Q: Let 2 < x < be such that x= -5 11 . Then ( 11 x 2 )( 6 x- 6 x)+ ( 11 x 2 )( 6 x+ 6 x) is equal to

Given, aligned & x=- 5 11 \\ & 1- ^2 x 2 2 x 2 =- 5 11 \\ & x 2 = 11 ,- 1 11 \\ & So, x 2 = 11 , As 4 < x 2 < 2 aligned aligned & As, x ( 2 , ) \\ & Now, ( 11 x 2 )( 6 x- 6 x)+ ( 11 x 2 )( 6 x+ 6 x) \\ & = \ 6 x 11 x 2 + 11 x 2 6 x \ \\ & = (6 x- 11 x 2 )+ (6 x- 11 x 2 ) \\ & = x 2 + x 2 \\ & = 1 2 3 + 11 2 3 \\ & = 11 +1 2 3 Option (2) is correct. aligned

Q: For any y ∈ ℝ , let cot - 1 y ∈ 0 , π and tan - 1 y ∈ - π 2 , π 2 . Then the sum of all the solutions of the equation tan - 1 6 y 9 - y 2 + cot - 1 9 - y 2 6 y =

Given, tan - 1 6 y 9 - y 2 + cot - 1 9 - y 2 6 y = 2 π 3 And 0 < | y | < 3 ⇒ y ∈ - 3 , 3 - 0 Now taking, Case-l: When 6 y 9 - y 2 > 0 ⇒ y > 0 tan − 1 ⁡ 6 y 9 − y 2 + tan − 1 ⁡ 6 y 9 − y 2 = 2 π 3 ⇒ 2 tan - 1 6 y 9 - y 2 = 2 π 3 ⇒ tan - 1 6 y 9 - y 2 = π 3 ⇒ 6 y 9 - y 2 = 3 ⇒ 6 y = 9 3 - 3 y 2 ⇒ 3 y 2 + 6 y - 9 3 = 0 ⇒ 3 y 2 + 9 y - 3 y - 9 3 = 0 ⇒

Q: Let tan - 1 x ∈ - π 2 , π 2 , for x ∈ ℝ . Then the number of real solutions of the equation 1 + cos 2 x = 2 tan - 1 tan x in the set - 3 π 2 , - π 2

Given, 1 + cos 2 x = 2 tan - 1 tan x ⇒ 2 cos 2 x = 2 tan - 1 tan x ⇒ cos x = tan - 1 tan x Now plotting the diagram of cos x & tan - 1 tan x we get, Now from above diagram we can say that, Number of solutions = Number of intersection points = 3

Q: Let M denote the determinant of a square matrix M . Let g : 0 , π 2 → ℝ be the function defined by g θ = f θ - 1 + f π 2 - θ - 1 where f θ =

Given, f θ = 1 2 1 sin θ 1 - sin θ 1 sin θ - 1 - sin θ 1 + sin π cos θ + π 4 tan θ - π 4 sin θ - π 4 - cos π 2 log e 4 π cot θ + π 4 log e π 4 tanπ ⇒ f θ = 1 2 1 sin θ 1 - sin θ 1 sin θ - 1 - sin θ 1 + 0 cos θ + π 4 tan θ - π 4 sin θ - π 4 0 log e 4 π - tan θ - π 4 - log e 4 π 0 Here we used cos θ + π 4 = - sin

Q: Consider the following lists: List- I List- II I x ∈ - 2 π 3 , 2 π 3 : cos x + sin x = 1 P has two elements II x ∈ - 5 π 18 , 5 π 18 : 3 tan 3 x = 1 Q has

Solving all questions one by one we get, (i) x ∈ - 2 π 3 , 2 π 3 , cos x + sin x = 1 cos x + sin x = 1 ⇒ sin π 4 + x = 1 2 ⇒ π 4 + x = n π + - 1 n π 4 ⇒ x = n π + - 1 n π 4 - π 4 So, x ∈ 0 , π 2 ∴ x has 2 elements. → P (ii) x ∈ - 5 π 18 , 5 π 18 : 3 tan 3 x = 1 Solving 3 tan 3 x = 1 ⇒ tan 3 x = 1 3 ⇒ 3 x = n π +

