Let L be a common tangent line to the curves Then the square of the slope of the line L is………….

Given curves x 2 9 + y 4 4 = 1 . . . . . . . . . ( i ) & x 2 + y 2 = 3 1 4 . . . . . . . . . . . ( i i ) Equation of any tangent to (i) be y = mx + 9 m 2 + 4 . . . . . . . . . . . . ( i i i ) For common tangent (iii) also should be tangent to (ii) so by condition of common tangency 9 m 2 + 4 = 3 1 4 ( 1 + m 2 ) OR 36m2 + 16 = 31 + 31m2 =>m2 = 3

If the arithmetic mean and geometric mean of the pth and qth terms of the sequence -16,8,-4,2,…. satisfy the equation 4x2 – 9x + 5 = 0, then p + q is equation to……….

Given sequence is -16, 8, -4, 2, ..... are in G.P. with first term a = -16 & common ratio r = − 1 2 Now t p = a r P − 1 = − 1 6 ( − 1 2 ) p − 1 & t q = a r q − 1 = − 1 6 ( − 1 2 ) q − 1 So A.M. = -8 [ ( − 1 2 ) p − 1 + ( − 1 2 ) q − 1 ] = α ( l e t ) Given equation is 4x2 – 9x + 5 = 0 gives x = 1 , 5 4 From roots we get possible value of b = 1 so 1 6 ( − 1 2 ) P + q − 2 2 = 1 O R ( − 1 2 ) p + q − 2 2 = 1 1 6 − ( − 1 2 ) 4 ⇒ p + q − 2 2 = 4 ⇒ p + q = 1 0

The total number of 4-digit numbers whose greatest common divisor with 18 is 3, is…….

18 = 32 * 2 For G.C.D to be 3. no. of four digits should be only multiple of 3, but not multiple of 9 & also should not be even. As we know no. of the form 9 k -> 1000 9 k + 1 -> 1000 9 k + 2 -> 1000 9 k + 3 -> 1000 -> Total no. = 2000 9 k + 4 -> 1000 9 k + 5 -> 1000 9 k + 6 ->1000 9 k + 7 -> 1000 9 k + 8 -> 1000 In which half will be even & half be odd so Required no. = 1000

The real value of α for which the expression 1−isin α1+isin α is purely real is: (a) ( n + 1 ) + π 2 (b) ( 2 n + 1 ) π 2 (c) n π (d) None of these, where n∈?

This is an Objective Type Questions as classified in NCERT Exemplar L e t z = 1 − i s i n α 1 + 2 i s i n α = ( 1 − i s i n α ) ( 1 − 2 i s i n α ) ( 1 + 2 i s i n α ) ( 1 − 2 i s i n α ) = 1 − 2 i s i n α − i s i n α + 2 i 2 s i n 2 α ( 1 ) 2 − ( 2 i s i n α ) 2 = 1 − 3 i s i n α − 2 s i n 2 α 1 − 4 i 2 s i n 2 α = ( 1 − 2 s i n 2 α ) − 3 i s i n α 1 + 4 s i n 2 α = 1 − 2 s i n 2 α 1 + 4 s i n 2 α − 3 s i n α 1 + 4 s i n 2 α . i Since, z is purely real, then 3 s i n α 1 + 4 s i n 2 α = 0 ⇒ s i n α = 0 S o , α = n π , n ∈ N . H e n c e , t h e c o r r e c t o p t i o n i s ( c ) .

What is the conjugate of 2−i(1−2i) z

This is Other Questions as classified in NCERT Exemplar G i v e n t h a t z = 2 − i ( 1 − 2 i ) 2 = 2 − i 1 + 4 i 2 − 4 i = 2 − i 1 − 4 − 4 i = 2 − i − 3 − 4 i = 2 − i − 3 − 4 i × − 3 + 4 i − 3 + 4 i = − 6 + 8 i + 3 i − 4 i 2 ( − 3 ) 2 − ( 4 i ) 2 = − 6 + 1 1 i + 4 9 − 1 6 i 2 = − 2 + 1 1 i 9 + 1 6 = − 2 2 5 + 1 1 2 5 i ∴ z ¯ = − 2 2 5 − 1 1 2 5 i H e n c e , z ¯ = − 2 2 5 − 1 1 2 5 i

For a positive integer n , find the value of ( 1 − i ) n 1 1 − i n .

