The equation of a circle is 3x2+3y2+6x−12y+6=0. Find its radius
1
√3
√11
√6
Correct answer is B
The equation of a circle is given as (x−a)2+(y−b)2=r2
Expanding this, we have x2+y2−2ax−2by+a2+b2=r2
Comparing with the given equation, 3x2+3y2+6x−12y+6=0≡x2+y2+2x−4y+2=0 (making the coefficients of x2 and y2 = 1 , we get that
−2a=2⟹a=−1
2b=4⟹b=2
r2−a2−b2=−2
∴
r = \sqrt{3}
x < -1, x < -\frac{1}{3}
x > -1, x > -\frac{1}{3}
x > \frac{1}{3}, x < -1
x < \frac{1}{3}, x > -1
Correct answer is B
3x^{2} + 4x + 1 > 0
3x^{2} + 3x + x + 1 > 0
3x(x + 1) + 1(x + 1) > 0
(3x + 1)(x + 1) > 0
3x + 1 > 0 \implies 3x > -1
x > -\frac{1}{3}
x + 1 > 0 \implies x > -1
x > -1, x > -\frac{1}{3}
Simplify \sqrt[3]{\frac{8}{27}} - (\frac{4}{9})^{-\frac{1}{2}}
\frac{-5}{6}
-\frac{4}{27}
0
\frac{2}{9}
Correct answer is A
\sqrt[3]{\frac{8}{27}} - (\frac{4}{9})^{\frac{-1}{2}}
\frac{2}{3} - (\frac{9}{4})^{\frac{1}{2}}
= \frac{2}{3} - \frac{3}{2}
= \frac{-5}{6}
A function is defined by f(x) = \frac{3x + 1}{x^{2} - 1}, x \neq \pm 1. Find f(-3).
-1\frac{1}{4}
-1
\frac{4}{5}
1
Correct answer is B
f(x) = \frac{3x + 1}{x^{2} - 1}
f(-3) = \frac{3(-3) + 1}{(-3)^{2} - 1} = \frac{-8}{8} = -1
Find the remainder when 5x^{3} + 2x^{2} - 7x - 5 is divided by (x - 2).
-51
-23
29
49
Correct answer is C
Using remainder theorem, the remainder when 5x^{3} + 2x^{2} - 7x -5 is divided by (x - 2) = f(2)
f(2) = 5(2^{3}) + 2(2^{2}) - 7(2) -5 = 40 + 8 - 14 - 5
= 29