What value of g will make the expression 4x2 - 18xy + g a perfect square?
9
\(\frac{9y^2}{4}\)
\(81y^2\)
\(\frac{18y^2}{4}\)
Correct answer is D
4x2 - 18xy + g = g \(\to\) (\(\frac{18y}{4}\))2
= \(\frac{18y^2}{4}\)
Make F the subject of the formula t = \(\sqrt{\frac{v}{\frac{1}{f} + \frac{1}{g}}}\)
\(\frac{gv-t^2}{gt^2}\)
\(\frac{gt^2}{gv-t^2}\)
\(\frac{v}{\frac{1}{t^2} - \frac{1}{g}}\)
\(\frac{gv}{t^2 - g}\)
Correct answer is B
t = \(\sqrt{\frac{v}{\frac{1}{f} + \frac{1}{g}}}\)
t2 = \(\frac{v}{\frac{1}{f} + \frac{1}{g}}\)
= \(\frac{vfg}{ftg}\)
\(\frac{1}{f} + \frac{1}{g}\) = \(\frac{v}{t^2}\)
= (g + f)t2 = vfg
gt2 = vfg - ft2
gt2 = f(vg - t2)
f = \(\frac{gt^2}{gv-t^2}\)
Find the minimum value of X2 - 3x + 2 for all real values of x
-\(\frac{1}{4}\)
-\(\frac{1}{2}\)
\(\frac{1}{4}\)
\(\frac{1}{2}\)
Correct answer is A
y = X2 - 3x + 2, \(\frac{dy}{dx}\) = 2x - 3
at turning pt, \(\frac{dy}{dx}\) = 0
∴ 2x - 3 = 0
∴ x = \(\frac{3}{2}\)
\(\frac{d^2y}{dx^2}\) = \(\frac{d}{dx}\)(\(\frac{d}{dx}\))
= 270
∴ ymin = 2\(\frac{3}{2}\) - 3\(\frac{3}{2}\) + 2
= \(\frac{9}{4}\) - \(\frac{9}{2}\) + 2
= -\(\frac{1}{4}\)
x = \(\frac{3}{2}\), y = \(\frac{3}{2}\)
x = \(\frac{1}{2}\), y = \(\frac{3}{2}\)
x = \(\frac{-1}{2}\), y = \(\frac{-3}{2}\)
x = \(\frac{1}{3}\), y = \(\frac{3}{2}\)
Correct answer is B
\(\frac{2}{x} - {\frac{3}{y}}\) = 2.....(1)
\(\frac{4}{x} + {\frac{3}{y}}\) = 10 ... (2)
(1) + (2):
\(\frac{6}{x}\) = 12 \(\to\) x = \(\frac{6}{12}\)
x = \(\frac{1}{2}\)
put x = \(\frac{1}{2}\) in equation (i)
= 4 - \(\frac{3}{y}\) = 2
= 4 - 2
= \(\frac{3}{y}\)
therefore y = \(\frac{3}{2}\)
If the function f(fx) = x3 + 2x2 + qx - 6 is divisible by x + 1, find q
-5
-2
2
5
Correct answer is A
x + 1 = 0, x = -1; f(x) = x3 + 2x2 + qx - 6
0 = -1 + 2 - q - 6
q = -5