11\(cm^{2}s^{-1}\)
22\(cm^{2}s^{-1}\)
33\(cm^{2}s^{-1}\)
44\(cm^{2}s^{-1}\)
Correct answer is B
With radius = 7cm, \(Area = \pi r^{2} = \frac{22}{7} \times 7^{2}\)
= \(154cm^{2}\)
The next second, radius = 7.5cm, \(Area = \pi r^{2} = \frac{22}{7} \times 7.5^{2}\)
= \(176cm^{2}\)
Change in area = \((176 - 154)cm^{2} = 22cm^{2}\)
\(\therefore\) The rate of increase = \(22cm^{2}s^{-1}\)
OR
\(Area (A) = \pi r^{2} \implies \frac{\mathrm d A}{\mathrm d r} = 2\pi r\)
Given \(\frac{\mathrm d r}{\mathrm d t} = 0.5\)
\(\frac{\mathrm d A}{\mathrm d r} \times \frac{\mathrm d r}{\mathrm d t} = \frac{\mathrm d A}{\mathrm d t}\)
\(\frac{\mathrm d A}{\mathrm d t} = 2\pi r \times 0.5 = 2 \times \frac{22}{7} \times 7 \times 0.5\)
= \(22cm^{2}s^{-1}\)
Find the minimum value of \(y = 3x^{2} - x - 6\).
\(-6\frac{1}{6}\)
\(-6\frac{1}{12}\)
\(-6\)
\(0\)
Correct answer is B
\(3x^{2} - x - 6 = y\)
\(a = 3, b = -1, c = -6\)
Minimum value of x = \(\frac{-b}{2a} = \frac{-(-1)}{6} = \frac{1}{6}\)
\(y(\frac{1}{6}) = 3(\frac{1}{6}^{2}) - \frac{1}{6} - 6 = -6\frac{1}{12}\)
Find the gradient to the normal of the curve \(y = x^{3} - x^{2}\) at the point where x = 2.
\(\frac{-1}{8}\)
\(\frac{1}{8}\)
\(\frac{-1}{24}\)
\(1\)
Correct answer is A
Given : \(y = x^{3} - x^{2}\)
\(\frac{\mathrm d y}{\mathrm d x} = 3x^{2} - 2x\)
\(\therefore \text{The gradient of the tangent at point (x = 2)} = 3(2^{2}) - 2(2) \)
= \(12 - 4 = 8\)
Recall, the tangent and the normal are perpendicular to each other and the product of the gradients of perpendicular lines = -1.
\(\implies \text{the gradient of the normal} = \frac{-1}{8}\)
Find \(\lim\limits_{x \to 3} \frac{2x^{2} + x - 21}{x - 3}\).
0
1
7
13
Correct answer is D
\(\lim\limits_{x \to 3} \frac{2x^{2} + x - 21}{x - 3}\)
\(2x^{2} + x - 21 = 2x^{2} - 6x + 7x - 21 \) (by factorizing)
= \((2x + 7)(x - 3)\)
\(\therefore \lim\limits_{x \to 3} \frac{2x^{2} + x - 21}{x - 3} \equiv \lim\limits_{x \to 3} \frac{(2x+7)(x-3)}{x-3}\)
\(\lim\limits_{x \to 3} (2x + 7) = 2(3) + 7 = 13\)
(4, 2)
(4, -2)
(-4, 2)
(-4, -2)
Correct answer is A
The point of intersection for the two lines exists at the point where the two lines are equal to each other. Make anyone of the variables the subject of the formula and equate the two lines to each other and solve for the coordinates of the point of intersection.
Given lines \(2y + 3x - 16 = 0\) and \(7y - 2x - 6 = 0\) , making x the subject of the formula:
Line 1 : \(2y + 3x - 16 = 0 \implies 3x = 16 - 2y\)
\(\therefore x = \frac{16}{3} - \frac{2}{3}y\)
Line 2 : \(7y - 2x - 6 = 0 \implies -2x = 6 - 7y\)
\(\therefore x = \frac{7}{2}y - 3\)
Equating them together and solving, we have:
\(\frac{16}{3} - \frac{2}{3}y = \frac{7}{2}y - 3 \implies \frac{16}{3} + 3 = \frac{7}{2}y + \frac{2}{3}y\)
\(\frac{25}{3} = \frac{25}{6}y \therefore y = 2\)
Putting y = 2 in the equation \(3x = 16 - 2y\), we have
\(3x = 16 - 2(2) = 16 - 4 = 12 \implies x = 4\)
The coordinate of P is (4, 2).