\(\frac{12}{5}\)
\(\frac{12}{5}\)
-2
2
Correct answer is D
In an AP, Tn = a + (n - 1)d
T6 = a + 5d = 11
The first term = a = 1
∴ T6 = 1 + 5d = 11
5d = 11 - 1
5d = 10
∴ d = 2
Find the range of values of x for which \(\frac{1}{x}\) > 2 is true
x < \(\frac{1}{2}\)
x < 0 or x < \(\frac{1}{2}\)
0 < x < \(\frac{1}{2}\)
1 < x < 2
Correct answer is C
\(\frac{1}{x}\) > 2 = \(\frac{x}{x^2}\) > 2
x > 2x2
= 2x2 < x
= 2x2 - x < 0
= x(2x - 10 < 0
Case 1(+, -) = x > 0, 2x - 1 < 0
x > 0, x < \(\frac{1}{2}\) (solution)
Case 2(-, 4) = x < 0, 2x - 1 > 0
x < 0, x , \(\frac{1}{2}\) = 0
Find P if \(\frac{x - 3}{(1 - x)(x + 2)}\) = \(\frac{p}{1 - x}\) + \(\frac{Q}{x + 2}\)
\(\frac{-2}{3}\)
\(\frac{-5}{3}\)
\(\frac{5}{3}\)
\(\frac{2}{3}\)
Correct answer is A
\(\frac{x - 3}{(1 - x)(x + 2)}\) = \(\frac{p}{1 - x}\) + \(\frac{Q}{x + 2}\)
Multiply both sides by LCM i.e. (1 - x(x + 2))
∴ x - 3 = p(x + 2) + Q(1 - x)
When x = +1
(+1) - 3 = p(+1 + 2) + Q(1 - 1)
-2 = 3p + 0(Q)
3p = -2
∴ p = \(\frac{-2}{3}\)
Solve for r in the following equation \(\frac{1}{r - 1}\) + \(\frac{2}{r + 1}\) = \(\frac{3}{r}\)
3
4
5
6
Correct answer is A
\(\frac{1}{r - 1}\) + \(\frac{2}{r + 1}\) = \(\frac{3}{r}\)
Multiply through by r(r -1) which is the LCM
= (r)(r + 1) + 2(r)(r - 1)
= 3(r - 1)(r + 1)
= r2 + r + 2r2 - 2r
3r2 - 3 = 3r2
r = 3r2 - 3
-r = -3
∴ r = 3
If a = 1, b = 3, solve for x in the equation \(\frac{a}{a - x}\) = \(\frac{b}{x - b}\)
\(\frac{4}{3}\)
\(\frac{2}{3}\)
\(\frac{3}{2}\)
\(\frac{3}{4}\)
Correct answer is C
\(\frac{a}{a - x}\) = \(\frac{b}{x - b}\)
\(\frac{1}{1 - x}\) = \(\frac{3}{x - 3}\)
∴ 3(1 - x) = x - 3
3 - 3x = x - 3
Rearrange 6 = 4x; x = \(\frac {6}{4}\)
= \(\frac{3}{2}\)