Find the coordinates of the centre of the circle 4x2+4y2−5x+3y−2=0.
(−54,34)
(38,−58)
(58,−38)
(54,−34)
Correct answer is C
Equation : (x−a)2+(y−b)2=r2
Expanding : x2+y2−2ax−2by+a2+b2=r2
Given, 4x2+4y2−5x+3y−2=0
Divide through by 4 to make the coefficient of x2 and y2 to be 1.
x2+y2−54x+34y−12=0
Comparing, 2a=54⟹a=58
2b=−34⟹b=−38
(a,b)=(58,−58)
Given that (1−314)(−6P)=(3−26), find the value of P.
-8
-5
4
-3
Correct answer is D
(1−314)(−6P)=(3−26)
⟹(1×−6)+(−3×P)=3
−6−3P=3⟹−3P=9
P=−3
12
16
6
4
Correct answer is D
F=ma
32=8m⟹m=4kg
Find the least value of the function f(x)=3x2+18x+32
5
4
-3
-2
Correct answer is A
f(x)=3x2+18x+32
dydx=6x+18=0
6x=−18⟹x=−3
f(−3)=3(−32)+18(−3)+32=27−54+32=5
Find the coordinates of the point on the curve y=x2+4x−2, where the gradient is zero.
(-2, 10)
(-2, 2)
(-2, -2)
(-2, -6)
Correct answer is D
y=x2+4x−2
dydx=2x+4=0
2x=−4⟹x=−2
y(−2)=(−22)+4(−2)−2=4−8−2=−6
\therefore (x, y) = (-2, -6)