12x−24x2
43x2−2x7x
4x2−27x+6
12x2−24x3−2x
Correct answer is D
ddx[log(4x3−2x)] ... (1)
Let u = 4x3 - 2x.
ddx(log(4x3−2x))=(ddu)(dudx)
ddu(logu) = 1u
dudx=12x2−2
∴
= \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.