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# Competition Level Test: Solution Of Triangles- 2

## 30 Questions MCQ Test Mathematics For JEE | Competition Level Test: Solution Of Triangles- 2

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This mock test of Competition Level Test: Solution Of Triangles- 2 for JEE helps you for every JEE entrance exam. This contains 30 Multiple Choice Questions for JEE Competition Level Test: Solution Of Triangles- 2 (mcq) to study with solutions a complete question bank. The solved questions answers in this Competition Level Test: Solution Of Triangles- 2 quiz give you a good mix of easy questions and tough questions. JEE students definitely take this Competition Level Test: Solution Of Triangles- 2 exercise for a better result in the exam. You can find other Competition Level Test: Solution Of Triangles- 2 extra questions, long questions & short questions for JEE on EduRev as well by searching above.
QUESTION: 1

### If 2cos θ < √3 and x ∈ [-π,π] then the solution set for ‘x’ is

Solution:

The value scheme for this is shown.

From the figure,

∴

QUESTION: 2

### If 2 sin x + 1 > 0 and x ∈ [0, 2π] , then the solution set for x is

Solution:

2 sin x + 1 > 0  ⇒
The value scheme for this is shown below.
From the figure,

∴

QUESTION: 3

### If cos x – sin x > 1 and 0 < x < 2π , then the solution set for x is

Solution:

The value scheme for this is shown below.

and un general,

∴

QUESTION: 4

The solution set of inequality

Solution:

From the graph of y = sin x, it obvious that, between 0 and 2π.

The requried solution set is

QUESTION: 5

The set of values of x for which

Solution:

But for this value of x,

which does not satisfy the given equation as it
reduces to indeterminate form.

QUESTION: 6

The solution of inequality

Solution:

From the graph of y = cos x, it is obvious that

The required dolution is

QUESTION: 7

The number of values of x lying in the interval (-π, π) which satisfy the equation

Solution:

⇒
⇒
or
⇒
∴ Number of solution = 4

QUESTION: 8

The equation esin x - e-sin x - 4 = 0 has

Solution:

The given equation can be written as

e2sin x - 4e sinx -1 = 0
⇒
⇒ sin x = In (2 +√5)
[In (2 - √5) not defined as 2-√5 is - ve]
Now, 2 + √5 > e ⇒ = In (2 + √5) > 1 = sin x > 1
which is not possible ∴ No real solution.

QUESTION: 9

If tan (π cos θ) = cot (π sin θ) , then the value(s) of

Solution:

We have, tan (π cos θ) = cot (π sin θ)

⇒
⇒
⇒
⇒
⇒
(for n = 0 and n = -1)

QUESTION: 10

If the equation cos 3x + cos 2x = sin 3x/2 + sin x/2, 0 < x < 2π then the number of value of x is

Solution:

⇒

⇒
or

QUESTION: 11

then x is equal to

Solution:

So that, the given equation can be written
⇒ y2 - 30y + 81 = 0
⇒
⇒
⇒
⇒

⇒

QUESTION: 12

The number of solutions of cos x.cos 2x.cos 3x = 1/4 in [0,π] is equal to

Solution:

cos x. cos 2x. cos 3x = 1/4

⇒ 4cos3x • cosx • cos2x =1
⇒ 2 (cos4x + cos2x) cos 2x =1
⇒ 2(2cos2x - 1 +cos2x)cos2x =1
⇒ 4cos3 2x + 2cos2 2x - 2 cos2x-1 = 0
⇒ = 2cos2x(2cos 2x+ 1)- (2cos 2x+ 1) = 0
⇒ (2cos 2x + 1) (2cos2 2x - 1) = 0
⇒ cos 4x•(2cos 2x+ 1) = 0
⇒ It cos 4x = 0 4x = (2n1 + 1)π/ 2
⇒
or
⇒
⇒
Thus, there are six solutions.

QUESTION: 13

Total number of solutions of sin4 x + cos4 x = sin x. cos x in [0, 2π] is equal to

Solution:

sin4 x + cos4 x= sin x. cos x
⇒ (sin2 x + cos2 x)2 - 2 sin2 x . cos2 x = sin x . cos x
⇒
⇒ sin2 2x + sin 2x - 2 = 0
⇒ (sin 2x + 2)  (sin 2x - 1) = 0 ⇒ sin 2x = 1
⇒
⇒
⇒
thus there are two solution.

