The solution of the quadratic inequality (x3 + x - 12) ≥ 0 is
x ≥ -3 or x ≤ 4
x ≥ 3 or x ≥ -4
x ≤ 3 or x v -4
x ≥ 3 or x ≤ -4
Correct answer is B
(x3 + x - 12) ≥ 0
(x + 4)(x - 3) ≥ 0
Either x + 4 ≥ 0 implies x ≥ -4
Or x - 3 ≥ 0 implies x ≥ 3
∴ x ≥ 3 or x ≥ -4
Make L the subjects of the formula if \(\sqrt{\frac{42w}{5l}}\)
\(\sqrt{\frac{42w}{5d}}\)
\(\frac{42W}{5d^2}\)
\(\frac{42}{5dW}\)
\(\frac{1}{d}\sqrt{\frac{42w}{5}}\)
Correct answer is B
\(\sqrt{\frac{42w}{5l}}\)
square both side of the equation
\(d^2 = \left(\sqrt{\frac{42W}{5l}}\right)^2\\
d^2 = \frac{42W}{5l}\\
5ld^2=42W\\
l = \frac{42W}{5d^2}\)
3/4
-9/2
45/4
-3/4
Correct answer is A
x⊕y = xy + x + y
= -3/4 (6) + (-3/4) + 6
= -9/2 - 3/4 + 6
= (-18-3+3+24) / 4
= 3/4
Determine the value of \(\int_0 ^{\frac{\pi}{2}
}(-2cos x)dx\)
-2
-1/2
-3
-3/2
Correct answer is A
\(\int_0 ^{\frac{\pi}{2}}(-2cos x)dx = [-2sin x + c]_0 ^{\frac{\pi}{2}}\\
=(-2sin\frac{\pi}{2}+c+2sin0-c)\\
=-2sin90+c+2sin0-c\\
=-2(1)+2(0)\\
=-2\)
Find the value of x for which the function f(x) = 2x3 - x2 - 4x + 4 has a maximum value
2/3
1
- 2/3
-1
Correct answer is B
f(x) = 2x3 - x2 - 4x – 4
f’(x) = 6x2 - 2x – 4
As f’(x) = 0
Implies 6x2 - 2x – 4 = 0
3x – x – 2 = 0 (By dividing by 2)
(3x – 2)(x + 1) = 0
3x – 2 = 0 implies x = -2/3
Or x + 1 = 0 implies x = -1
f’(x) = 6x2 - 2x – 4
f’’(x) = 12x – 2
At max point f’’(x) < 0
∴f’’(x) = 12x – 2 at x = -1
= 12(-1) – 2
= -12 – 2 = -14
∴Max at x = 1