Mathematics Questions and Answers

\(\frac{d}{dx} [\log (4x^3 - 2x)]\) ... (1)

Let u = 4x\(^3\) - 2x.

\(\frac{\mathrm d}{\mathrm d x} (\log (4x^3 - 2x)) = (\frac{\mathrm d}{\mathrm d u})(\frac{\mathrm d u}{\mathrm d x})\)

\(\frac{\mathrm d}{\mathrm d u} (\log u)\) = \(\frac{1}{u}\)

\(\frac{\mathrm d u}{\mathrm d x} = 12x^2 - 2\)

\(\therefore \frac{d}{dx} [\log (4x^3 - 2x)] = \frac{12x^2 - 2}{u}\)

= \(\frac{12x^2 - 2}{4x^3 - 2x}\)

\(y = 6x^3 + 2x^{-2} - x^{-3}\)

\(\frac{\mathrm d y}{\mathrm d x} = 18x^2 - 4x^{-3} + 3x^{-4}\)

\(\begin{vmatrix} 2 & -4 \\ x & 9 \end{vmatrix} = 58\)

\(\implies (2 \times 9) - (-4 \times x) = 58\)

\(18 + 4x = 58 \implies 4x = 58 - 18 = 40\)

\(x = 10\)

Q(m, n) and R(n, -4)

Midpoint : P(2, m)

\(\implies (\frac{m + n}{2}, \frac{n - 4}{2}) = (2, m)\)

\(m + n = 2 \times 2 \implies m + n = 4 ... (i)\)

\(n - 4 = 2 \times m \implies n - 4 = 2m ... (ii)\)

Solving (i) and (ii) simultaneously,

m = 0 and n = 4.

The equation of a straight line is given as \(y = mx + b\) 

where m = the slope of the line

b = intercept

Given points A(3, 12) and B(5, 22), the slope = \(\frac{22 - 12}{5 - 3}\)

= \(\frac{10}{2}\) = 5

Hence, the equation of the line is \(y = 5x + 2\).