-8
-5
-4
-3
Correct answer is B
\(\begin{vmatrix} 2 & -3 \\ 1 & 4 \end{vmatrix} \begin{vmatrix} -6 \\ k \end{vmatrix} \begin{vmatrix} 3 \\ -26 \end{vmatrix} = 15\)
\(\begin{vmatrix} 2[-6] & - 3k \\ 1[-6] & + 4k \end{vmatrix} = \begin{vmatrix} 3 \\ -26 \end{vmatrix}\)
\(\begin{vmatrix} -12 & - 3k \\ -6 & + 4k \end{vmatrix} = \begin{vmatrix} 3 \\ -26 \end{vmatrix}\)
-12 - 3k = 3
-3k = 3 + 12
k = \(\frac{15}{-3}\)
k = -5
Evaluate\({1_0^∫} x^2(x^3+2)^3\)
\(\frac{56}{12}\)
\(\frac{65}{12}\)
12
65
Correct answer is B
\({1_0^∫} x^2(x^3+2)^3\)dx
let \( u = x^3 + 2, du = 3x^2dx\)
when x = 1, u = 3
when x = 0, u = 2
dx = \(\frac{du}{3x^2}\)
\({3_2^∫}\) \(\frac{x^2[u]^3}{3x^2}\)
\({3_2^∫}\) \(\frac{u^3}{3}\) du
= \(\frac{u^4}{3*4}\)\(_2\)3
\(\frac{1}{12} [u^4]\)\(_2\)3
\(\frac{1}{12} [3^4 - 2^4]\)
\(\frac{1}{12}[81 - 16]\)
\(\frac{65}{12}\)
If \(x^2+y^2+-2x-6y+5 =0\), evaluate dy/dx when x=3 and y=2.
2
-2
-4
4
Correct answer is A
\(x^2+y^2+-2x-6y+5 =0\)
When differentiated:
\(x^2+y^2+-2x-6y+5 =0\) → 2x + 2y - 2 - 6 = 0
where x=3 and y=2
2[3] + 2[2] - 8 = 0
6 + 4 - 8 = 2
Given that \(\frac{8x+m}{x^2-3x-4} ≡ \frac{5}{x+1} + \frac{3}{x-4}\)
23
17
-17
18
Correct answer is C
\(\frac{8x+m}{x^2-3x-4} ≡ \frac{5}{x+1} + \frac{3}{x-4}\)
\(\frac{8x+m}{x^2-3x-4}\) ≡ \(\frac{5(x-1)+ 3(x+4)}{x^2-3x-4}\)
multiplying both sides by x2-3x-4
8x+m ≡ 5(x-4)+3(x+1)
8x + m ≡ 5x - 20 + 3x + 3
8x - 5x - 3x + m = -20 + 3
m = -17
Differentiate \(\frac{5x^ 3+x^2}{x}\), x ≠ 0 with respect to x.
10x + 1
10x + 2
x(15x + 1)
x(15x + 2)
Correct answer is A
\(\frac{5x^ 3+x^2}{x}\) → \(\frac{5x^ 3}{x} + \frac{x^2}{x}\)
5x\(^2\) + x
Then dy/dx = 10x + 1