Correct Answer: B. \(\frac{65}{12}\)

Explanation

\({1_0^∫} x^2(x^3+2)^3\)dx

let \( u = x^3 + 2, du = 3x^2dx\)

when x = 1, u = 3

when x = 0, u = 2

dx = \(\frac{du}{3x^2}\)

\({3_2^∫}\) \(\frac{x^2[u]^3}{3x^2}\)

\({3_2^∫}\) \(\frac{u^3}{3}\) du

= \(\frac{u^4}{3*4}\)\(_2\)3

\(\frac{1}{12} [u^4]\)\(_2\)3

\(\frac{1}{12} [3^4 - 2^4]\)

\(\frac{1}{12}[81 - 16]\)

\(\frac{65}{12}\)

Correct Answer: D. x = 0 or x = 3

Explanation

\(3^{2x-2} - 28(3^{x-2}) + 3 = 0\)

\(\frac{3^{2x}}{3^2} - \frac{28.3^x}{3^2} + 3 = 0\)

\(\frac{3^{2x}}{9} - \frac{28.3^x}{9} + 3 = 0\)

let p = 3\(^x\)

\(\frac{p^{2}}{9} - \frac{{28p}}{9} + 3 = 0\)

multiply through by 9

p\(^2\) - 28p + 27 = 0

p\(^2\) - p - 27p + 27 = 0

p (p - 1) - 27(p - 1) = 0

(p-1)(p-27) = 0

p = 1 or 27

when p = 1

p = 3\(^x\)

3\(^x\) = 1

3\(^x\) = 3\(^0\)

x = 0

when p = 27

3x = 27

3x = 33

x = 3

Correct Answer: B. -5

Explanation

\(\begin{vmatrix} 2 & -3 \\ 1 & 4 \end{vmatrix} \begin{vmatrix} -6 \\ k \end{vmatrix} \begin{vmatrix} 3 \\ -26 \end{vmatrix} = 15\)

\(\begin{vmatrix} 2[-6] & - 3k \\ 1[-6] & + 4k \end{vmatrix} = \begin{vmatrix} 3 \\ -26 \end{vmatrix}\)

\(\begin{vmatrix} -12 & - 3k \\ -6 & + 4k \end{vmatrix} = \begin{vmatrix} 3 \\ -26 \end{vmatrix}\)

-12 - 3k = 3

-3k = 3 + 12

k = \(\frac{15}{-3}\)

k = -5

Correct Answer: C. 36m

Explanation

from the equation of motion

u = 5m/s, a = 2m/s\(^2\), t = 4s

s = ut + \(\frac{1}{2} at^2\)

s = 5*4 + \(\frac{1}{2} 2*4^2\)

s = 20 + 16

s = 36m

Explanation

p = \(1 - \frac{6Σ(d_i)^2}{n(n^2-1}\)

p = Spearman's rank correlation coefficient

di = difference between the two ranks of each observation

n = number of observation

P = \(1 - \frac{6(4+1+1+0+16+0+0+0)}{8(8^2-1)}\)

p = \(1 - \frac{6(22)}{8(63)}\)

p = \(1 - \frac{132}{504}\)

p = \(1 - \frac{11}{42}\)

p = \(\frac{11}{42}\) or 0.7381

Correct Answer: A. (7, -2)

Explanation

Let (x1, y1) be the image of the point (x, y) under the given transformations.

x1 = 3x - y

y1 = x + 4y

\(\begin{vmatrix} 3 & -1 \\ 1 & 4 \end{vmatrix} \begin{vmatrix} x \\ y = \end{vmatrix} \begin{vmatrix} x_1 \\ y_1 \end{vmatrix}\)

\(\begin{vmatrix} 3 & -1 \\ 1 & 4 \end{vmatrix} \begin{vmatrix} 2 \\ 1 = \end{vmatrix} \begin{vmatrix} 7 \\ -2 \end{vmatrix}\)

Correct Answer: D. 2?3

Explanation

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

x1 = x, x2 = -x, y1 = 4, y2 = 3, d = 7

7 = \(\sqrt{(-x -x)^2 + (3 - 4)^2}\)

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

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

square both sides

7\(^2\) = 4x\(^2\) + 1

collect like terms

4x\(^2\) = 49 - 1

4x\(^2\) = 48

x\(^2\) = \(\frac{48}{4}\)

x\(^2\) = 12

x = √12

x = 2√3

Explanation

\(2^{(2y+1)} - 5(2^y) + 2\) = 0

Let p = 2\(^y\)

\(2^{2y} (2^1) - 5(2^y)\) + 2 = 0

2p\(^2\) - 5p + 2 = 0

2p\(^2\) - p - 4p + 2 = 0

p (2p - 1) - 2(2p - 1) = 0

(p - 2)(2p - 1) = 0

p = 2 or \(\frac{1}{2}\)

p = 2\(^y\)

when p = 2

2\(^y\) = 2

y = 1

when p = \(\frac{1}{2}\)

2\(^y\) = \(\frac{1}{2}\)

2\(^y\) = 2\(^{-1}\)

y = -1

y = -1 or 1

Correct Answer: A. 10

Explanation

\((x + \frac{2}{x^2})^5\)

n = 5, r = 4, p = x and q = \(\frac{2}{x^2}\)

5C\(_4\)x\(^4\) (\(\frac{2}{x^2}\))1 = 5C\(_4\) \(\frac{2x^4}{x^2}\)

5C\(_4\) 2x\(^2\) = \(\frac{5!}{[5-4]!4!}\) * 2x\(^2\)

\(\frac{5*4!}{4!} * 2x^2\) = 5 * 2x\(^2\) = 10x\(^2\)

The coefficient is 10.

Correct Answer: B. (\(\frac{1}{4}\), \(\frac{3}{4}\))

Explanation

2x\(^2\) + 2y\(^2\) - x - 3y - 41

standard equation of circle

(x-a)\(^2\) + (x-b)\(^2\) = r\(^2\)

General form of equation of a circle.

x\(^2\) + y\(^2\) + 2gx + 2fy + c = 0

a = -g, b = -f., r2 = g2 + f2 - c

the centre of the circle is (a,b)

comparing the equation with the general form of equation of circle.

2x\(^2\) + 2y\(^2\) - x - 3y - 41

= x\(^2\) + y\(^2\) + 2gx + 2fy + c

2x\(^2\) + 2y\(^2\) - x - 3y - 41 = 0

divide through by 2

g = \(\frac{-1}{4}\) ; 2g = \(\frac{-1}{2}\)

f = \(\frac{-3}{4}\) ; 2f = \(\frac{-3}{2}\)

a = -g → - \(\frac{-1}{4}\) ; = \(\frac{1}{4}\)

b = -f → - (\frac{-3}{4}\) = (\frac{3}{4}\)

therefore the centre is (\(\frac{1}{4}\), \(\frac{3}{4}\))