Test: Single Correct MCQs: Trigonometric Functions & Equations | JEE Advanced


29 Questions MCQ Test Mathematics For JEE | Test: Single Correct MCQs: Trigonometric Functions & Equations | JEE Advanced


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QUESTION: 1

If tanθ = - , then sinθ is (1979)

Solution:

II quad or IV quad

QUESTION: 2

If α + β + γ = 2π, then (1979)

Solution:

 

QUESTION: 3

Given A = sin 2θ + cos4θ then for all real values of θ                      (1980)

Solution:

 

 

But   

 

QUESTION: 4

The equation     has

Solution:

The given equation is

where  

LHS  = 

Q 1 + cos x < 2 and sin2 x £ 1 for 

And  R.H.S. =   

given equation is not possible for any real  value of x.

QUESTION: 5

The general solution of the trigonometric equation sin x+cos x = 1 is given by : (1981 - 2 Marks)

Solution:

sin x + cos x = 1

(the set of integers)

where n = 0, ± 1, ± 2, . . . .

QUESTION: 6

The value of th e expression   cos ec 20° – sec 20° is equal to     (1988 - 2 Marks)

Solution:

The given expression is

 

 

QUESTION: 7

The general solution of sin x – 3 sin 2x + sin 3x = cos x – 3 cos 2x + cos 3x is (1989 - 2 Marks)

Solution:

The given equation is sin x – 3 sin 2x + sin 3x = cos x – 3 cos 2x + cos 3x
⇒ 2 sin 2x cos x – 3 sin 2x = 2 cos 2x cos x – 3 cos 2x
⇒ sin 2x (2 cos x – 3) = cos 2x (2 cos x – 3)
⇒ sin 2x = cos 2x (as cos x ≠ 3/2)

⇒ tan 2x = 1 ⇒ 2x = nπ + π/4 ⇒

QUESTION: 8

The equation  (cos p – 1) x 2 + (cos p)x + sin p = 0 In the variable x, has real roots. Then p can take any value in the interval (1990 -  2 Marks) 

Solution:

The given equation is (cos p – 1) x2 + (cos p) x + sin p = 0 For this equation to have real roots D≥ 0
⇒ cos2 p – 4 sin p (cos p – 1)≥ 0
⇒ cos2 p – 4 sin p cos p + 4 sin2 p + 4 sin p – 4 sin2 p≥ 0

⇒ (cos p – 2 sin p)2 + 4 sin p (1– sin p)≥ 0
For ever y real value of p (cosp – 2sinp)2≥ 0 and 1– sin p ≥ 0

QUESTION: 9

Number of solutions of the equation tan x + sec x= 2 cosx lying in the interval [0, 2π] is :   (1993)

Solution:

The given eq is, tan x + sec x = 2 cos x

⇒ sin x + 1 = 2 cos2 x ⇒ 2 sin2 x + sin x – 1 = 0

⇒ (2 sin x –1)  (sin x +1 ) = 0 ⇒

But for given eq. is not defined,

∴ only 2 solutions.

QUESTION: 10

Let     then (sec2x – tan2x) equals             (1994)

Solution:

 

 

QUESTION: 11

Let n be a positive in teger such that       Then (1994)

Solution:

For n = 2 the given equation is not satisfied.
Considering n > 1 and n ≠ 2

QUESTION: 12

If ω is an imaginary cube root of unity then the value of          (1994)

Solution:

QUESTION: 13

3 (sin x - cosx)4 + 6 (sin x + cosx)2 +  4 (sin6 x + cos6x) =            (1995S)

Solution:

3 (sin x – cos x)4 + 6 (sin x + cos x)2 + 4 (sin6 x + cos6 x)
=  3 (1 – sin 2x)2 + 6 (1 + sin 2x) + 4[(sin2 x + cos2x)3 – 3 sin2 x cos2x (sin2 x + cos2 x)]

=  3 – 6 sin 2x + 3 sin2 2x + 6 + 6 sin 2x

= 13 + 3 sin2 2x – 3 sin2 2x = 13

QUESTION: 14

The general values of θ satisfying the equation 2sin2θ – 3sinθ – 2 = 0 is         (1995S)

Solution:

The given equation is  2 sin2 θ- 3 sinθ- 2 = 0

⇒ (2 sinθ + 1) (sinθ – 2) = 0

⇒ sinθ      [∴ sinθ - 2=0 is not possible]

⇒ sinθ = sin(-π/ 6) = sin (7π/ 6)

⇒ θ= nπ + (-1)n(- π/6) =np + (-1)n7π/6

⇒ Thus, θ = nπ + (-1)n7π/ 6

QUESTION: 15

 is true if and only if                         (1996 - 1 Mark)

Solution:

We have

But 

⇒ x – y = 0 [ as perfect square of real number can never be negative] Also then x ≠ 0 as then sec2 θ will become indeterminate.

QUESTION: 16

In a triangle PQR, ∠R =π /2 . If tan (P/2) and tan (Q/2) are the roots of the equation ax2 + bx + c = 0 (a ≠ 0) then. (1999 - 2 Marks)

Solution:

Given that in 

Also tan P/2 and tan Q/2 are roots of the equation ax2 + bx + c = 0 (a ≠ 0)

∴ tan P/2 + tan Q/2 = 

Now consider, 

 

⇒ a – c = – b ⇒ a + b = c

QUESTION: 17

Let f (θ) = sinθ(sinθ + sin3θ). Then f(θ) is           (2000S)

Solution:

f(θ) = sinθ (sinθ + sin 3θ)

= (sin θ + 3 sinθ - 4 sin θ). sinθ

= (4 sinθ - 4 sin 3θ) sinθ = sin2θ (4 - 4 sin2θ)

= 4 sin2θ (1– sin2θ)

= 4 sin2θ cos2θ = (2 sinθ cosθ)2 = (sin 2θ)2 ≥ 0

which is true for all θ.