QUESTION: 14

The number of solutions of the pair of equations 2 sin2 θ - cos2θ = 0 and 2 cos2 θ - 3sin θ = 0 in the interval [0, 2π] is

Solution:

Given equation is
2 sin2 θ - cos 2 θ = 0
⇒ 2 sin2 θ - (1 - 2sin2 θ) = 0
⇒ 4 sin2 θ = 1
⇒           .......(1)
Also, 2cos2 θ - 3sin θ = 0

⇒ sin θ 1/2 (∵ sin θ ≠ -2)     .......(2)
From Eqs, (i) and (ii), we get sin θ = 1/2 satisfying both the given equation

QUESTION: 15

The set of values of q satisfying the inequation 2 sin2 θ - 5 sin θ + 2 > 0 , where 0 < θ < 2π is

Solution:

Since, 2sin2 x θ - 5 sin θ + 2 > 0

QUESTION: 16

The number of values of x in the interval [0, 3π] satisfying the equation 2 sin2 x + 5 sin x - 3 = 0 is

Solution:

Since, 2 sin2 x + 5 sin x - 3 = 0
⇒ (2 sin x - 1) (sin x + 3) = 0
⇒ sin x = 1/2  (∵ sin x ≠ = -3)

From figure, it is clear that the number of intersection points of y = 1/2 and y = sin x is 4.
Hence, number of solutions of given equation in [0,3π] is 4.

QUESTION: 17

The solutions of the equation 4 cos2 x + 6 sin2 x = 5 are

Solution:

Given, 4 cos2 x + 6 sin2 x = 5
⇒  4 (cos2 x + sin2 x)+2 sin2 x = 5
2 sin2 x = 5 - 4

QUESTION: 18

The most general solution of

Solution:

Given equation can be written as

QUESTION: 19

The general solution of sin x - cos x = √2, for any integer n is

Solution:

Given  that, sin x - cos x = √2,

⇒
⇒
⇒
⇒

QUESTION: 20

The equation √3 sin x + cos x = 4 has

Solution:

Since, we get know a cos x + b sin x = c has solution.,

Hence, given equation is

∴ Given equation has no solution.

QUESTION: 21

Solution:

QUESTION: 22

The equation 3 sin2x + 10 cos x – 6 = 0 is satisfied, if

Solution:

Given, 3 sin2x + 10 cos x - 6 = 0
⇒ 3 (1 - cos2 x) + 10 cos x - 6 = 0
⇒ 3 - 3cosx + 9 cos x + cos x - 6 = 0
⇒ - 3 cos x (cos x - 3) + 1 (cos x - 3) = 0
⇒ (cos x - 3) (1 - 3 cos x) = 0
⇒ cos x = 1/3
⇒

QUESTION: 23

then the general value of θ is

Solution:

⇒

QUESTION: 24

The number of solutions for the equation sin 2x + cos 4x = 2 is

Solution:

Given equation is
sin 2x + cos 4x = 2
which is only possible,
if sin 2x = cos 4x = 1
when sin 2x = 1

For which, cos 4x = -1
∴ Given equation has no solution.

QUESTION: 25

If sin 3θ = sin θ, how many solutions exist such that -2π < θ <2π ?

Solution:

We have, sin 3θ =  sinθ
⇒ sin 3θ - sinθ  = 0

⇒ cos 2θ . sin θ = 0
⇒ cos 2θ = 0 or sin θ = 0

Thus, total number of solution = 11

QUESTION: 26

The most general value of θ satisfying the equation sin θ = sin α and cos θ = cos α is

Solution:

Given. sin θ = sin α
⇒  θ = nπ + (-1)n α
⇒ θ = 2nπ+α, (2n+ 1) + π - α
and cosθ = cos α
⇒ θ = 2nπ+α, 2nπ - α
∴ θ  = 2nπ+α, satisfies both equations.

QUESTION: 27

The general value of θ in the equation cos θ = 1/√2, tan θ = -1 is

Solution:

Here ‘θ’ lies in the 4th quadrant.

⇒ General value of

QUESTION: 28

The number of values of x in [0, 2π] satisfying the equation 3 cos 2x -10 cos x + 7 = 0 is

Solution:

Given, 3 cos 2x - 10 cos x + 7 = 0
⇒ 3 (2 cos2 = x - 1) - 10 cos x + 7 = 0
⇒ 6 cos2 x - 10 cos x + 4 = 0
⇒ 3 cos2 x - 5 cos x + 2 = 0
⇒ (3 cos x - 2) (cos x - 1) = 0
⇒ cos x = 1 or  cos x = 2/3,
⇒ x = 0, 2π

Hence, four values exist, when x ∈ [0, 2π].

QUESTION: 29

If sinx = 1 + cos3x, then x is equal to

Solution:

sin6x = 1 + cos4x < 1
Hence, sin6 x = 1 + cos4 3x < 1 is possible only when
sin6 x = 1 + cos4 3x = 1

Common value of x is (2n + 1) π/2,

The required solution

QUESTION: 30

If equation sin4x = 1 + tan8x, then x is

Solution:

sin4x = 1 + tan8x only when
sin4x = 1 and 1 + tan8x = 1
⇒ sin2x = 1  and tan8x = 0
Which is never possible, since sin x and tan x vanish simultaneously.
Therefore, the given equation has no solution.