This is a short answer type question as classified in NCERT Exemplar W e h a v e ( 1 − i ) n ( 1 − 1 i ) n = [ ( 1 − i ) ( 1 − 1 i ) ] n = [ ( 1 − i ) ( 1 − 1 i × i i ) ] n = [ ( 1 − i ) ( 1 − i i 2 ) ] n = [ ( 1 − i ) ( 1 + i ) ] n [ ? i 2 = − 1 ] = [ 1 − i 2 ] n = [ 1 + 1 ] n = 2 n H e n c e , ( 1 − i ) n ( 1 − 1 i ) n = 2 n .

If ( 1 + i ) 3 1 − i 3 − ( 1 − i ) 3 1 + i 3 = x + i y , then find ( x , y ) .

This is a short answer type question as classified in NCERT Exemplar W e h a v e ( 1 + i 1 − i ) 3 − ( 1 − i 1 + i ) 3 = x + i y ⇒ [ 1 + i 1 − i × 1 + i 1 + i ] 3 − [ 1 − i 1 + i × 1 − i 1 − i ] 3 = x + i y ⇒ [ 1 + i 2 + 2 i 1 − i 2 ] 3 − [ 1 + i 2 − 2 i 1 − i 2 ] 3 = x + i y ⇒ [ 1 − 1 + 2 i 1 + 1 ] 3 − [ 1 − 1 − 2 i 1 + 1 ] 3 = x + i y ⇒ [ 2 i 2 ] 3 − [ 2 i 2 ] 3 = x + i y ⇒ ( i ) 3 − ( − i ) 3 = x + i y ⇒ i 2 . i + i 2 . i = x + i y ⇒ − i − i = x + i y ⇒ 0 − 2 i = x + i y C o m p a r i n g t h e r e a l a n d i m a g i n a r y p a r t s , w e g e t x = 0 , y = − 2 . H e n c e , ( x , y ) = ( 0 , − 2 ) .

Show that ? z − 2 z + 3 ? = 2 represents a circle. Find its centre and radius.

This is a long answer type question as classified in NCERT Exemplar G i v e n t h a t : | z − 2 z − 3 | = 2 L e t z = x + i y ∴ | x + i y − 2 x + i y − 3 | = 2 ⇒ | ( x − 2 ) + i y ( x − 3 ) + i y | = 2 ⇒ | ( x − 2 ) + i y | | ( x − 3 ) + i y | = 2 [ ? | a b | = | a | | b | ] ⇒ | ( x − 2 ) + i y | = 2 | ( x − 3 ) + i y | ⇒ ( x − 2 ) 2 + y 2 = 2 ( x − 3 ) 2 + y 2 [ ? | x + i y | = x 2 + y 2 ] S q u a r i n g b o t h s i d e s ( x − 2 ) 2 + y 2 = 4 [ ( x − 3 ) 2 + y 2 ] ⇒ x 2 + 4 − 4 x + y 2 = 4 [ x 2 + 9 − 6 x + y 2 ] ⇒ x 2 + y 2 − 4 x + 4 = 4 x 2 + 4 y 2 − 2 4 x + 3 6 ⇒ 3 x 2 + 3 y 2 − 2 0 x + 3 2 = 0 ⇒ x 2 + y 2 − 2 0 3 x + 3 2 3 = 0 H e r e , g = − 1 0 3 , f = 0 r = g 2 + f 2 − c = 1 0 0 9 − 0 − 3 2 3 = 4 9 = 2 3 H e n c e , t h e r e q u i r e d e q u a t i o n o f a c i r c l e i s x 2 + y 2 − 2 0 3 x + 3 2 3 = 0 C e n t r e = ( − g , − f ) = ( 1 0 3 , 0 ) a n d r = 2 3 .

l i m n → ∞ t a n { ∑ r = 1 n t a n − 1 ( 1 1 + r + r 2 ) } is equal to___________.

