Find the quadratic equation whose roots are c and -c
x2 - c2 = 0
x2 + 2cx = 0
x2 + 2cx + c2 = 0
x2 - 2cx + c2 = 0
Correct answer is A
If p = \(\frac{1}{2}\) and \(\frac{1}{p - 1} = \frac{2}{p + x}\), find the value of x
-2\(\frac{1}{2}\)
-1\(\frac{1}{2}\)
1\(\frac{1}{2}\)
2\(\frac{1}{2}\)
Correct answer is B
p = \(\frac{1}{2}; \frac{1}{p - 1} = \frac{2}{p + x}\)
\(\frac{1}{\frac{1}{2} - 1} = \frac{2}{\frac{1}{2} + x}\)
\(\frac{1}{\frac{1 - 2}{2}} = \frac{2}{\frac{1 + 2x}{2}}\)
\(\frac{1}{-\frac{1}{2}} = \frac{2}{\frac{1 + 2x}{2}}\)
-2 = \(\frac{4}{1 + 2x} -2(1 + 2x) = 4\)
1 + 2x = \(\frac{4}{-2}\)
1 + 2x = -2
2x = -2 - 1
2x = -3
x = -\(\frac{3}{2}\)
x = -1\(\frac{1}{2}\)
14.67cm
22.67cm
29.33cm
37.33cm
Correct answer is B
perimeter of minor segment = Length of arc xy + chord xy
where lxy = \(\frac{120}{360} \times 2x \times \frac{22}{7} \times 7\)
= 14.67cm
perimeter of minor segment = 14.67 + 8 = 22.67cm
3
6
9
12
Correct answer is A
Total surface area of cube = 6s2
6s2 = x
s2 = x = \(\frac{x}{6}\).....(1)
volume of a cube = s2 = \(\frac{x}{2}\)
s2 = \(\frac{x}{2}\)......(2)
put(1) into (2)
s(\(\frac{x}{6}\)) = \(\frac{x}{2}\)
s = \(\frac{x}{2} \times \frac{6}{x}\)
= 3cm
11o
20o
22o
75o
Correct answer is D
Length of an arc = \(\frac{\theta}{360} \times 2\pi r\)
18.33 = \(\frac{\theta}{360} \times 2 \times \frac{22}{7} \times 14\)
\(\theta = \frac{18.33 \times 360 \times 17}{2 \times 22 \times 14}\)
= 75o (approx.)