If \(4x^{2} + 5kx + 10\) is a perfect square, find the value of k
\(\frac{5\sqrt{10}}{4}\)
\(4\sqrt{10}\)
\(5\sqrt{10}\)
\(\frac{4\sqrt{10}}{5}\)
Correct answer is D
\(4x^{2} + 5kx + 10 = (2x + \sqrt{10})^{2}\)
Expanding the right hand side equation, we have
\(4x^{2} + 4x\sqrt{10} + 10\)
Comparing with the left hand side, we have
\(5k = 4\sqrt{10} \implies k = \frac{4}{5}\sqrt{10}\)
6
5
4
3
Correct answer is D
\((\sqrt{x} + 1) * (\sqrt{x} - 1) = 4 \implies \frac{\sqrt{x} + 1}{\sqrt{x} - 1} + \frac{\sqrt{x} - 1}{\sqrt{x} + 1} = 4\)
\(\frac{(\sqrt{x} + 1)(\sqrt{x} + 1) + (\sqrt{x} - 1)(\sqrt{x} - 1)}{(\sqrt{x} - 1)(\sqrt{x} + 1)}\)
= \(\frac{x + 2\sqrt{x} + 1 + x - 2\sqrt{x} + 1}{x - 1} \implies \frac{2x + 2}{x - 1} = 4\)
\(2x + 2 = 4x - 4 \therefore 4x - 2x = 2x = 2 + 4= 6\)
\(x = 3\)
Given that \(f(x) = 3x^{2} - 12x + 12\) and \(f(x) = 3\), find the values of x.
1, 3
-1, -3
1, -3
-1, 3
Correct answer is A
\(f(x) = 3x^{2} - 12x + 12\) and \(f(x) = 3\)
\(\therefore f(x) = 3 = 3x^{2} - 12x + 12 \implies 3x^{2} - 12x + 12 - 3 = 0\)
\(3x^{2} - 12x + 9 = 0; 3x^{2} - 9x - 3x + 9 = 0\)
\(3x(x - 3) - 3(x - 3) = (3x - 3)(x - 3) = 0\)
3x - 3 = 0 or x - 3 = 0
\(x = 1 or 3\)
Find the domain of \(g(x) = \frac{4x^{2} - 1}{\sqrt{9x^{2} + 1}}\)
\({x : x \in R, x = \frac{1}{2}}\)
\(x: x \in R, x\neq \frac{1}{3}\)
\(x : x \in R, x = \frac{1}{3}\)
\(x: x \in R\)
Correct answer is D
The domain of a function refers to the regions where the function is defined or has a value on a particular region.
\(\frac{4x^{2} - 1}{\sqrt{9x^{2} + 1}}\) has a domain defined on all set of real numbers because the function is defined on the set of real numbers when the denominator \(\sqrt{9x^{2} + 1} \geq 0\).
\(\sqrt{9x^{2} + 1} \geq 0 \implies 9x^{2} + 1 \geq 0\) which because of the square sign has a value for all values of x, be it negative or positive.
Simplify \(\frac{\sqrt{3}}{\sqrt{3} -1} + \frac{\sqrt{3}}{\sqrt{3} + 1}\)
\(\frac{1}{2}\)
3
\(2\sqrt{3}\)
6
Correct answer is B
\(\frac{\sqrt{3}}{\sqrt{3} - 1} + \frac{\sqrt{3}}{\sqrt{3} + 1}\)
= \(\frac{\sqrt{3}(\sqrt{3} + 1) + \sqrt{3}(\sqrt{3} - 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)}\)
= \(\frac{3 + \sqrt{3} + 3 - \sqrt{3}}{3 + \sqrt{3} - \sqrt{3} - 1}\)
= \(\frac{6}{2} = 3\)