∑ r = 1 n t a n − 1 1 1 + r + r 2 = ∑ r = 1 n t a n − 1 ( r + 1 ) 1 + r ( r + 1 ) = t a n − 1 2 − t a n − 1 + . . . . . . + t a n − 1 ( n + 1 ) − t a n − 1 n For n ∞ value = π 2 − π 4 = π 4 ∴ t a n ( π 4 ) = 1

If the least and the largest real values of a, for which the equation z + |z−1|+2i=0(z∈C and i=−1) a has a solution, are p and q respectively, then 4(p2+q2) is equal to

z + α | z − 1 | + 2 i = 0 Let z = x + iy ⇒ x + i ( y + 2 ) = − α ( x − 1 ) 2 + y 2 for α ∈ R y + 2 = 0 ⇒ y = − 2 x 2 = α 2 [ ( x − 1 ) 2 + 4 ] = 0 1 α 2 ≥ 4 5 ⇒ α 2 ≤ 5 4 ⇒ − 5 2 ≤ α ≤ 5 2 4 ( p 2 + q 2 ) = 4 ( 5 4 + 5 4 ) = 1 0

For integers n and r, let The maximum value of k for which the sum ∑ i = 0 k ( 1 0 i ) ( 1 5 k − i ) + ∑ i = 0 k + 1 ( 1 2 i ) ( 1 3 k + 1 − i )

∑ i = 0 k ( 1 0 i ) ( 1 5 K − i ) + ∑ i = 0 k + 1 ( 1 2 i ) ( 1 3 k + 1 − i ) = Equating the coefficient of x k i n ( 1 + x ) 2 5 = 2 5 C k ..............(i) = Equating the coefficient of x k + 1 i n ( 1 + x ) 1 2 ( x + 1 ) 1 3 From (i) and (ii) Hence 2 6 ≥ k + 1 ⇒ k ≤ 2 5 But maximum value of k is not defined bonus

The area of the region : R =  { ( x , y ) : 5 x 2 ≤ y ≤ 2 x 2 + 9 } i s :

1 2 3 square units

6 3 square units

1 1 3 square units

The vector equation of the plane passing through the intersection of the planes r → . ( i ^ + j ^ + k ^ ) = 1  and r → . ( i ^ − 2 j ^ ) = − 2 ,  and the point (1,0,2) is:

r → . ( i ^ + 7 j ^ + 3 k ^ ) = 7

r → . ( i ^ + 7 j ^ + 3 k ^ ) = 7 3

r → . ( 3 i ^ + 7 j ^ + 3 k ^ ) = 7

r → . ( i ^ − 7 j ^ + 3 k ^ ) = 7 3

The angle of elevation of a jet plane from a point A on the ground is 60°. After a flight of 20 seconds at the speed of 432 km/hour, the angle of elevation changes to 30°. If the jet plane is flying at a constant height, then its height is:

3600  3 m

2 4 0 0 3 m

The triangle of maximum area that can be inscribed in a given circle of radius ‘r’ is:

An equilateral triangle having each of its side of length  3 r

An isosceles triangle with base equal to 2r.

A right angle triangle having two of its sides of length 2r and r.

An equilateral triangle of height 2 r 3 .

The sum of the series ∑ n = 1 ∞ n 2 + 6 n + 1 0 ( 2 n + 1 ) !  is equal to:

4 1 8 e − 1 9 8 e − 1 − 1 0

4 1 8 e + 1 9 8 e − 1 − 1 0

− 4 1 8 e + 1 9 8 e − 1 − 1 0

Consider the following system of equations: x + 2y – 3z = a 2x + 6y – 11z = b x – 2y + 7z = c, where a, b and c are real constants. Then the system of equations:

Has infinite number of solution when 5a = 2b +c

Has no solution for all a, b, and c

Has a unique solution when 5a = 2b + c

Has a unique solution for all a, b and c

If f : R ->R is a function defined by f(x) = [ x − 1 ] c o s ( 2 x − 1 2 ) π ,  where[.] denotes the greatest integer function, then f is:

Discontinuous at all integral value values of x except at x = 1

Let F 1 (A,B,C) = ( A ∧ ∼ B ) ∨ [ ∼ C ∧ ( A ∨ B ) ] ∨ ∼ A a n d F 2 ( A , B ) = ( A ∨ B ) ∨ ( B → ∼ A )   be two logical expressions. Then:

F 1 is not a tautology but F 2 is a tautology

Both F 1 and F 2 are not tautologies

F 1 and F 2 both are tautologies

F 1 a tautology but F 2 is not a tautology

The system of linear equations 3x -2y -kz = 10 2x -4y -2z = 6 x + 2y -z = 5m is inconsistent if

k = 3, m =  4 5

The function f(x)  = 4 x 3 − 3 x 2 6 − 2 s i n x + ( 2 x − 1 ) c o s x :

Increases in  ( − ∞ , 1 2 ]

Decreases in  [ 1 2 , ∞ )

Maths NCERT Exemplar Solutions Class 11th Chapter Five: Overview, Questions, Preparation

Q: If ?z1?=?z2?=1 , is it necessary that z1=z2 ?

