Given that \(f '(x) = 3x^{2} - 6x + 1\) and f(3) = 5, find f(x).
\(f(x) = x^{3} - 3x^{2} + x + 20\)
\(f(x) = x^{3} - 3x^{2} + x + 31\)
\(f(x) = x^{3} - 3x^{2} + x + 2\)
\(f(x) = x^{3} - 3x^{2} + x - 13\)
Correct answer is C
\(f ' (x) = 3x^{2} - 6x + 1\)
\(f(x) = \int (3x^{2} - 6x + 1) \mathrm {d} x\)
= \(x^{3} - 3x^{2} + x + c\)
\(f(3) = 5 = 3^{3} - 3(3^{2}) + 3 + c\)
\(27 - 27 + 3 + c = 5 \implies 3 + c = 5\)
\(c = 2\)
\(f(x) = x^{3} - 3x^{2} + x + 2\)
2
\(\frac{1}{2}\)
\(-\frac{1}{2}\)
-2
Correct answer is B
\((1 + kx)^{8} = ^{8}C_{0}(1^{8})(kx)^{0} + ^{8}C_{1}(1^{7})(kx)^{1} + ...\)
The 5th term = \(^{8}C_{5 - 1}(1^{4})(kx)^{4}\)
= \(^{8}C_{4} (kx)^{4}\)
\(\implies 70k^{4} = \frac{35}{8}\)
\(k^{4} = \frac{\frac{35}{8}}{70}\)
\(k^{4} = \frac{1}{16}\)
\(k = \frac{1}{2}\)
Find the derivative of \(3x^{2} + \frac{1}{x^{2}}\)
\(6x + 2x^{2}\)
\(6x + \frac{1}{2x}\)
\(6x - \frac{2}{x^{3}}\)
\(6x - \frac{1}{2x}\)
Correct answer is C
\(y = 3x^{2} + \frac{1}{x^{2}} \equiv y = 3x^{2} + x^{-2}\)
\(\frac{\mathrm d y}{\mathrm d x} = 6x - 2x^{-3}\)
= \(6x - \frac{2}{x^{3}}\)
7
2
-2
-7
Correct answer is B
\(a * b = a^{2} + b + ab\)
\(5 * x = 5^{2} + x + 5x = 37\)
\(25 + 6x = 37 \implies 6x = 12\)
\(x = 2\)
Find the locus of points which is equidistant from P(4, 5) and Q(-6, -1).
3x - 5y + 13 = 0
3x - 5y - 7 = 0
5x - 3y + 7 = 0
5x + 3y - 1 = 0
Correct answer is D
Midpoint between P(4, 5) and Q(-6, -1) = \(\frac{4 - 6}{2}, \frac{5 - 1}{2} = (-1, 2)\)
Gradient of PQ = \(\frac{5 - (-1)}{4 - (-6)} = \frac{3}{5}\)
Gradient of line perpendicular to PQ = \(\frac{-1}{\frac{3}{5}} = -\frac{5}{3}\)
\(line = \frac{y - 2}{x + 1} = \frac{-5}{3}\)
\(3y - 6 = -5x - 5\)
\(5x + 3y - 6 + 5 = 0 \implies 5x + 3y - 1 = 0\)