$\begin{matrix} L i s t - I & L i s t - I I P . & Z_{k} = c o s (\frac{2 k π}{2016}) + I s i n (\frac{2 k π}{2016}) k = 1, 2, 3, . . ., 2015 t h e n 1 - \sum_{k = 1}^{2015} c o s (\frac{2 k π}{2016}) = & 1. & 0 Q . & I f (a^{2} - a + k) (11 b^{2} - 4 b + 2) = \frac{9}{2} h a v e e x a c t l y o n e o r d e r e d p a i r (a, b) t h e n k = & 2. & 2 I n a t r i a n g l e A B C, e q u a t i o n s o f m e d i a n s A D a n d B E a r e 2 x + 3 y = 6, 3 x R . & - 2 y = 10 r e s p e c t i v e l y a n d A D = 6, B E = 11 & 3. & 3 a n d a r e a o f t r i a n g l e A B C i s 11 k t h e n K = S . & y (x) = a c o s I n x + b s i n I n x, (x > 0) a n d \frac{1}{y (x)} (x^{2} \frac{d^{2} y (x)}{d x^{2}} + x \frac{d y (x)}{d x}) + 1 = & 4. & 4 \end{matrix}$

P-2,Q-3, R-1, S-4

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P-2,Q-3, R-4, S-1

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P-3,Q-2, R-1, S-4

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P-3,Q-2, R-4, S-1

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Solution

The correct option is B
P-2,Q-3, R-4, S-1

$P : z_{0} + z_{1} + . . . . + z_{2015} = 0$

$\Rightarrow z_{1} + z_{2} + . . . . + z_{2015} = - 1$

$\Rightarrow$ $= \sum_{r = 0}^{2015}$ $c o s (\frac{2 k π}{2015}) = - 1$

$Q : \frac{4 k - 1}{4}$ $= \frac{9}{2}$ $\times \frac{11}{18}$ $= \frac{11}{4}$ $\Rightarrow k = 3$

$R : A G = 4$

$A r e a = 2 (Δ A B E)$ $= 2 \times \frac{1}{2} \times 4 \times 11 = 44$

$S : x^{2} y_{2} + x y_{1} + y = 0$

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Similar questions

$\begin{matrix} L i s t - I & L i s t - I I P . & I f_{α} = \frac{π}{7} t h e n \frac{1}{c o s α} + \frac{2 c o s α}{c o s 2 α} = & 1. & 2 Q . & u n d e r s e t x \to \infty L t [(x - 1) (x - 2) (x + 3) (x + 5) (x + 10)]^{\frac{1}{5}} - x = & 2. & 3 R . & \int_{- 2}^{2} \frac{3 x^{2} d x}{1 + e^{x}} = & 3. & 4 S . & L e t f (x) = x^{3} + x^{2} + 2 x - 1. T h e m i n i m u m i n t e g r a l v a l u e o f x i f x s a t i s f i e s f (f (x)) > f (2 x + 1) i s & 4. & 8 \end{matrix}$

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.

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

Q. Let A and B be two independent events such that

P (A) = \frac{1}{3}

and

P (B) = \frac{1}{4}

\begin{matrix} L i s t I & L i s t I I P . & P (A \cup B) i s e q u a l t o & 1. & \frac{1}{2} Q . & P (\frac{A}{A \cup B}) i s e q u a l t o & 2. & \frac{2}{3} R . & P (\frac{B}{A^{'} \cap B^{'}}) i s e q u a l t o & 2. & \frac{1}{3} S . & P (\frac{A}{B}) i s e q u a l t o & 4. & 0 \end{matrix}

Q. Match the electronic configurations of the elements given in List - I with their correct characteristic(s) (i.e., properties for given configuration) given in List - II.

\begin{matrix} List-I & List - II P. & 1 s^{2} & 1. & Element shows highest negative oxidation state Q. & 1 s^{2} 2 s^{2} 2 p^{5} & 2. & Element shows highest first ionisation enthalpy R. & 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{5} & 3. & Element shows highest electronegativity on Pauling scale S. & 1 s^{2} 2 s^{2} 2 p^{2} & 4. & Element shows maximum electron gain enthalpy (most exothermic) \end{matrix}

List-II gives distance between pair of points given in List-I. Match them correctly.
$\begin{matrix} L i s t - I & L i s t I I (P) & (- 5, 7), (- 1, 3) & (1) & 5 (Q) & (5, 6), (1, 3) & (2) & \sqrt{8} (R) & (\sqrt{3} + 1, 1), (0, \sqrt{3}) & (3) & \sqrt{6} (S) & (0, 0), (- \sqrt{3}, \sqrt{3}) & (4) & 4 \sqrt{2} \end{matrix}$

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The correct option is B P-2,Q-3, R-4, S-1 P:z0+z1+....+z2015=0 ⇒z1+z2+....+z2015=–1 ⇒ =∑2015r=0 cos(2kπ2015)=–1 Q:4k−14 =92 ×1118 =114 ⇒k=3 R:AG=4 Area =2(ΔABE) =2×12×4×11=44 S:x2y2+xy1+y=0