This is Other Questions as classified in NCERT Exemplar L e t z 1 = x 1 + y 1 i a n d z 2 = x 2 + y 2 i ∴ | x 1 + y 1 i | = | x 2 + y 2 i | ⇒ x 1 2 + y 1 2 = x 2 2 + y 2 2 ⇒ x 1 2 + y 1 2 = x 2 2 + y 2 2 ⇒ x 1 2 = x 2 2 a n d y 1 2 = y 2 2 ⇒ x 1 = ± x 2 a n d y 1 = ± y 2 S o , z 1 = x 1 + y 1 i a n d z 2 = ± x 2 ± y 2 i ∴ z 1 ≠ z 2 H e n c e , i t i s n o t n e c e s s a r y t h a t z 1 = z 2 .

Q: If ?z1?=?z2?=1 , is it necessary that z1=z2 ?

This is Other Questions as classified in NCERT Exemplar L e t z 1 = x 1 + y 1 i a n d z 2 = x 2 + y 2 i ∴ | x 1 + y 1 i | = | x 2 + y 2 i | ⇒ x 1 2 + y 1 2 = x 2 2 + y 2 2 ⇒ x 1 2 + y 1 2 = x 2 2 + y 2 2 ⇒ x 1 2 = x 2 2 a n d y 1 2 = y 2 2 ⇒ x 1 = ± x 2 a n d y 1 = ± y 2 S o , z 1 = x 1 + y 1 i a n d z 2 = ± x 2 ± y 2 i ∴ z 1 ≠ z 2 H e n c e , i t i s n o t n e c e s s a r y t h a t z 1 = z 2 .

Updated on Aug 28, 2025 14:22 IST

By Raj Pandey

Table of content

Complex Numbers and Quadratic Equations Short Answer Type Questions
Complex Numbers and Quadratic Equations Long Answer Type Questions
Complex Numbers and Quadratic Equations Other Questions
Complex Numbers and Quadratic Equations Objective Type Questions
JEE MAINS 2 April 2025
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JEE Mains 2021
JEE Mains 2020

Complex Numbers and Quadratic Equations Short Answer Type Questions

1. For a positive integer $n$ , find the value of ${(1 - i)}^{n} \frac{1}{1 - i^{n}}$ .

Sol.

\begin{array}{l} W e h a v e {(1 - i)}^{n} {(1 - \frac{1}{i})}^{n} \\ = {[(1 - i) (1 - \frac{1}{i})]}^{n} = {[(1 - i) (1 - \frac{1}{i} \times \frac{i}{i})]}^{n} = {[(1 - i) (1 - \frac{i}{i^{2}})]}^{n} \\ = {[(1 - i) (1 + i)]}^{n} [∵ i^{2} = - 1] \\ = {[1 - i^{2}]}^{n} = {[1 + 1]}^{n} = 2^{n} \\ H e n c e, {(1 - i)}^{n} {(1 - \frac{1}{i})}^{n} = 2^{n} . \end{array}

2. Evaluate $\sum_{n = 1} {(1 + i)}^{3}$ , where $n \in ℕ$ .

Sol.

\begin{array}{l} W e h a v e \sum_{n = 1}^{13} (i^{n} + i^{n + 1}) \\ = (i + i^{2}) + (i^{2} + i^{3}) + (i^{3} + i^{4}) + (i^{4} + i^{5}) + (i^{5} + i^{6}) + (i^{6} + i^{7}) + (i^{7} + i^{8}) + (i^{8} + i^{9}) + (i^{9} + i^{10}) \\ (i^{10} + i^{11}) + (i^{11} + i^{12}) + (i^{12} + i^{13}) + (i^{13} + i^{14}) \\ = i + 2 (i^{2} + i^{3} + i^{4} + i^{5} + i^{6} + i^{7} + i^{8} + i^{9} + i^{10} + i^{11} + i^{12} + i^{13}) + i^{14} \\ = i + 2 [- 1 - i + 1 + i - 1 - i + 1 + i - 1 - i + 1 + i] + (- 1) \\ = i + 2 (0) - 1 \Rightarrow - 1 + i \\ H e n c e, \sum_{n = 1}^{13} (i^{n} + i^{n + 1}) = - 1 + i . \end{array}

Commonly asked questions

For a positive integer $n$ , find the value of ${(1 - i)}^{n} \frac{1}{1 - i^{n}}$ .

