Evaluate \(\lim \limits_{x \to 1} \frac{1 - x}{x^{2} - 3x + 2}\)
1
\(\frac{1}{2}\)
0
-1
Correct answer is A
\(\lim \limits_{x \to 1} \frac{1 - x}{x^{2} - 3x + 2}\)
\(\frac{1 - x}{x^{2} - 3x + 2} = \frac{-(x - 1)}{(x - 1)(x - 2)}\)
= \(\frac{-1}{x - 2}\)
\(\lim \limits_{x \to 1} \frac{1 - x}{x^{2} - 3x + 2} = \lim \limits_{x \to 1} \frac{-1}{x - 2}\)
= \(\frac{-1}{1 - 2} = \frac{-1}{-1} = 1\)
\(\frac{3}{2}\)
\(\frac{2}{3}\)
\(-\frac{2}{3}\)
\(-\frac{3}{2}\)
Correct answer is D
\(\log_{0.25} 8 = x\)
\(8 = 0.25^{x}\)
\(2^{3} = (2^{-2})^{x} \implies 3 = -2x\)
\(x = -\frac{3}{2}\)
Find the sum of the exponential series \(96 + 24 + 6 +...\)
144
128
72
64
Correct answer is B
\(S_{\infty} = \frac{a}{1 - r}\) (for an exponential series)
\(r = \frac{24}{96} = \frac{6}{24} = \frac{1}{4}\)
\(S_{\infty} = \frac{96}{1 - \frac{1}{4}} = \frac{96}{\frac{3}{4}}\)
= \(\frac{96 \times 4}{3} = 128\)
\(\pm 16\)
\(\pm 8\)
\(\pm 4\)
\(\pm 2\)
Correct answer is C
\(2x^{2} + kx + 5 = 0\)
\(\alpha + \beta = \frac{-b}{a} = \frac{-k}{2}\)
\(\alpha \beta = \frac{c}{a} = \frac{5}{2}\)
\(\alpha^{2} + \beta^{2} = (\alpha + \beta)^{2} - 2\alpha \beta\)
\(-1 = (\frac{-k}{2})^{2} - 2(\frac{5}{2})\)
\(-1 = \frac{k^{2}}{4} - 5 \implies \frac{k^{2}}{4} = 4\)
\(k^{2} = 16 \therefore k = \pm 4\)
Find the equation of the line passing through (0, -1) and parallel to the y- axis.
y = -1
y = 0
x = 0
x = -1
Correct answer is C
No explanation has been provided for this answer.