List−IList−II P.Zk=cos(2kπ2016)+I sin(2kπ2016)k=1,2,3,...,2015 then 1−∑2015k=1cos(2kπ2016)=1.0Q.If(a2−a+k)(11b2−4b+2)=92 have exactly one ordered pair (a,b)then k=2.2 In a triangle ABC,equations of medians AD and BE are 2x+3y=6,3x R.−2y=10 respectively and AD=6,BE=113.3 and area of triangle ABC is 11k then K= S.y(x)=a cos Inx+b sin Inx,(x>0)and 1y(x)(x2d2y(x)dx2+xdy(x)dx)+1=4.4

List−IList−II P.Zk=cos(2kπ2016)+I sin(2kπ2016)k=1,2,3,...,201

1.	List−IList−II P.Zk=cos(2kπ2016)+I sin(2kπ2016)k=1,2,3,...,2015 then 1−∑2015k=1cos(2kπ2016)=1.0Q.If(a2−a+k)(11b2−4b+2)=92 have exactly one ordered pair (a,b)then k=2.2 In a triangle ABC,equations of medians AD and BE are 2x+3y=6,3x R.−2y=10 respectively and AD=6,BE=113.3 and area of triangle ABC is 11k then K= S.y(x)=a cos Inx+b sin Inx,(x>0)and 1y(x)(x2d2y(x)dx2+xdy(x)dx)+1=4.4
Answer» $\begin{matrix} L i s t - I & L i s t - I I P . & Z_{k} = c o s (\frac{2 k π}{2016}) + I s i n (\frac{2 k π}{2016}) k = 1, 2, 3, . . ., 2015 t h e n 1 - \sum_{k = 1}^{2015} c o s (\frac{2 k π}{2016}) = & 1. & 0 Q . & I f (a^{2} - a + k) (11 b^{2} - 4 b + 2) = \frac{9}{2} h a v e e x a c t l y o n e o r d e r e d p a i r (a, b) t h e n k = & 2. & 2 I n a t r i a n g l e A B C, e q u a t i o n s o f m e d i a n s A D a n d B E a r e 2 x + 3 y = 6, 3 x R . & - 2 y = 10 r e s p e c t i v e l y a n d A D = 6, B E = 11 & 3. & 3 a n d a r e a o f t r i a n g l e A B C i s 11 k t h e n K = S . & y (x) = a c o s I n x + b s i n I n x, (x > 0) a n d \frac{1}{y (x)} (x^{2} \frac{d^{2} y (x)}{d x^{2}} + x \frac{d y (x)}{d x}) + 1 = & 4. & 4 \end{matrix}$