QUESTION: 18

The number of distinct real roots of   = 0 in the interval is       (2001S)

Solution:

To simplify the det. Let sin x = a; cos x = b the equation becomes

= 0 Operating C2 – C1; C3 – C2 we get

⇒ a (a – b)2 – (b – a) [b(a – b) – b (b – a)] = 0
⇒ a (a – b)2 – 2b (b – a) (a – b) = 0
⇒ (a – b)2 (a – 2b) = 0
⇒ (a = b) or a =2b

⇒ tan x = 1 or tan x = 2. But we have -π/ 4 ≤ x ≤π/4

⇒ tan ( -π/4) ≤ tan x ≤ tan (π/4) ⇒ -1 ≤ tan x≤1

∴ tan x = 1 ⇒ x = π/4   ∴ Only one real root is there.

QUESTION: 19

The maximum value of (cot α1). (cot α2 ) .... (cot αn), under the restrictions   and  1 is                                      (2001S)

Solution:

 We are given that (cot α1). (cot α2 ) .... (cot αn) = 1
⇒ (cos α1) (cos α2) .... (cos αn)  
= (sin α1)  (sin α2) ....(sin αn) ....(1)
Let y = (cos α1) (cos α2) .... (cos αn)  (to be max.)
Squaring both sides,
we get y2 = (cos2 α1)  ( cos2 α2) .... (cos2 αn)          
= cos α1 sin α1 cos α2 sin α2 .... cos αn sin αn(Using (1))

[sin 2α1 sin 2α2 .... sin2 αn ]

∴ Max. value of y is 1/2n/2.

QUESTION: 20

If α + β = π/2 and β + γ = α, then tan α equals                    (2001S)  

Solution:

Given that α + β = π/2 ⇒ α = π/2 -β



 

QUESTION: 21

The number of integral values of k for which the equation 7 cos x + 5 sin x = 2k + 1 has a solution is (2002S)

Solution:

We know that

NOTE THIS STEP

(considering only integral values)
⇒ k can take 8 integral values.

QUESTION: 22

Given both θ and φ are acute an gles and     then the value of θ + φ belongs to (2004S)

Solution:

Given that sinθ = 1 /2 and cosφ = 1 /3 and θ and φ both are acute angles

 

QUESTION: 23

cos(α – β) = 1 an d cos(α + β) = 1/e where α, β ∈ [–π, π].Pairs of α, β which satisfy both the equations is/are (2005S)

Solution:

Given that cos (α - β) = 1 and cos (α + β) = 1/e

where  α,β ∈ [-π,π]

Now, cos (α - β) = 1 ⇒ α - β = 0  ⇒ α = β

Now, cos (α + β) = 1/e ⇒ cos 2α = 1/e

⇒ There will be two values of 2α satisfying cos2α = 1/e in [0, 2π ] and two in [– 2π , 0].

⇒ There will be four values of α in [-π,π] and correspondingly four values of β. Hence there are four sets of (α, β).

QUESTION: 24

The values of θ ∈ (0, 2π) for which 2 sin2θ – 5 sinθ + 2 > 0, are (2006 - 3M, –1)

Solution:

2 sin2θ - 5 sinθ + 2 > 0

⇒ (sinθ - 2) (2 sinθ - 1) > 0

         

From graph, we get 

QUESTION: 25

 Let and   then (2006 - 3M, –1)

Solution:

 tanθ < 1 and cotθ >1

Let  tanθ = 1 - x and cotθ = 1+y

Where x, y > 0 and are very small, then

NOTE THIS STEP

QUESTION: 26

The number of solutions of the pair of equations
2sin2θ – cos2θ = 0
2cos2θ – 3sinθ = 0
in the interval [0, 2π] is (2007 - 3 Marks)

Solution:

2 sin2θ - cos 2θ=0 ⇒1 - 2 cos 2θ = 0

 

 ....(1)

where q ∈ [0, 2π] Also 2 cos 2θ - 3 sinθ=0 ⇒ 2 sin2θ + 3 sinθ – 2 = 0

⇒ (2 sinθ – 1) (sinθ + 2) = 0

[∵ sinθ ≠-2]

.... (2),    where q ∈ [0, 2π]

Combining (1) and (2), we ge

∴ Two solutions are there.

QUESTION: 27

For x ∈( 0,p ) , the equation sinx + 2sin 2x – sin 3x = 3 has (JEE Adv. 2014)

Solution:

sin x + 2sin 2x – sin 3x = 3
⇒ sin x + 4sin x cos x – 3sin x + 4sin3 x = 3
⇒ sin x (–2 + 2cos x + 4sin2 x) = 3
⇒ sin x (– 2 + 2cos x + 4 – 4cos2 x) = 3

⇒ 2 + 2 cos x – 

⇒ 

LHS  and RHS ≥ 3

∴  Equation has no solution.

QUESTION: 28

 Let  The sum of all distinct solutions of the equation

cot x) = 0 in the set S is equal to (JEE Adv. 2016)

Solution:

sec x + cosec x + 2 (tan x – cot x) = 0

= cos2x – sin2 x

 

 

For x ∈ S, n = 0

 

∴ Sum of all values of x  =

QUESTION: 29

The value of is equal to         (JEE Adv. 2016)

Solution: