Given that \(f(x) = 5x^{2} - 4x + 3\), find the coordinates of the point where the gradient is 6.
(4,1)
(4,-2)
(1,4)
(1,-2)
Correct answer is C
\(f(x) = 5x^{2} - 4x + 3\)
\(f'(x) = 10x - 4 = 6 \implies 10x = 10; x=1\)
When x = 1, f(x) = y = \(5(1^{2}) - 4(1) + 3 = 4\)
The coordinates are (1,4)
\(x^{2} + y^{2} + 8x - 10y + 21 = 0\)
\(x^{2} + y^{2} + 8x - 10y - 21 = 0\)
\(x^{2} + y^{2} - 8x - 10y - 21 = 0\)
\(x^{2} + y^{2} - 8x - 10y + 21 = 0\)
Correct answer is D
On the y-intercept, x=0. At this point, y = 5(0) - 2y = -6; y = 3, so the y- intercept has coordinates (0,3).
The equation of the circle is given as \((x-a)^{2} + (y-b)^{2} = r^{2}\), where (a,b) is the centre and r is the radius.
The radius of the circle through the y- intercept = \(\sqrt{(4-0)^{2} + (5-3)^{2}} = \sqrt{20}\)
The equation of the circle is \((x-4)^{2} + (y-5)^{2} = 20\); expanding, we have
\(x^{2} + y^{2} + 8x - 10y + 21 = 0\)
\(\frac{130}{221}\)
\(\frac{140}{221}\)
\(\frac{140}{204}\)
\(\frac{220}{23}\)
Correct answer is B
\(\cos(x+y) = \cos x\cos y - \sin x\sin y\)
Given \(\sin\) of an angle implies we have the value of the opposite and hypotenuse of the right-angled triangle. We find the adjacent side using Pythagoras' theorem.
\(Adj^{2} = Hyp^{2} - Opp^{2}\)
For triangle with angle x, \(adj = \sqrt{13^{2} - 5^{2}} = \sqrt{144} = 12\)
For triangle with angle y, \(adj = \sqrt{17^{2} - 8^{2}} = \sqrt{225} = 15\)
\(\therefore \cos x = \frac{12}{13}; \cos y = \frac{15}{17}\)
\(\cos(x+y) = (\frac{12}{13}\times\frac{15}{17}) - (\frac{5}{13}\times\frac{8}{17}) = \frac{180}{221} - \frac{40}{221}\)
= \(\frac{140}{221}\)
If \(B = \begin{pmatrix} 2 & 5 \\ 1 & 3 \end{pmatrix}\), find \(B^{-1}\).
\(A = \begin{pmatrix} -3 & -5 \\ 1 & 2 \end{pmatrix}\)
\(A = \begin{pmatrix} 3 & -5 \\ 1 & 2 \end{pmatrix}\)
\(A = \begin{pmatrix} 3 & -5 \\ -1 & 2 \end{pmatrix}\)
\(A = \begin{pmatrix} -3 & 5 \\ 1 & -2 \end{pmatrix}\)
Correct answer is C
\(B.B^{-1} = 1\), let \(B^{-1} = \begin{pmatrix} a & b \\ c & d \end{pmatrix}\)
\(\begin{pmatrix} 2 & 5 \\ 1 & 3 \end{pmatrix}\)\(\begin{pmatrix} a & b \\ c & d \end{pmatrix}\) = \(\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\)
Multiplying \(B \times B^{-1}\), we have the following equations:
\(2a+5c = 1......... (1)\); \(a+3c = 0 ........(2)\)
\(2b+5d = 0 ..........(3)\); \(b+3d = 1..........(4)\)
Solving the equations simultaneously, we have
\(a = 3; b = -5; c = -1; d = 2 \implies B^{-1} = \begin{pmatrix} 3 & -5 \\ -1 & 2 \end{pmatrix}\)
\(\frac{-9}{8}\)
\(\frac{-7}{8}\)
\(\frac{7}{8}\)
\(\frac{9}{8}\)
Correct answer is B
\(\frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\alpha^{2} + \beta^{2}}{\alpha\beta}\)
\(\alpha + \beta = \frac{-b}{a}\); \(\alpha\beta = \frac{c}{a}\)
\(\alpha^{2} + \beta^{2} = (\alpha + \beta)^{2} - 2(\alpha\beta)\)
From the equation, a = 2, b = -3, c =4
\(\alpha + \beta = \frac{-(-3)}{2} = \frac{3}{2}\)
\(\alpha\beta = \frac{4}{2} = 2\)
\(\alpha^{2} + \beta^{2} = (\frac{3}{2})^{2} - 2(2)) = \frac{9}{4} - 4 = \frac{-7}{4}\)
\(\implies \frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\frac{-7}{4}}{2} = \frac{-7}{8}\)