This section contains four questions- each having two matching lists- Choices for the correct combination of elements from List - I and List - II are givenas options A - B - C and D- out of which one is correct-WhineWhineP- - Z - k -cos lfrac2 k pi 2016- i -sin frac2 k pi 2016 k -1-2-3- - - 2015 then - 1 - - 0 1- 1- sum- k -1-2015cos 2 k pi 2016-WhineQ- -Ifa- 2-a-k11 b- 2-4 b-2-frac92- 2 - - 2- WlineR- - In a triangle ABC- equations of medians - 3-83- ext AD -6- BE -11 and area of triangle ABC is 11 K then K - WhineS- -text y x - a cosln x - b sinln x - x -0 and - 4 - - 4 - frac 1 y x x - 2 - frac d - 2 y x dx - 2- x frac - dy x dx -1-Whine array A- P-2 Q-3 R-1 S-4B- P-2 Q-3 R-4 S-1C- P-3 Q-2 R-1 S-4D- P-3 Q-2 R-4 S-1

The correct option is B P 2 Q 3 R 4 S 1 P : z_0 + z_1 + ..... + z_2015 = 0 ⇒ z_1 + z_2 + ....+ z_2015 = 1 ⇒∑_r=0^2015 cos (2k π/2016) = 1 Q : 4k 1/4 = 9/2 ...

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Byju's Answer

Standard XII

Mathematics

Chain Rule of Differentiation

This section ...

Question

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.

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

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

$P : z_{0} + z_{1} + . . . . . + z_{2015} = 0 \Rightarrow z_{1} + z_{2} + . . . . + z_{2015} = - 1 \Rightarrow \sum_{r = 0}^{2015} c o s (\frac{2 k π}{2016}) = - 1 Q : \frac{4 k - 1}{4} = \frac{9}{2} \times \frac{11}{18} = \frac{11}{4} \Rightarrow k = 3 R : A G = 4 A r e a = 2 (Δ A B E) = 2 \times \frac{1}{2} \times 4 \times 11 = 44$

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Similar questions

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

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.

$\begin{matrix} List - I & List - II P . & I f α = \frac{π}{7} t h e n \frac{1}{c o s α} + \frac{2 c o s α}{c o s 2 α} & 1. & 2 Q . & L t x \to \infty [(x - 1) (x - 2) (x + 3) (x + 5) (x + 10)]^{\frac{1}{5}} - x = & 2. & 3 R . & \int_{- 2}^{2} \frac{3 x^{2} d x}{1 + e^{x}} = & 3. & 4 S . & Let f(x) = x^{3} + x^{2} + 2 x - 1. The minimum integral value of x & 4. & 8 if x satisfies f(f(x)) > f(2x + 1) is \end{matrix}$

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

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}

\begin{matrix} L i s t I & L i s t I I P . & A r e a b o u n d e d b y y = x | x |, x - a x i s, & 1. & \frac{10}{3} x = 1 a n d x = - 1 Q . & A r e a b o u n d e d b y x - y + 2, & 2. & \frac{64}{3} x = \sqrt{y} a n d x = 0 R . & A r e a e n c l o s e d b y y^{2} = x & 3. & \frac{2}{3} a n d y = | x | S . & A r e a b o u n d e d b y y^{2} = x & 4. & \frac{1}{6} y = 4 a n d x = 0 \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=0cos(2kπ2016)=−1Q:4k−14=92×1118=114⇒k=3R:AG=4Area=2(ΔABE)=2×12×4×11=44