Evaluate without using tables sin(-1290º)
32
-32
22
1
12
Correct answer is E
sin(-1290º) = -sin(1290º)
sin([3*360] + 210)
where sin 360 = 0 and sin 210 = 180 + 30 ⇔ -30º
-sin ([3* 0] + [-30])
-sin(-30)
sin30 = 12
k = -7, 1 = -15
k = -15, 1 = -7
k = 153 , 1 = -7
k = 73 , 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 (a2−1a)(a43+a23)a2−1a2
a23
a-13
a2 + 1
a
a13
Correct answer is E
(a2−1a)(a43+a23)a2−1a2
= (a2−1a)(a2+1a23)a4−1a2
= a4−1a53 x a2a4−1
= a13
Without using tables, simplify 1n√216−1n√125−1n√82(1n3−1n5)
-3
3
35
−32
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