Q: Let f : 0 , 2 → ℝ be the function defined by f x = 3 − sin 2 π x sin π x − π 4 − sin 3 π x + π 4 . If α , β ∈ 0 , &

Let π x − π 4 = θ ⇒ π x = θ + π 4 ∵ f x ≥ 0 ⇒ 3 − sin 2 θ + π 2 sin θ − sin 3 θ + π ≥ 0 ⇒ 3 − cos 2 θ sin θ + sin 3 θ ≥ 0 ⇒ 2 + 2 sin 2 θ sin θ + 3 sin θ − 4 sin 3 θ ≥ 0 ⇒ sin θ 5 − 2 sin 2 θ ≥ 0 ⇒ sin θ ≥ 0 ⇒ θ ∈ 0 , π (Since, x ∈ 0 , &#16

Q: Let f x = sin ⁡ π cos ⁡ x and g x = cos ⁡ 2 π sin ⁡ x be two functions defined for x > 0 . Define the following sets whose elements are written in the increasing

I f x = sin ⁡ π cos ⁡ x and X = x : f x = 0 f x = 0 ⇒ sin ⁡ π cos ⁡ x = 0 ⇒ π cos ⁡ x = m π , m ∈ I ⇒ cos ⁡ x = m cos ⁡ x = 0 , 1 , - 1 ⇒ x = n π 2 ⇒ X = n π 2 , n ∈ I ⇒ X = ± π 2 , ± π , ± 3 π 2 , ± 2 π … … I - P , Q I I g x = cos ⁡ 2 π sin ⁡ x an

Question 1

Match each entry in List-I to the correct entry in List-II and choose the correct option. List-I List-II (P) The number of elements in the set \ x [- , ] : ^6 x + ^4 x = 1\ (1) is 1 (Q) The number of

Accepted Answer

For (P): ^6 x + ^4 x = 1 ^6 x + (1 - ^2 x)^2 = 1 ^6 x + 1 - 2 ^2 x + ^4 x = 1 ^2 x ( ^4 x + ^2 x - 2) = 0 ^2 x ( ^2 x - 1)( ^2 x + 2) = 0 This gives ^2 x = 0 or ^2 x = 1. In the interval [- , ], the solutions are x \ - , - 2 , 0, 2 , \ . Number of solutions = 5. For (Q): ^2 x + ^6 x = 1 1 - ^2 x + ^6 x = 1 ^2 x ( ^4 x - 1) = 0 This gives ^2 x = 0 or ^2 x = 1. In the interval [- 2 , 2 ], the solutions are x \ - 2 , 0, 2 \ . Number of solutions = 3. For (R): ^2 ( x 2 ) - ^2 x = 1 2 1 + x 2 - (1 -

Question 2

Passage: Consider the curve C_1 given by y = e^ -x for x [0, 10 ], and the curve C_2 given by y = e^ -x ( x + x) for x [0, 10 ]. Let n be the total number of points of intersection of the curves C_1 a

Accepted Answer

Equating the equations of the curves C_1 and C_2: e^ -x = e^ -x ( x + x) Since e^ -x 0 for all real x, dividing both sides by e^ -x gives: x + x = 1 Multiplying both sides by 1 2 : 1 2 x + 1 2 x = 1 2 (x + 4 ) = 1 2 The general solution is given by: x + 4 = 2m + 4 or x + 4 = 2m + 3 4 for integer m This simplifies to: x = 2m or x = 2m + 2 Given the interval x [0, 10 ], the valid values for x are: For x = 2m : x = 0, 2 , 4 , 6 , 8 , 10 (6 solutions) For x = 2m + 2 : x = 2 , 5 2 , 9 2 , 13 2 , 17 2

Question 3

Let f: R R be a function defined by f(x)= \ array cc x^2 ( x^2 ), & if x 0 \ 0, & if x=0 array . Then which of the following statements is TRUE?

