 \times 4 - 2(5)}{5 + 4}\)

= \(\frac{18}{9}\)

\(x = 2\)

\(y = \frac{4(-5) + 5(7)}{5 + 4}\)

= \(\frac{15}{9}\)

\((x, y) = (2, 1\frac{2}{3})\)

Explanation

(a)

(b) Lower quartile \(Q_{1}\) = 45.6cm

Upper quartile \(Q_{3}\) = 52.0cm

Semi- interquartile range = \(\frac{Q_{3} - Q_{1}}{2}\)

= \(\frac{52 - 45.6}{2}\)

= \(\frac{6.4}{2}\)

= \(3.2cm\)

Correct Answer: D. \(\begin{pmatrix} \frac{5}{3} &, 0 \end {pmatrix}\)

Explanation

y = 8x + 5

m = 8

y = 3x\(^2\) - 2x - 5

\(\frac{dy}{dx}\) = 6x - 2x - 5

\(\frac{6x}{6} = \frac{10}{6}\)

x = \(\frac{5}{3}\)

y = 0

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

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

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