Evaluate $\sum_{n = 1} {(1 + i)}^{3}$ , where $n \in ?$ .

If $\frac{{(1 + i)}^{3}}{1 - i^{3}} - \frac{{(1 - i)}^{3}}{1 + i^{3}} = x + i y$ , then find $(x, y)$ .

Complex Numbers and Quadratic Equations Long Answer Type Questions

1. If $\overset{ˉ}{∣ z} + 1 ∣ = z + 2 (1 + i)$ , then find $z$ .

Sol.

\begin{array}{l} G i v e n t h a t : | z + 1 | = z + 2 (1 + i) \\ L e t z = x + i y \\ S o, | x + i y + 1 | = (x + i y) + 2 (1 + i) \\ \Rightarrow | (x + 1) + i y | = x + i y + 2 + 2 i \\ \Rightarrow | (x + 1) + i y | = (x + 2) + (y + 2) i \\ \Rightarrow \sqrt[]{{(x + 1)}^{2} + y^{2}} = (x + 2) + (y + 2) i [∵ | x + i y | = \sqrt[]{x^{2} + y^{2}}] \\ S q u a r i n g b o t h s i d e s \\ {(x + 1)}^{2} + y^{2} = {(x + 2)}^{2} + {(y + 2)}^{2} . i^{2} + 2 (x + 2) (y + 2) i \\ \Rightarrow x^{2} + 1 + 2 x + y^{2} = x^{2} + 4 + 4 x - y^{2} - 4 y - 4 + 2 (x + 2) (y + 2) i \\ C o m p a r i n g t h e r e a l a n d i m a g i n a r y p a r t s, w e g e t \\ x^{2} + 1 + 2 x + y^{2} = x^{2} + 4 x - y^{2} - 4 y a n d 2 (x + 2) (y + 2) = 0 \\ \Rightarrow 2 y^{2} - 2 x + 4 y + 1 = 0 \dots (i) \\ a n d (x + 2) (y + 2) = 0 \dots (i i) \\ x + 2 = 0 o r y + 2 = 0 \\ ∴ x = - 2 o r y = - 2 \\ N o w p u t x = - 2 i n e q n . (i) \\ \Rightarrow 2 y^{2} - 2 \times (- 2) + 4 y + 1 = 0 \\ \Rightarrow 2 y^{2} + 4 + 4 y + 1 = 0 \\ \Rightarrow 2 y^{2} + 4 y + 5 = 0 \\ b^{2} - 4 a c = {(4)}^{2} - 4 \times 2 \times 5 = 16 - 40 = - 24 < 0 n o r e a l r o o t s . \\ N o w p u t y = - 2 i n e q n . (i) \\ \Rightarrow 2 {(- 2)}^{2} - 2 x + 4 (- 2) + 1 = 0 \\ \Rightarrow 8 - 2 x - 8 + 1 = 0 \Rightarrow x = \frac{1}{2} a n d y = - 2 \\ H e n c e, t h e v a l u e o f z = x + i y = (\frac{1}{2} - 2 i) . \end{array}

2. If $arg (z - 1) = arg (z + 3 i)$ , then find $x - 1 : y$ , where $z = x + i y$ .

Sol.

\begin{array}{l} G i v e n t h a t : a r g (z - 1) = a r g (z + 3 i) \\ \Rightarrow a r g (x + y i - 1) = a r g (x + y i + 3 i) \\ \Rightarrow a r g [(x - 1) + y i] = a r g [x + (y + 3) i] \\ \Rightarrow t a n^{- 1} \frac{y}{x - 1} = t a n^{- 1} \frac{y + 3}{x} \\ \Rightarrow \frac{y}{x - 1} = \frac{y + 3}{x} \\ \Rightarrow x y = (x - 1) (y + 3) \Rightarrow x y = x y + 3 x - y - 3 \\ \Rightarrow 3 x - y = 3 \Rightarrow 3 x - 3 = y \\ \Rightarrow 3 (x - 1) = y \Rightarrow \frac{(x - 1)}{y} = \frac{1}{3} \Rightarrow x - 1 : y = 1 : 3 \\ H e n c e, x - 1 : y = 1 : 3 . \end{array}

Commonly asked questions

If $arg (z - 1) = arg (z + 3 i)$ , then find $x - 1 : y$ , where $z = x + i y$ .

