k = -7, 1 = -15
k = -15, 1 = -7
k = \(\frac{15}{3}\) , 1 = -7
k = \(\frac{7}{3}\) , 1 = -17
Correct answer is A
If k = -7 is put as -15, the equation x4 - kx3 + 10x2 + 1x - 3 becomes x4 - (7x3) + 10x2 + (15)-3 = x4 + 7x3 + 10x2 - 15x - 3
This equation is divisible by (x - 1) and (x + 2) with the remainder as 27
k = -7, 1 = -15
Simplify \(\frac{(a^2 - \frac{1}{a}) (a^{\frac{4}{3}} + a^{\frac{2}{3}})}{a^2 - \frac{1}{a}^2}\)
a\(\frac{2}{3}\)
a-\(\frac{1}{3}\)
\(a^{2}\) + 1
a
a\(\frac{1}{3}\)
Correct answer is E
\(\frac{(a^2 - \frac{1}{a})(a^{\frac{4}{3}} + a^{\frac{2}{3}})}{a^2 - \frac{1}{a}^2}\)
= \(\frac{(\frac{a^2 - 1}{a})(\frac{a^2 + 1}{a^{\frac{2}{3}}})}{a^{4} - \frac{1}{a^2}}\)
= \(\frac{a^4 - 1}{a^{\frac{5}{3}}}\) x \(\frac{a^2}{a^4 - 1}\)
= a\(\frac{1}{3}\)
Without using tables, simplify \(\frac{1n \sqrt{216} - 1n \sqrt{125} - 1n\sqrt{8}}{2(1n3 - 1n5)}\)
-3
3
\(\frac{3}{5}\)
\(\frac{-3}{2}\)
1n 6 - 2 1n 5
Correct answer is C
No explanation has been provided for this answer.
The locus of all points having a distance of 1 unit from each of the two fixed points a and b is
A line parallel to the line ab
A line perpendicular to the line ab through the mid-point of ab
A circle through a and b with centre at the mid-point of ab
A circle with centre at a and passes through b
A circle in a plane perpendicular to ab and centre at the mid-point of the line ab
Correct answer is B
The locus of all points having a distance of 1 unit from each of the two fixed points. a and b is: a perpendicular to the (ab) through the mid-points ab
If (25)x - 1 = 64(\(\frac{5}{2}\))6, then x has the value
7
4
32
64
5
Correct answer is B
(25)x - 1 = 64(\(\frac{5}{2}\))6
(\(\frac{5}{2}\))6 = (\(\frac{5^6}{64}\))
25x - 1 = 64 x (\(\frac{5^6}{64}\))
52x - 2 = 56
2x - 2 = 6x
= \(\frac{8}{2}\)
= 4