Accepted Answer

Option-1 : f ( x )= x ^2 x ^2 f(x)=0 x^2 =0 x^2 =n , n N x ^2= 1 n x = 1 n 1 n 1 10^ 10 10^ 10 n n 10^ 20 , finite number of solutions Option-2 : x = 1 n 1 n > 1 > n n 10^ 10 n >10^ 20 , Infinite number of solutions Option-4 : 1 ^2 < 1 n < 1 n ( , ^2 ) n ( ^2, ^4 ), Definitely more than 25 solutions

Question 4

Let 2 < x < be such that x= -5 11 . Then ( 11 x 2 )( 6 x- 6 x)+ ( 11 x 2 )( 6 x+ 6 x) is equal to

Accepted Answer

Given, aligned & x=- 5 11 \ & 1- ^2 x 2 2 x 2 =- 5 11 \ & x 2 = 11 ,- 1 11 \ & So, x 2 = 11 , As 4 < x 2 < 2 aligned aligned & As, x ( 2 , ) \ & Now, ( 11 x 2 )( 6 x- 6 x)+ ( 11 x 2 )( 6 x+ 6 x) \ & = \ 6 x 11 x 2 + 11 x 2 6 x \ \ & = (6 x- 11 x 2 )+ (6 x- 11 x 2 ) \ & = x 2 + x 2 \ & = 1 2 3 + 11 2 3 \ & = 11 +1 2 3 Option (2) is correct. aligned

Question 5

For any y &#8712; &#8477; , let cot - 1 y &#8712; 0 , &#960; and tan - 1 y &#8712; - &#960; 2 , &#960; 2 . Then the sum of all the solutions of the equation tan - 1 6 y 9 - y 2 + cot - 1 9 - y 2 6 y =

Accepted Answer

Given, tan - 1 6 y 9 - y 2 + cot - 1 9 - y 2 6 y = 2 &#960; 3 And  0 &#60; | y | &#60; 3 &#8658; y &#8712; - 3 , 3 - 0 Now taking,  Case-l: When  6 y 9 - y 2 &#62; 0 &#8658; y &#62; 0 tan &#8722; 1 &#8289; 6 y 9 &#8722; y 2 + tan &#8722; 1 &#8289; 6 y 9 &#8722; y 2 = 2 &#960; 3 &#8658; 2 tan - 1 6 y 9 - y 2 = 2 &#960; 3 &#8658; tan - 1 6 y 9 - y 2 = &#960; 3 &#8658; 6 y 9 - y 2 = 3 &#8658; 6 y = 9 3 - 3 y 2 &#8658; 3 y 2 + 6 y - 9 3 = 0 &#8658; 3 y 2 + 9 y - 3 y - 9 3 = 0 &#8658;

Question 6

Let tan - 1 x &#8712; - &#960; 2 , &#160; &#960; 2 , for x &#8712; &#8477; . Then the number of real solutions of the equation 1 + cos 2 x = 2 tan - 1 tan x in the set - 3 &#960; 2 , &#160; - &#960; 2

Accepted Answer

Given, 1 + cos 2 x = 2 tan - 1 tan x &#8658; 2 cos 2 x = 2 tan - 1 tan x &#8658; cos x = tan - 1 tan x Now plotting the diagram of  cos x &#160; &#38; &#160; tan - 1 tan x  we get, Now from above diagram we can say that, Number of solutions = Number of intersection points = 3

Question 7

Let M denote the determinant of a square matrix M . Let g : 0 , &#960; 2 &#8594; &#8477;  be the function defined by g &#952; = f &#952; - 1 + f &#960; 2 - &#952; - 1  where  f &#952; =

Accepted Answer

Given, f &#952; = 1 2 1 sin &#952; 1 - sin &#952; 1 sin &#952; - 1 - sin &#952; 1 + sin &#960; cos &#952; + &#960; 4 tan &#952; - &#960; 4 sin &#952; - &#960; 4 - cos &#960; 2 log e 4 &#960; cot &#952; + &#960; 4 log e &#960; 4 tan&#960; &#8658; f &#952; = 1 2 1 sin &#952; 1 - sin &#952; 1 sin &#952; - 1 - sin &#952; 1 + 0 cos &#952; + &#960; 4 tan &#952; - &#960; 4 sin &#952; - &#960; 4 0 log e 4 &#960; - tan &#952; - &#960; 4 - log e 4 &#960; 0 Here we used  cos &#952; + &#960; 4 = - sin

