\(f(x) = p + qx\), where p and q are constants. If f(1) = 7 and f(5) = 19, find f(3).
13
15
17
26
Correct answer is A
\(f(x) = p + qx\)
\(f(1) = p + q(1) \implies p + q = 7 .... (1)\)
\(f(5) = p + 5q = 19 .....(2)\)
Solving for p and q using simultaneous equation, p = 4, q = 3
\(f(3) = 4 + 3(3) = 13\)
The equation of a circle is \(3x^{2} + 3y^{2} + 6x - 12y + 6 = 0\). Find its radius
1
\(\sqrt{3}\)
\(\sqrt{11}\)
\(\sqrt{6}\)
Correct answer is B
The equation of a circle is given as \((x - a)^{2} + (y - b)^{2} = r^{2}\)
Expanding this, we have \(x^{2} + y^{2} - 2ax - 2by + a^{2} + b^{2} = r^{2}\)
Comparing with the given equation, \(3x^{2} + 3y^{2} + 6x -12y + 6 = 0 \equiv x^{2} + y^{2} + 2x - 4y + 2 = 0\) (making the coefficients of \(x^{2}\) and \(y^{2}\) = 1 , we get that
\(-2a = 2 \implies a = -1\)
\(2b = 4 \implies b = 2\)
\(r^{2} - a^{2} - b^{2} = -2\)
\(\therefore r^{2} - (-1)^{2} - (2)^{2} = -2 \implies r^{2} = -2 + 1 + 4 = 3\)
\(r = \sqrt{3}\)
\(x < -1, x < -\frac{1}{3}\)
\(x > -1, x > -\frac{1}{3}\)
\(x > \frac{1}{3}, x < -1\)
\(x < \frac{1}{3}, x > -1\)
Correct answer is B
\(3x^{2} + 4x + 1 > 0 \)
\(3x^{2} + 3x + x + 1 > 0\)
\(3x(x + 1) + 1(x + 1) > 0\)
\((3x + 1)(x + 1) > 0\)
\(3x + 1 > 0 \implies 3x > -1 \)
\(x > -\frac{1}{3}\)
\(x + 1 > 0 \implies x > -1\)
\(x > -1, x > -\frac{1}{3}\)
Simplify \(\sqrt[3]{\frac{8}{27}} - (\frac{4}{9})^{-\frac{1}{2}}\)
\(\frac{-5}{6}\)
\(-\frac{4}{27}\)
\(0\)
\(\frac{2}{9}\)
Correct answer is A
\(\sqrt[3]{\frac{8}{27}} - (\frac{4}{9})^{\frac{-1}{2}} \)
\(\frac{2}{3} - (\frac{9}{4})^{\frac{1}{2}}\)
= \(\frac{2}{3} - \frac{3}{2}\)
= \(\frac{-5}{6}\)
A function is defined by \(f(x) = \frac{3x + 1}{x^{2} - 1}, x \neq \pm 1\). Find f(-3).
\(-1\frac{1}{4}\)
\(-1\)
\(\frac{4}{5}\)
\(1\)
Correct answer is B
\(f(x) = \frac{3x + 1}{x^{2} - 1}\)
\(f(-3) = \frac{3(-3) + 1}{(-3)^{2} - 1} = \frac{-8}{8} = -1\)