This section contains four questions, each having two matching lists.

1.	This section contains four questions, each having two matching lists. Choices for the correct combination of elements from List – I and List – II are given as options (A), (B), (C) and (D), out of which one is correct. List - IList - IIP.Zk=cos(2kπ2016)+i sin(2kπ2016)k=1,2,3,....,2015 then1.01−∑2015k=1cos(2kπ2016)=Q.If(a2−a+k)(11b2−4b+2)=922.2have exactly one ordered pair (a,b) then k =R.In a triangle ABC, equations of medians3.3AD and BE are 2x + 3y = 6, 3x - 2y = 10 respectively andAD = 6, BE = 11 and area of triangle ABC is 11K then K = S.y(x) = a cos lnx + b sin lnx, (x > 0) and4.41y(x)(x2d2y(x)dx2+xdy(x)dx)+1=
Answer» This section contains four questions, each having two matching lists. Choices for the correct combination of elements from List – I and List – II are given as options (A), (B), (C) and (D), out of which one is correct. $\begin{matrix} List - I & List - II P . & Z_{k} = c o s (\frac{2 k π}{2016}) + i s i n (\frac{2 k π}{2016}) k = 1, 2, 3, . . . ., 2015 t h e n & 1. & 0 1 - \sum_{k = 1}^{2015} c o s (\frac{2 k π}{2016}) = Q . & I f (a^{2} - a + k) (11 b^{2} - 4 b + 2) = \frac{9}{2} & 2. & 2 have exactly one ordered pair (a,b) then k = R . & In a triangle ABC, equations of medians & 3. & 3 AD and BE are 2x + 3y = 6, 3x - 2y = 10 respectively and AD = 6, BE = 11 and area of triangle ABC is 11K then K = S . & y(x) = a cos lnx + b sin lnx, (x > 0) and & 4. & 4 \frac{1}{y (x)} (x^{2} \frac{d^{2} y (x)}{d x^{2}} + x \frac{d y (x)}{d x}) + 1 = \end{matrix}$

This section contains four questions, each having two matching lists. Choices for the correct combination of elements from List – I and List – II are given as options (A), (B), (C) and (D), out of which one is correct. List - IList - IIP.Zk=cos(2kπ2016)+i sin(2kπ2016)k=1,2,3,....,2015 then1.01−∑2015k=1cos(2kπ2016)=Q.If(a2−a+k)(11b2−4b+2)=922.2have exactly one ordered pair (a,b) then k =R.In a triangle ABC, equations of medians3.3AD and BE are 2x + 3y = 6, 3x - 2y = 10 respectively andAD = 6, BE = 11 and area of triangle ABC is 11K then K = S.y(x) = a cos lnx + b sin lnx, (x > 0) and4.41y(x)(x2d2y(x)dx2+xdy(x)dx)+1=

Answer»

This section contains four questions, each having two matching lists. Choices for the correct combination of elements from List – I and List – II are given as options (A), (B), (C) and (D), out of which one is correct.

$\begin{matrix} List - I & List - II P . & Z_{k} = c o s (\frac{2 k π}{2016}) + i s i n (\frac{2 k π}{2016}) k = 1, 2, 3, . . . ., 2015 t h e n & 1. & 0 1 - \sum_{k = 1}^{2015} c o s (\frac{2 k π}{2016}) = Q . & I f (a^{2} - a + k) (11 b^{2} - 4 b + 2) = \frac{9}{2} & 2. & 2 have exactly one ordered pair (a,b) then k = R . & In a triangle ABC, equations of medians & 3. & 3 AD and BE are 2x + 3y = 6, 3x - 2y = 10 respectively and AD = 6, BE = 11 and area of triangle ABC is 11K then K = S . & y(x) = a cos lnx + b sin lnx, (x > 0) and & 4. & 4 \frac{1}{y (x)} (x^{2} \frac{d^{2} y (x)}{d x^{2}} + x \frac{d y (x)}{d x}) + 1 = \end{matrix}$

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