\(\frac{1}{5}\)
\(\frac{2}{5}\)
\(\frac{8}{15}\)
\(\frac{14}{15}\)
Correct answer is D
P(at least one green wrapper) = 1 - P(no green wrapper)
= \(1 - (\frac{2}{6} \times \frac{1}{5})\)
= \(1 - \frac{1}{15}\)
= \(\frac{14}{15}\)
If \(f(x) = mx^{2} - 6x - 3\) and \(f'(1) = 12\), find the value of the constant m.
9
3
-3
-4
Correct answer is A
\(f(x) = mx^{2} - 6x - 3\)
\(f '(x) = \frac{\mathrm d y}{\mathrm d x} = 2mx - 6\)
\(f'(1) = 2m - 6 = 12\)
\(2m = 18 \implies m = 9\)
Evaluate \(\lim \limits_{x \to 3} \frac{x^{2} - 2x - 3}{x - 3}\)
4
3
2
0
Correct answer is A
\(\lim \limits_{x \to 3} \frac{x^{2} - 2x - 3}{x - 3}\)
\(\frac{x^{2} - 2x - 3}{x - 3} = \frac{x^{2} - 3x + x - 3}{x - 3}\)
\(\frac{(x - 3)(x + 1)}{x - 3} = x + 1\)
\(\lim \limits_{x \to 3} \frac{x^{2} - 2x - 3}{x - 3} \equiv \lim \limits_{x \to 3} (x + 1)\) (L'Hopital rule)
\(\lim \limits_{x \to 3} (x + 1) = 3 + 1 = 4\)
Calculate in surd form, the value of \(\tan 15°\).
\(2 + \sqrt{3}\)
\(1 + \sqrt{3}\)
\(\sqrt{3} - 1\)
\(2 - \sqrt{3}\)
Correct answer is D
\(\tan 15 = \tan (60 - 45)\)
\(\tan (x - y) = \frac{\tan x - \tan y}{1 + \tan x \tan y}\)
\(\tan (60 - 45) = \frac{\tan 60 - \tan 45}{1 + \tan 60 \tan 45}\)
= \(\frac{\sqrt{3} - 1}{1 + (\sqrt{3} \times 1)}\)
= \(\frac{\sqrt{3} - 1}{1 + \sqrt{3}}\)
Rationalizing by multiplying denominator and numerator by \(1 - \sqrt{3}\),
\(\tan 15 = 2 - \sqrt{3}\)
2
5
8
13
Correct answer is B
\((m_{1} v_{1} + m_{2} v_{2}) = (m_{1} + m_{2})v\) (Inelastic momentum)
\(8x + (5 \times 2) = (8 + 5) \times 3.85\)
\(8x + 10 = 13 \times 3.85 = 50.05\)
\(8x = 50.05 - 10 = 40.05\)
\(x \approxeq 5 m/s\)