If $\overset{?}{? z} + 1 ? = z + 2 (1 + i)$ , then find $z$ .

Show that $? \frac{z - 2}{z + 3} ? = 2$ represents a circle. Find its centre and radius.

Complex Numbers and Quadratic Equations Other Questions

1. What is the conjugate of $\frac{2 - i}{(1 - 2 i)}$ z

Sol.

\begin{array}{l} G i v e n t h a t z = \frac{2 - i}{{(1 - 2 i)}^{2}} = \frac{2 - i}{1 + 4 i^{2} - 4 i} = \frac{2 - i}{1 - 4 - 4 i} \\ = \frac{2 - i}{- 3 - 4 i} = \frac{2 - i}{- 3 - 4 i} \times \frac{- 3 + 4 i}{- 3 + 4 i} \\ = \frac{- 6 + 8 i + 3 i - 4 i^{2}}{{(- 3)}^{2} - {(4 i)}^{2}} = \frac{- 6 + 11 i + 4}{9 - 16 i^{2}} \\ = \frac{- 2 + 11 i}{9 + 16} = \frac{- 2}{25} + \frac{11}{25} i \\ ∴ \bar{z} = \frac{- 2}{25} - \frac{11}{25} i \\ H e n c e, \bar{z} = \frac{- 2}{25} - \frac{11}{25} i \end{array}

2. If $∣ z_{1} ∣ = ∣ z_{2} ∣ = 1$ , is it necessary that $z_{1} = z_{2}$ ?

Sol.

\begin{array}{l} L e t z_{1} = x_{1} + y_{1} i a n d z_{2} = x_{2} + y_{2} i \\ ∴ | x_{1} + y_{1} i | = | x_{2} + y_{2} i | \\ \Rightarrow \sqrt[]{x_{1}^{2} + y_{1}^{2}} = \sqrt[]{x_{2}^{2} + y_{2}^{2}} \\ \Rightarrow x_{1}^{2} + y_{1}^{2} = x_{2}^{2} + y_{2}^{2} \Rightarrow x_{1}^{2} = x_{2}^{2} a n d y_{1}^{2} = y_{2}^{2} \\ \Rightarrow x_{1} = \pm x_{2} a n d y_{1} = \pm y_{2} \\ S o, z_{1} = x_{1} + y_{1} i a n d z_{2} = \pm x_{2} \pm y_{2} i \\ ∴ z_{1} \neq z_{2} \\ H e n c e, i t i s n o t n e c e s s a r y t h a t z_{1} = z_{2} . \end{array}

Commonly asked questions

What is the conjugate of $\frac{2 - i}{(1 - 2 i)}$ z

If $? z_{1} ? = ? z_{2} ? = 1$ , is it necessary that $z_{1} = z_{2}$ ?

Complex Numbers and Quadratic Equations Objective Type Questions

1. $s i n x + i c o s 2 x$ and $c o s x - i s i n 2 x$ are conjugate to each other for:

(a) $x = n π$

(b) $x = n + \frac{π}{2} + \frac{π}{2}$

(c) $x = 0$

(d) no value of $x$

Sol.

\begin{array}{l} L e t z = s i n x + i c o s 2 x \\ \bar{z} = s i n x - i c o s 2 x \\ B u t w e a r e g i v e n t h a t \bar{z} = c o s x - i s i n 2 x \\ ∴ s i n x - i c o s 2 x = c o s x - i s i n 2 x \\ C o m p a r i n g t h e r e a l a n d i m a g i n a r y p a r t s, w e g e t \\ s i n x = c o s x a n d c o s 2 x = s i n 2 x \\ \Rightarrow t a n x = 1 a n d t a n 2 x = 1 \\ \Rightarrow t a n x = t a n \frac{π}{4} a n d t a n 2 x = t a n \frac{π}{4} \\ ∴ x = n π + \frac{π}{4}, n \in I a n d 2 x = n π + \frac{π}{4} \\ \Rightarrow x = 2 x \Rightarrow 2 x - x = 0 \Rightarrow x = 0 \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}

