Let f(x) = 2x + 4 and g(x) = 6x + 7 here g(x) > 0. Solve the inequality \(\frac{f(x)}{g(x)}\) < 1
x < - \(\frac{3}{4}\)
x > - \(\frac{4}{3}\)
x > - \(\frac{3}{4}\)
x > - 12
Correct answer is C
\(\frac{f(x)}{g(x)}\) < 1
∴ \(\frac{2x +4}{6x + 7}\) < 1
= 2x + 4 < 6x + 7
= 6x + 7 > 2x + 4
= 6x - 2x > 4 - 7
= 4x > -3
∴ x > -\(\frac{3}{4}\)
3
2
1
-1
Correct answer is A
5 + 2r = k(r + 1) + L(r - 2)
but r - 2 = 0 and r = 2
9 = 3k
k = 3
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
= -\(\frac{1}{4}\)