If y = 3t3 + 2t2 - 7t + 3, find \(\frac{dy}{dt}\) at t = -1
-1
1
-2
2
Correct answer is C
y = 3t3 + 2t2 - 7t + 3
\(\frac{dy}{dt}\) = 9t2 + 4t - 7
When t = -1
\(\frac{dy}{dt}\) = 9(-1)2 + 4(-1) - 7
= 9 - 4 -7
= 9 - 11
= -2
What is the value of sin(-690)?
\(\frac{\sqrt{3}}{2}\)
-\(\frac{\sqrt{3}}{2}\)
\(\frac{-1}{2}\)
\(\frac{1}{2}\)
Correct answer is D
Sin(-690o) = Sin(-360 -3300
sin - 360 = sin 0
∴ sin(-690o) = sin(330o)
Negative angles are measured in clockwise direction
The acute angle equivalent of sin(-330o) = sin(30o)
sin(-330o) = sin(30o)
= \(\frac{1}{2}\)
= 0.5
\(\frac{5√3m}{3}\)
5√3m
10√3m
\(\frac{10√3m}{3}\)
Correct answer is C
RQP = 90o
PQS = 30o but
RQS = 90o = \(\frac{RS}{10}\)
RS = 10 tan 60o
RS = 10 tan 60o
RS = 10√3m
If M(4, q) is the mid-point of the line joining L(p, -2) and N(q, p). Find the values of p and q
P = 2, q = 4
p = 3, q = 1
p = 5, q = 3
p = 6, q = 2
Correct answer is D
\(\frac{p + q}{2}\) = 4
p + q = 8 ....(i)
\(\frac{p - 2}{2}\) = q
p - 2q 2 ....(ii)
q = 2, p = 6
3q = 6
the perpendicular bisector of XY
a right-angled triangle
a circle
a semi circle
Correct answer is D
Since XY is a fixed line and
XPY = 90o P is on one side of XY
P1P2P3......Pn are all possible cases where
XPY = 90o the only possible tendency is a semicircle because angles in semicircle equals 90o