2. The real value of $α$ for which the expression $\frac{1 - i s i n α}{1 + i s i n α}$ is purely real is:

(a) $(n + 1) + \frac{π}{2}$

(b) $\frac{(2 n + 1) π}{2}$

(c) $n π$

(d) None of these, where $n \in ℕ$

Sol.

\begin{array}{l} L e t z = \frac{1 - i s i n α}{1 + 2 i s i n α} = \frac{(1 - i s i n α) (1 - 2 i s i n α)}{(1 + 2 i s i n α) (1 - 2 i s i n α)} \\ = \frac{1 - 2 i s i n α - i s i n α + 2 i^{2} s i n^{2} α}{{(1)}^{2} - {(2 i s i n α)}^{2}} \\ = \frac{1 - 3 i s i n α - 2 s i n^{2} α}{1 - 4 i^{2} s i n^{2} α} = \frac{(1 - 2 s i n^{2} α) - 3 i s i n α}{1 + 4 s i n^{2} α} \\ = \frac{1 - 2 s i n^{2} α}{1 + 4 s i n^{2} α} - \frac{3 s i n α}{1 + 4 s i n^{2} α} . i \\ Since, z i s p u r e l y r e a l, t h e n \\ \frac{3 s i n α}{1 + 4 s i n^{2} α} = 0 \Rightarrow s i n α = 0 \\ S o, α = n π, n \in N . \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}

Commonly asked questions

If $z$ is a complex number, then

(a) $∣ z ∣^{2} > ∣ z ∣^{2}$

(b) $∣ z ∣^{2} = ∣ z ∣^{2}$

(c) $∣ z ∣^{2} < ∣ z ∣^{2}$

(d) $∣ z ∣^{2} \geq ∣ z ∣^{2}$

$s i n x + i c o s 2 x$ and $c o s x - i s i n 2 x$ are conjugate to each other for:

(a) $x = n π$

(b) $x = n + \frac{π}{2} + \frac{π}{2}$

(c) $x = 0$

(d) no value of $x$

The real value of $α$ for which the expression $\frac{1 - i s i n α}{1 + i s i n α}$ is purely real is:

(a) $(n + 1) + \frac{π}{2}$

(b) $\frac{(2 n + 1) π}{2}$

(c) $n π$

(d) None of these, where $n \in ?$

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Commonly asked questions

Let $λ$ be an integer. IF the shortest distance between the lines $x - λ = 2 y - 1 = - 2 z$ and $x = y + 2 λ = z - λ i s \frac{\sqrt[]{7}}{2 \sqrt[]{2}}$ , then the value of $| λ | i s . . . . . . . . . . . . . . . . . . .$

For integers n and r, let

The maximum value of k for which the sum

$\sum_{i = 0}^{k} (\begin{matrix} 10 & i \end{matrix}) (\begin{matrix} 15 & k - i \end{matrix}) + \sum_{i = 0}^{k + 1} (\begin{matrix} 12 & i \end{matrix}) (\begin{matrix} 13 & k + 1 - i \end{matrix})$

Let $i = \sqrt[]{- 1} .$ If $\frac{{(- 1 + i \sqrt[]{3})}^{21}}{{(1 - i)}^{24}} + \frac{{(1 + i \sqrt[]{3})}^{21}}{{(1 + i)}^{24}} = k,$ and n = $[| k |]$ be the greatest integral part of $| k | .$ Then $\sum_{j = 0}^{n + 5} {(j + 5)}^{2} - \sum_{j = 0}^{n + 5} (j + 5)$ is equal to………………..

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Commonly asked questions

$\underset{n \to \infty}{l i m} t a n {\sum_{r = 1}^{n} t a n^{- 1} (\frac{1}{1 + r + r^{2}})}$ is equal to___________.

If the least and the largest real values of a, for which the equation z + $| z - 1 | + 2 i = 0 (z \in C a n d i = \sqrt[]{- 1})$ a has a solution, are p and q respectively, then $4 (p^{2} + q^{2})$ is equal to

Let M be any 3 × 3 matrix with entires from the set {0, 1, 2}. The maximum number of such matrices, for which the sum of diagonal elements M^TM is seven is____________.

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