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
9/16
81/16
9
9 \(\frac{9}{16}\)
Correct answer is D
GP : 36, P, \(\frac{q}{4}\), q, ... p + q = ?
| Recall, | common | ratio, | r | = | Tn
Tn-1 |
= | T2
T1 |
= | T3
T2 |
= | T4
T3 |
| ∴ | P
36 |
= | 9
4 |
÷ | p | ; | p\(^2\) | = | 9
4 |
x | 36 | ; | p\(^2\) | = | 81 |
| p | = | 9 | ∴ | r | = | T2
T1 |
= | 9
36 |
= | 1
4 |
| Also | r | = | T4
T3 |
= | q | ÷ | 9
4 |
∴ \(\frac{1}{4}\) = q ÷ \(\frac{9}{4}\) ;
\(\frac{9}{4}\) = 4q
| 16q | = | 9 | , | q | = | 9
16 |
∴ | p | + | q | = | 9 | + | 9
16 |
= | 9 | 9
16 |
Evaluate \(∫^0_{-1}\) (x + 1)(x - 2) dx
7/6
5/6
-5/6
-7/6
Correct answer is D
\(∫^0_{-1}\) (x + 1)(x - 2) dx
= \(∫^0_{-1}\) \(x^2 - x - 2\) dx
Integrated \(x^2 - x - 2\) = \(\frac{x^3}{3} - \frac{x^2}{2} -2\)