Question 8

Consider the following lists:   List- I   List- II I x &#8712; - 2 &#960; 3 , 2 &#960; 3 : cos x + sin x = 1 P has two elements II x &#8712; - 5 &#960; 18 , 5 &#960; 18 : 3 tan 3 x = 1 Q has

Accepted Answer

Solving all questions one by one we get, (i)  x &#8712; - 2 &#960; 3 , 2 &#960; 3 , cos x + sin x = 1 cos x + sin x = 1 &#8658; sin &#960; 4 + x = 1 2 &#8658; &#960; 4 + x = n &#960; + - 1 n &#960; 4 &#8658; x = n &#960; + - 1 n &#960; 4 - &#960; 4 So,  x &#8712; 0 , &#960; 2 &#8756; &#160; &#160; x  has  2  elements. &#8594; P (ii)   x &#8712; - 5 &#960; 18 , 5 &#960; 18 : 3 tan 3 x = 1 Solving  3 tan 3 x = 1 &#8658; tan 3 x = 1 3 &#8658; 3 x = n &#960; +

Question 9

Let f : 0 , &#160; 2 &#8594; &#8477; be the function defined by f x = 3 &#8722; sin 2 &#960; x sin &#960; x &#8722; &#960; 4 &#8722; sin 3 &#960; x + &#960; 4 . If &#945; , &#160; &#946; &#8712; 0 , &

Accepted Answer

Let &#960; x &#8722; &#960; 4 = &#952; &#8658; &#960; x = &#952; + &#960; 4 &#8757; f x &#8805; 0 &#8658; 3 &#8722; sin 2 &#952; + &#960; 2 sin &#952; &#8722; sin 3 &#952; + &#960; &#8805; 0 &#8658; 3 &#8722; cos 2 &#952; sin &#952; + sin 3 &#952; &#8805; 0 &#8658; 2 + 2 sin 2 &#952; sin &#952; + 3 sin &#952; &#8722; 4 sin 3 &#952; &#8805; 0 &#8658; sin &#952; 5 &#8722; 2 sin 2 &#952; &#8805; 0 &#8658; sin &#952; &#8805; 0 &#8658; &#952; &#8712; 0 , &#160; &#960; (Since,  x &#8712; 0 , &#16

Question 10

Let f x = sin &#8289; &#960; cos &#8289; x and g x = cos &#8289; 2 &#960; sin &#8289; x be two functions defined for x &#62; 0 . Define the following sets whose elements are written in the increasing

Accepted Answer

I f x = sin &#8289; &#960; cos &#8289; x &#160; and X = x : f x = 0 f x = 0 &#8658; sin &#8289; &#960; cos &#8289; x = 0 &#8658; &#960; cos &#8289; x = m &#960; , &#160; m &#8712; I &#8658; cos &#8289; x = m cos &#8289; x = 0 , &#160; 1 , &#160; - 1 &#8658; x = n &#960; 2 &#8658; X = n &#960; 2 , &#160; n &#8712; I &#8658; X = &#177; &#960; 2 , &#160; &#177; &#960; , &#177; 3 &#960; 2 , &#177; &#160; 2 &#960; &#8230; &#8230; I - P , &#160; Q I I &#160; g x = cos &#8289; 2 &#960; sin &#8289; x an

Question 11

Let f x = sin &#8289; &#960; cos &#8289; x and g x = cos &#8289; 2 &#960; sin &#8289; x be two functions defined for x &#62; 0 . Define the following sets whose elements are written in the increasing

Accepted Answer

I f x = sin &#8289; &#960; cos &#8289; x &#160; and X = x : f x = 0 f x = 0 sin &#8289; &#960; cos &#8289; x = 0 &#960; cos &#8289; x = n &#960; cos &#8289; x = n cos &#8289; x = 0 , &#160; 1 , &#160; - 1 &#8658; x = n &#960; 2 X = n &#960; 2 , &#160; n &#8712; N &#8658; X = &#960; 2 , &#160; &#960; , 3 &#960; 2 , &#160; 2 &#960; &#8230; &#8230; I - P , &#160; Q I I &#160; g x = cos &#8289; 2 &#960; sin &#8289; x and Z = x : g x = 0 cos &#8289; 2 &#960; sin &#8289; x = 0 2 &#960; sin &#8289; x =

Trigonometric Equations — Previous Year Questions & Solutions