\(\frac{d}{dx} [\log (4x^3 - 2x)]\) is equal to
\(\frac{12x - 2}{4x^2}\)
\(\frac{43x^2 - 2x}{7x}\)
\(\frac{4x^2 - 2}{7x + 6}\)
\(\frac{12x^2 - 2}{4x^3 - 2x}\)
Correct answer is D
\(\frac{d}{dx} [\log (4x^3 - 2x)]\) ... (1)
Let u = 4x\(^3\) - 2x.
\(\frac{\mathrm d}{\mathrm d x} (\log (4x^3 - 2x)) = (\frac{\mathrm d}{\mathrm d u})(\frac{\mathrm d u}{\mathrm d x})\)
\(\frac{\mathrm d}{\mathrm d u} (\log u)\) = \(\frac{1}{u}\)
\(\frac{\mathrm d u}{\mathrm d x} = 12x^2 - 2\)
\(\therefore \frac{d}{dx} [\log (4x^3 - 2x)] = \frac{12x^2 - 2}{u}\)
= \(\frac{12x^2 - 2}{4x^3 - 2x}\)
If \(y = 6x^3 + 2x^{-2} - x^{-3}\), find \(\frac{\mathrm d y}{\mathrm d x}\).
\(\frac{\mathrm d y}{\mathrm d x} = 15x^2 - 4x^{-2} - 3x^{-2}\)
\(\frac{\mathrm d y}{\mathrm d x} = 6x + 4x^{-1} - 3x^{-4}\)
\(\frac{\mathrm d y}{\mathrm d x} = 18x^2 - 4x^{-3} + 3x^{-4}\)
\(\frac{\mathrm d y}{\mathrm d x} = 12x^2 + 4x^{-1} - 3x^{-2}\)
Correct answer is C
\(y = 6x^3 + 2x^{-2} - x^{-3}\)
\(\frac{\mathrm d y}{\mathrm d x} = 18x^2 - 4x^{-3} + 3x^{-4}\)
If \(\begin{vmatrix} 2 & -4 \\ x & 9 \end{vmatrix} = 58\), find the value of x.
10
30
14
28
Correct answer is A
\(\begin{vmatrix} 2 & -4 \\ x & 9 \end{vmatrix} = 58\)
\(\implies (2 \times 9) - (-4 \times x) = 58\)
\(18 + 4x = 58 \implies 4x = 58 - 18 = 40\)
\(x = 10\)
If P(2, m) is the midpoint of the line joining Q(m, n) and R(n, -4), find the values of m and n.
m = 0, n = 4
m = 4, n = 0
m = 2, n = 2
m = -2, n = 4
Correct answer is A
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.
y = 5x + 2
y = 5x + 3
y = 12x + 2
y = 22x + 3
Correct answer is A
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 \(y = 5x + 2\).