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MCQ Class 11 Mathematics Trigonometric Functions with Answers - Set - 4
Question 1:
The value of cos(π/7) × cos(2π/7) × cos(4π/7) is
(a) -1/2
(b) -1/4
(c) -1/6
(d) -1/8
Correct Answer – (D)
We know that cos A × cos 2A × cos 2² A × ……………… × cos 2n-1 A = sin (2ⁿ A)/{2ⁿ × sin A} ……………1
Given, cos(π/7) × cos(2π/7) × cos(4π/7)
= cos(π/7) × cos(2π/7) × cos(2² π/7)
= [sin (2³ × π/7) ]/{2³ × sin (π/7)} ……………..from equation 1
= [sin (8π/7) ]/{8 × sin (π/7)}
= [sin (π + π/7) ]/{8 × sin (π/7)}
= -sin (π/7)/{8 × sin (π/7)}
= -1/8
So, cos(π/7) × cos(2π/7) × cos(4π/7) = -1/8
Question 2 :
In a ΔABC, (b + c) cos A + (c + a) cos B + (a + b) cos C is equal to
(a) a + b + c
(b) 0
(c) none of these
(d) Rr
Correct Answer – (A)
Given (b + c) cos A + (c + a) cos B + (a + b) cos C
= b × cos A + c × cos A + c × cos B + a × cos B + a × cos C + b × cos C
= (b × cos C + c × cos B) + (c × cos A + a × cos C) + (b × cos A + a × cos B)
= a + b + c {since b × cos C + c × cos B = a, c × cos A + a × cos C = b, b × cos A + a × cos B = c}
Question 3 :
The value of cos 4A – cos 4B is
(a) (cos A – cos B) × (cos A + cos B) × (cos A – sin B) × (cos A + sin B)
(b) 2(cos A – cos B) × (cos A + cos B) × (cos A – sin B) × (cos A + sin B)
(c) 4(cos A – cos B) × (cos A + cos B) × (cos A – sin B) × (cos A + sin B)
(d) 8(cos A – cos B) × (cos A + cos B) × (cos A – sin B) × (cos A + sin B)
Correct Answer – (D)
Given, cos 4A – cos 4B
= 2cos² 2A – 1 – (2cos2 2B – 1) {since 2cos² x – 1 = cos 2x}
= 2cos² 2A – 1 – 2cos² 2B + 1
= 2cos² 2A – 2cos² 2B
= 2(cos² 2A – cos² 2B)
= 2(cos 2A – cos 2B) × (cos 2A + cos 2B)
= 2{2cos² A – 1 – (2cos² B – 1)} × {2cos² A – 1 + 1 – 2sin² B} {since 1 – 2sin² x = cos 2x}
= 2{2cos² A – 1 – 2cos² B + 1} × {2cos² A – 1 + 1 – 2sin² B}
= 2{2cos² A – 2cos² B} × {2cos² A – 2sin² B}
= 2 × 2 × 2{cos² A – cos² B} × {cos² A – sin² B}
= 8(cos A – cos B) × (cos A + cos B) × (cos A – sin B) × (cos A + sin B)
So, cos 4A – cos 4B = 8(cos A – cos B) × (cos A + cos B) × (cos A – sin B) × (cosA + sin B)
Question 4 :
The value of (sin7x + sin5x) /(cos7x + cos5x) + (sin9x + sin3x) / (cos9x + cos3x) is
(a) tan6x
(b) 2 tan6x
(c) 3 tan6x
(d) 4 tan6x
Correct Answer – (B)
Given, (sin7x + sin5x) /(cos7x + cos5x) + (sin9x + sin3x) / (cos9x + cos3x)
⇒ [{2×sin(7x + 5x)/2 × cos(7x – 5x)/2}/{2 × cos(7x + 5x)/2 × cos(7x – 5x)/2}] +
[{2×sin(9x + 3x)/2 × cos(9x – 3x)/2}/{2 × cos(9x + 3x)/2 × cos(9x – 3x)/2}]
⇒ [{2 × sin6x × cosx}/{2 × cos6x × cosx}] + [{2 × sin6x × cosx}/{2 × cos6x × cosx}]
⇒ (sin6x/cos6x) + (sin6x/cos6x)
⇒ tan6x + tan6x
⇒ 2 tan6x
Question 5 :
The equation (cos p – 1) x² + cos p × x + sin p = 0, where x is a variable, has real roots. Then the interval of p may be any one of the following:
(a) (0, π)
(b) (−π/2, π/2)
(c) (0, π)
(d) (−π, 0)
Correct Answer – (A)
The equation (cos p – 1)
x² + cos p × x + sin p = 0, where x is a variable, has real roots.
