Find the coordinates of the centre of the circle \(3x^{2}+3y^{2} - 4x + 8y -2=0\)
(-2,4)
(\(\frac{-2}{3}, \frac{4}{3}\))
(\(\frac{2}{3}, \frac{-4}{3}\))
(2, -4)
Correct answer is C
The equation for a circle with centre coordinates (a, b) and radius r is
\((x-a)^{2} + (y-b)^{2} = r^{2}\)
Expanding the above equation, we have
\(x^{2} - 2ax +a^{2} + y^{2} - 2by + b^{2} - r^{2} = 0\) so that
\(x^{2} - 2ax + y^{2} - 2by = r^{2} - a^{2} - b^{2}\)
Taking the original equation given, \(3x^{2} + 3y^{2} - 4x + 8y = 2\) and making the coefficients of \(x^{2}\) and \(y^{2}\) = 1,
\(x^{2} + y^{2} - \frac{4x}{3} + \frac{8y}{3} = \frac{2}{3}\), comparing, we have
\(2a = \frac{4}{3}; 2b = \frac{-8}{3}\)
\(\implies a = \frac{2}{3}; b = \frac{-4}{3}\)
If \(x^{2} - kx + 9 = 0\) has equal roots, find the values of k.
3, 4
±3
±5
±6
Correct answer is D
For equal roots, we have that \(b^{2} = 4ac\), so, given a=1, b = -k and c = 9,
\((-k)^{2} = 4\times1\times9 \implies k^{2} = 36\)
\(k = \sqrt{36} = \pm6\)
Simplify \(\frac{1}{(1-\sqrt{3})^{2}}\)
\(1- \frac{1}{2}\sqrt{3}\)
\(1+ \frac{1}{2}\sqrt{3}\)
\(\sqrt{3}\)
\(1+\sqrt{3}\)
Correct answer is B
\(\frac{1}{(1-\sqrt{3})^{2}}\)
\((1-\sqrt{3})^{2} = (1-\sqrt{3})(1-\sqrt{3})\)
\(1 - 2\sqrt{3} + 3 = 4 - 2\sqrt{3}\)
\(\frac{1}{4-2\sqrt{3}}\)
After rationalising (multiplying the denominator and numerator with \(4+2\sqrt{3}\), we have
\(\frac{4+2\sqrt{3}}{4} = 1 + \frac{1}{2}\sqrt{3}\)
\(-4\sqrt{3}\)
\(\frac{-4\sqrt{3}}{3}\)
\(\frac{-3\sqrt{3}}{4}\)
\(\frac{-3\sqrt{3}}{4}\)
Correct answer is B
\(a \Delta b\) = \(\frac{a+b}{\sqrt{ab}}\)
\(-3\Delta -1\) = \(\frac{-3 + -1}{\sqrt{-3\times -1}}\)
\(\frac{-4}{\sqrt{3}}\), rationalising, we have
\(\frac{-4 \times \sqrt{3}}{\sqrt{3}\times \sqrt{3}} = \frac{-4\sqrt{3}}{3}\)
If \(log_{y}\frac{1}{8}\) = 3, find the value of y.
-2
-\(\frac{1}{2}\)
\(\frac{1}{2}\)
2
Correct answer is C
\(log_{y}\frac{1}{8} = 3 \implies y^{3} = \frac{1}{8}\) (Laws of logarithm)
\(y^{3} = \frac{1}{2^{3}} = (\frac{1}{2})^{3}\)
Equating both sides, we have
\(y = \frac{1}{2}\)