Trigonometric Functions Online Objective Test | Trigonometric Functions Online Objective Test

# Trigonometric Functions # CBSE 11th Mathematics Prepare / Learn

Q1.

If A = $\sin^{2} θ + \cos^{4} θ$ then for all real values of $θ$

θ

A. 1≤A≤2

B. 3/4≤A≤1

C. 34≤A≤1316

D. 1316≤A≤1

Answer : Option B
Explaination / Solution:

# Trigonometric Functions # CBSE 11th Mathematics Prepare / Learn

Q2. he general solution of tan 3x = 1 is (n ∈ I)

A. nπ/3+π/12

B. nπ±π4

C. nπ

D. nπ+π4

Answer : Option A
Explaination / Solution:

# Trigonometric Functions # CBSE 11th Mathematics Prepare / Learn

Q3. If the radius of the circumcircle of an isosceles triangle PQR is equal to PQ ( = PR), then the angle P is

$s i n \frac{π}{14} \sin \frac{3 π}{14} \sin \frac{5 π}{14} \sin \frac{7 π}{14} \sin \frac{9 π}{14} \sin \frac{11 π}{14} \sin \frac{13 π}{14}$

A. 2π/3

B. π2

C. π6

D. π3

Answer : Option A
Explaination / Solution:

# Trigonometric Functions # CBSE 11th Mathematics Prepare / Learn

Q4. The maximum value of sin

(x + \frac{π}{6}) + \cos (x + \frac{π}{6})

in the interval

(0, \frac{π}{2})

is attained at

A. π/2

B. π/3

C. π/6

D. π/12

Answer : Option D
Explaination / Solution:

# Trigonometric Functions # CBSE 11th Mathematics Prepare / Learn

Q5. The solution of tan 2 θ tan θ = 1 is

A. (2n±1)π6

B. (4n+1)π6

C. (2n+1)π3

D. (2n+1)π6

Answer : Option D
Explaination / Solution:

# Trigonometric Functions # CBSE 11th Mathematics Prepare / Learn

Q6.

sin $(180 + ϕ)$ sin $(180 - ϕ)$ $c o s e c^{2} ϕ$

c o s e c^{2} ϕ

A. none of these.

B. -1

C. 0

D. 1

Answer : Option B
Explaination / Solution:

# Trigonometric Functions # CBSE 11th Mathematics Prepare / Learn

Q7. The general solution of the equation sin x + cos x = 2 is

A. 2nπ+π2or2nπ,n∈I

B. nπ,n∈I

C. 2nπ+π2or2nπ,n∈I

D. no solution

Answer : Option D
Explaination / Solution:

# Trigonometric Functions # CBSE 11th Mathematics Prepare / Learn

Q8. Which of the following is not possible ?

\cos θ = \frac{1 + t^{2}}{1 - t^{2}}, t \neq 0

\cos θ = \frac{1 + t^{2}}{1 - t^{2}}, t \neq 0

B. secθ=52

C. tanθ=100

D. sinθ=57

Answer : Option A
Explaination / Solution:

he value of $\cos θ$ lies between - 1 and 1

But when t=2 the value of $\cos θ = \frac{1 + t^{2}}{1 - t^{2}}, t \neq 0$ is $\frac{5}{- 3}$ which is more than 1 , so it is not possible.

# Trigonometric Functions # CBSE 11th Mathematics Prepare / Learn

Q9.

If ( 1 + tan $θ$ ) ( 1 + tan $ϕ$ ) = 2, then $θ$ + $ϕ$ =

A. 75∘

B. 60∘

C. 45∘

D. 30∘

Answer : Option C
Explaination / Solution:

# Trigonometric Functions # CBSE 11th Mathematics Prepare / Learn

Q10.

The set of values of x for which $\frac{\tan 3 x - \tan 2 x}{1 + \tan 3 x \tan 2 x} = 1$ is

A. {2nπ+π4:n=1,2,3...}

B. nπ+π4,nϵZ

C. {π4}

D. ϕ

Answer : Option B
Explaination / Solution:

\frac{\tan 3 x - \tan 2 x}{1 + \tan 3 x \tan 2 x} = 1 \Rightarrow t a n (x) = 1 \Rightarrow t a n (x) = t a n (\frac{π}{4}) \Rightarrow x = n π + \frac{π}{4}, n ϵ Z

[∵ t a n (A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}]

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