Correct Answer: C. 4.7m²

Explanation

(a)

(b)

(c)(i) 2x\(^{2}\) + x = 10 \(\equiv\) 2x\(^{2}\) + x - 10 = 0

Thus, the solution of 2x\(^{2}\) + x - 10 = 0 are the values of x at which the curve cuts the x- axis.

x = -2.5 or x = 2.

(ii) 2x\(^{2}\) + x - 10 = 2x

To solve the given equation, we first draw the graph of y = 2x: when x = -2, y = 2(-2) = -4; when x = 3, y = 2(3) = 6.

The values of x at which y = 2x\(^{2}\) + x - 10 and y = 2x intersect give the solution of 2x\(^{2}\) + x - 10 = 2x.

These are x = -2 or x = 2.5.

Correct Answer: C. 3

Explanation

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

\(\begin{vmatrix} k & k \\ 4 & k \end{vmatrix} = (k^{2} - 4k)\)

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

\(\therefore (k^{2} - 4k) + (2k + 3) = k^{2} - 2k + 3 = 6\)

\(k^{2} - 2k - 3 = 0\), factorising, we have \(k + 1 = 0\) or \(k - 3 = 0\)

Since k > 0, k = 3.

Correct Answer: C. {3, 9, 15}

Correct Answer: C. 120

Explanation

In a circular seating arrangement, we fix the position of one person and then rotate the others, so we have

\((6-1)! = 5! = 5\times4\times3\times2 = 120\)

Correct Answer: A. 3

Correct Answer: C. 9

Explanation

Total no of balls = 30

Let x = no. of red balls

Pr(red) = \(\frac{x}{30}\)

Pr(black) = \(\frac{3}{10} = \frac{9}{30}\)

Pr(white) = \(\frac{2}{5} = \frac{12}{30}\)

No. of black balls = 9

No. of white balls = 12

9 = 12 + x = 30

x = 30 - 21

x = 9

No. of red balls = 9

Correct Answer: A. 61.5%

Explanation

(a) \(\sqrt{50} - 3\sqrt{2}(2\sqrt{2} - 5) - 5\sqrt{32}\)

\(\sqrt{25 \times 2} - (3 \times 2)\sqrt{2 \times 2} + (3 \times 5)\sqrt{2} - 5\sqrt{16 \times 2}\)

= \(5\sqrt{2} - 6(2) + 15\sqrt{2} - 5(4\sqrt{2})\)

= \(5\sqrt{2} - 12 + 15\sqrt{2} - 20\sqrt{2}\)

= \((5 + 15 - 20)\sqrt{2} - 12\)

= -12.

(b) \(\frac{1}{2} \log_{10} (\frac{25}{4}) - 2 \log_{10} (\frac{4}{5}) + \log_{10} (\frac{320}{125})\)

= \(\log_{10} (\frac{25}{4})^{\frac{1}{2}} - \log_{10} (\frac{4}{5})^{2} + \log_{10} (\frac{320}{125})\)

= \(\log_{10} (\frac{5}{2}) - \log_{10} (\frac{16}{25}) + \log_{10} (\frac{320}{125})\)

= \(\log_{10} (\frac{\frac{5}{2} \times \frac{320}{125}}{\frac{16}{25}}\)

= \(\log_{10} (\frac{\frac{160}{25}}{\frac{16}{25}}\)

= \(\log_{10} 10 = 1\)

Correct Answer: B. 6y + x - 31 = 0

Explanation

y = \(3x^2 + 2\)

\(y^1 = \frac{dy}{dx} = 6x\)

Evaluating this derivative at x = 1 (since the point of interest is (1, 5)) gives us the slope of the tangent line at that point:

\(^mtangent = y^1(1) = 6 (1) = 6\)

Slope of the Normal Line \( ^mnorma l= - \frac{1}{^mtangent}\)

\(^mnormal = - \frac{1}{6}\)

\(y−y_1= ^mnormal⋅(x−x_1)\)

=y-5=-\(\frac{1}{6}(x-1)\)

=y-5=-\(\frac{1}{6}x+\frac{1}{6}\)

=y=-\(\frac{1}{6}x+\frac{1}{6}+5\)

Multiply through by 6

=6y=-x+1+30

∴6y+ x - 31=0