0
8√3
13
2−4√3
Correct answer is B
(1+2√3)2=1+4√3+12=13+4√3
(1−2√3)2=1−4√3+12=13−4√3
13+4√3−(13−4√3)=13+4√3−13+4√3
= 8√3
Find the maximum value of 2+sin(θ+25).
1
2
3
4
Correct answer is C
sinθ≤1
i.e Maximum value of sinθ∀θ=1.
Therefore, 2+sinθ≤2+1=3
(−1415)ms−1
(−21)ms−1
(4−9)ms−1
(14−9)ms−1
Correct answer is C
u=(−53)ms−1
a=(3−4)ms−2;t=3secs
v=u+at⟹v=(−53)+(3−4)×3
= (−53)+(9−12)=(4−9)ms−1\)