Correct Answer: D. \(\frac{y}{4x+x-3}\)

Explanation

2y\(^2\) + 7 = 3y - xy, find \(\frac{dy}{dx}\)

4y \(\frac{dy}{dx}\) + 0 = 3 \(\frac{dy}{dx}\) - x \(\frac{dy}{dx}\) + y

{4x+x-3} \(\frac{dy}{dx}\) = y

= \(\frac{y}{4x+x-3}\)

Correct Answer: D. \((\frac{\pi}{3}, \frac{5\pi}{3})\)

Explanation

\(2\cos x - 1 = 0 \implies 2\cos x = 1\)

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

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

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

Correct Answer: D. \(\frac{11}{24}\)

Explanation

n = 3, \(\frac{1}{(n-1)!} - \frac{1}{(n+1)!} = \frac{1}{(3-1)!} - \frac{1}{(3+1)!}\)

= \(\frac{1}{2} - \frac{1}{24} = \frac{12 -1}{24}\)

= \(\frac{11}{24}\)

Correct Answer: B. 72°

Explanation

\(a . b = |a||b| \cos \theta\)

\(\begin{pmatrix} 13 \\ 1 \end{pmatrix}. \begin{pmatrix} 1 \\ 4 \end{pmatrix} = 13 \times 1 + 1 \times 4 = 13 + 4 = 17\)

\(17 = (\sqrt{13^{2} + 1^{2}})(\sqrt{1^{2} + 4^{2}}) \cos \theta\)

\(17 = (\sqrt{170})(\sqrt{17}) \cos \theta\)

\(\cos \theta = \frac{17}{17\sqrt{10}} = \frac{\sqrt{10}}{10} = 0.3162\)

\(\theta = \cos^{-1} 0.3162 = 72°\)

Correct Answer: B. \(2,\frac{3}{2}\)

Explanation

\(p = \begin{bmatrix} x&4\\3&7\end{bmatrix}, Q =\begin{bmatrix} x&3\\1&2x\end{bmatrix}\)

\(p = \begin{bmatrix} x&4\\3&7\end{bmatrix}=7x-12, Q =\begin{bmatrix} x&3\\1&2x\end{bmatrix}=2x^2-3\)

|Q| = |P| + 3 (Given)

\(= 2\times2 - 3 = 7x - 12 + 3\)

\(= 2\times2 - 3 - 7x + 12 - 3 = 0\)

\(= 2\times2 - 7x + 6 = 0\)

\(= 2\times2 - 4x - 3x + 6 = 0\)

\(= 2x(x - 2) - 3(x - 2) = 0\)

\(= (x - 2)(2x - 3) = 0\)

\(\therefore x=2,\frac{3}{2}\)

Correct Answer: B. \(x > -1, x > -\frac{1}{3}\)

Explanation

\(3x^{2} + 4x + 1 > 0 \)

\(3x^{2} + 3x + x + 1 > 0\)

\(3x(x + 1) + 1(x + 1) > 0\)

\((3x + 1)(x + 1) > 0\)

\(3x + 1 > 0 \implies 3x > -1 \)

\(x > -\frac{1}{3}\)

\(x + 1 > 0 \implies x > -1\)

\(x > -1, x > -\frac{1}{3}\)

Correct Answer: B. \(\begin{vmatrix} 4 & 3 \\ 6 & 4 \end{vmatrix}\)

Explanation

\( P^{2} = \begin{vmatrix} 1 & 1 \\ 2 & 1 \end{vmatrix} \begin{vmatrix} 1 & 1 \\ 2 & 1 \end{vmatrix}\)

\(\begin{vmatrix} 1 \times 1 + 1 \times 2 & 1 \times 1 + 1 \times 1 \\ 2 \times 1 + 1 \times 2 & 2 \times 1 + 1 \times 1 \end{vmatrix}\)

= \(\begin{vmatrix} 3 & 2 \\ 4 & 3 \end{vmatrix} + \begin{vmatrix} 1 & 1 \\ 2 & 1 \end{vmatrix}\)

= \(\begin{vmatrix} 4 & 3 \\ 6 & 4 \end{vmatrix}\)

Correct Answer: C. x = 2

Explanation

The vertical line \(x = \frac{-b}{2a}\) is the axis of symmetry of the curve.

\(y = x^{2} - 4x - 12\)

\(\text{Axis of symmetry} = x = \frac{-(-4)}{2(1)} = \frac{4}{2} = 2\)

Correct Answer: D. \(y = x - 30\)

Correct Answer: A. 5.5m

Explanation

From S = 12t + \(\frac{5}{2t^2}\) - t\(^3\);

Distance traveled in 2 seconds

S = 12[2] + \(\frac{5}{2[2]^2}\) - 2\(^3\).

= 24 + 10 - 8 = 26m

Distance traveled in 3 seconds

S = 12[3] + \(\frac{5}{2[3]^2}\) - 3\(^3\)

= 36 + 22.5 - 27 = 31.5m

Distance traveled in the 3rd second

= 31.5m - 26m

: S = 5.5m