If \(3x - \frac{1}{4})^{\frac{1}{2}} > \frac{1}{4} - x \), then the interval of values of x is
x > \(\frac{1}{3}\)
x < \(\frac{1}{3}\)
x < \(\frac{1}{4}\)
x < \(\frac{9}{16}\)
x > \(\frac{9}{16}\)
Correct answer is E
\(3x - (\frac{1}{4})^{-\frac{1}{2}} > \frac{1}{4} - x\)
= \(3x - 4^{\frac{1}{2}} > \frac{1}{4} - x\)
= \(3x - 2 > \frac{1}{4} - x\)
= \(3x + x > \frac{1}{4} + 2 \implies 4x > \frac{9}{4}\)
\(x > \frac{9}{16}\)
The minimum point on the curve y = x2 - 6x + 5 is at
(1, 5)
(2, 3)
(-3, -4)
(3, -4)
Correct answer is D
Given the curve \(y = x^{2} - 6x + 5\)
At minimum or maximum point, \(\frac{\mathrm d y}{\mathrm d x} = 0\)
\(\frac{\mathrm d y}{\mathrm d x} = 2x - 6\)
\(2x - 6 = 0 \implies x = 3\)
Since 3 > 0, it is a minimum point.
When x = 3, \(y = 3^{2} - 6(3) + 5 = -4\)
Hence, the turning point has coordinates (3, -4).
\(\sqrt{3}\) units
3 units
\(\sqrt{41 + 20 \sqrt{2}}\) units
\(\sqrt{41 - 20 \sqrt{2}}\) units
Correct answer is C
Force to the east = 5 units
force to the North - east = 4 units.
Resultant of the two forces is the square root
52 + 42 = 41 and plus the sum of its resistance
5 x 4\(\sqrt{2}\) = 20\(\sqrt{2}\)
= \(\sqrt{41 + 20 \sqrt{2}}\) units
Simplify \(\frac{a - b}{a + b}\) - \(\frac{a + b}{a - b}\)
\(\frac{4ab}{a - b}\)
\(\frac{-4ab}{a^2 - b^2}\)
\(\frac{-4ab}{a^{-2} - b}\)
\(\frac{4ab}{a^{-2} - b^{-2}}\)
Correct answer is B
\(\frac{a - b}{a + b}\) - \(\frac{a + b}{a - b}\) = \(\frac{(a - b)^2}{(a + b)}\) - \(\frac{(a + b)^2}{(a - b0}\)
applying the principle of difference of two sqrt. Numerator = (a - b) + (a + b) (a - b) - (a + b)
= (a = b + a = b)(a = b - a = b)
2a(-2b) = -4ab
= \(\frac{-4ab}{(a + b)(a - b)}\)
= \(\frac{-4ab}{a^2 - b^2}\)
125o
82o
135o
105o
60o
Correct answer is B
Sum of interior angles of a polygon = (2n - 4) x 90
where n is the no. of sides
sum of interior angles of the quad. = ({2 x 4} -4) x 90
(8 - 4) x 90 = 4 x 90o
= 360o
(p + 10)o + (p - 30)o + (2p - 45)o + (p + 15)o = 360o
p + 10o + p - 30o + 2p - 45o + p + 15 = 360o
5p = 360o + 50o
= 82o