Now, for real roots,
Discriminant ≥ 0
⇒ cos² p – 4(cosp – 1)sinp ≥ 0
⇒ (cosp – 2sinp)² – 4sin² p + 4sinp ≥ 0
⇒ (cosp – 2sinp)² + 4sin p(1 – sinp) ≥ 0 ………..1
Now, 1 – sinp ≥ 0
⇒ For all real p such that 0 < p < π
So that 4sin p(1 – sinp) ≥ 0
So, p ∈ (0, π)
MCQ Class 11 Mathematics Trigonometric Functions with Answers
Question 6 :
tan² θ = 1 – a² then the value of sec θ + tan³ θ × cosec θ is
(a) (2 – a²)
(b) (2 – a²)1/2
(c) (2 – a²)3/2
(d) None of these
Correct Answer – (C)
Answer: (c) (2 – a²)3/2
Given, tan² θ = 1 – a²
⇒ tan θ = √(1 – a²)
From the figure and apply Pythagorus theorem,
AC² = AB² + BC²
⇒ AC² = {√(1 – a²)}² + 12
⇒ AC² = 1 – a² + 1
⇒ AC² = 2 – a²
⇒ AC = √(2 – a²)
Now, sec θ = √(2 – a²)
cosec θ = √(2 – a²)/√(1 – a²)
and tan θ = √(1 – a²)
Given, sec θ + tan³ θ × cosec θ
= √(2 – a²) + {(1 – a²)3/2 × √(2 – a²)/√(1 – a²)}
= √(2 – a²) + {(1 – a²) × (1 – a²) × √(2 – a²)/√(1 – a²)}
= √(2 – a²) + (1 – a²) × √(2 – a²)
= √(2 – a²) × (1 + 1 – a²)
= √(2 – a²) × (2 – a²)
= (2 – a²)3/2
So, sec θ + tan³ θ × cosec θ = (2 – a²)3/2
Question 7 :
The value of cos 420° is
(a) 0
(b) 1
(c) 1/2
(d) √3/2
Correct Answer – (C)
cos 420° = cos(360° + 60° ) = cos 60° = 1/2
Question 8 :
If x > 0 then the value of f(x) = -3 × cos√(3 + x + x²) lie in the interval
(a) [-1, 1]
(b) [-2, 2]
(c) [-3, 3]
(d) None of these
Correct Answer – (C)
Given x > 0 then 3 + x + x² > 0
Now, -1 ≤ cos√(3 + x + x² ) ≤ 1 {Since -1 ≤ cosx ≤ 1}
⇒ 3 ≥ -3 × cos√(3 + x + x² ) ≥ -3 {Multiply by -3}
⇒ -3 ≤ f(x) ≤ 3
⇒ f(x) ∈ [-3, 3]
Question 9 :
The value of (sec 8A – 1)/(sec 4A – 1) is
(a) 0
(b) 1
(c) tan 8A/tan 2A
(d) tan 2A/tan 8A
Correct Answer – (C)
Given, (sec 8A – 1)/(sec 4A – 1)
= (1/cos 8A – 1)/(1/cos 4A – 1)
= {(1 – cos 8A)/cos 8A}/{(1 – cos 4A)/cos 4A}
= {(1 – cos 8A) × cos 4A}/{(1 – cos 4A) × cos 8A}
= (2sin² 4A × cos 4A}/{2sin² 2A × cos 8A} {since cos 2A = 1 – 2sin² A}
= (2sin 4A × sin 4A × cos 4A}/{2sin 2A × sin 2A × cos 8A}
= (sin 8A × sin 4A}/{2sin 2A × sin 2A × cos 8A} {since sin 2A = 2×sin A × cos A}
= (sin 8A × 2sin 2A × cos 2A}/{2sin 2A × sin 2A × cos 8A}
= (sin 8A × cos 2A}/{sin 2A × cos 8A}
= (sin 8A/cos 8A)/(sin 2A/cos 2A)
= tan 8A/tan 2A
So, (sec 8A – 1)/(sec 4A – 1) = tan 8A/tan 2A
Question 10 :
The least values of cos² θ + sec² θ is
(a) 0
(b) 1
(c) 2
(d) more than 2
Correct Answer – (C)
If a × cos² θ + b × sec² θ is given,
then the least value = 2√ab
Now given, cos² θ + sec² θ
Here, a = 1, b = 1
Now, least value = 2√(1 × 1) = 2 × 1 = 2
- NCERT Solutions Class 11 Mathematics Relations and Functions : Exercise 2.1
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