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is equal to (1984 )
We have,
(1 + cosπ/ 8) (1 + cos 3π/8) 1 + cos 5π/8) (1 + cos 7π/8)
= (1 + cosπ/8) (1 + cos 3π/8)(1 + cos(p  3π/8)) (1 + cos(ππ/8))
= (1 + cos π/8) (1 + cos3π/8) 1  cos3π/8 (1  cosπ/8)
= (1  cos^{2}π/8) (1  cos^{2} 3π/8) = sin^{ 2} π/8 sin^{2} 3π/8
∴ (c) is the correct answer.
The expression 3  is equal to
The given expression is
The number of all possible triplets (a_{1}, a_{2}, a_{3}) such that a_{1} + a_{2} cos(2x) + a_{3}sin^{2}(x) = 0 for all x is (1987  2 Marks)
Since a_{1} + a_{2} cos 2x + a_{3 }sin^{2}x = 0 for all x
Putting x = 0 and x = π/2 , we get a_{1} + a_{2} = 0 ....(1)
and a_{1} – a_{2}+ a_{3} = 0 ....(2)
⇒ a_{2 }= – a_{1} and a_{3} = – 2a_{1}
∴ The given equation becomes
a_{1}– a_{1} cos 2x – 2a_{1} sin^{2} x = 0, ∀ x
⇒ a_{1} (1– cos 2x – 2 sin^{2} x) = 0, ∀ x
⇒ a_{1} (2 sin^{2} x– 2 sin^{2} x) = 0, ∀ x
The above is satisfied for all values of a_{1}.
Hence infinite number of triplets (a_{1}, – a_{1}, – 2a_{1}) are possible.
The values of θ lying between θ = 0 and θ = π/2 and satisfying the equation (1988  2 Marks)
= 0 are
(a,c) We have
Operating
Operating
Expanding along R_{1 }we get [1 + 4 sin 4θ + 1]= 0
Let 2sin^{2}x + 3sinx – 2 > 0 and x^{2} – x – 2 < 0 (x is measured in radians). Then x lies in the interval (1994)
2 sin^{2} x + 3 sin x – 2 > 0
(2 sin x – 1 ) (sin x + 2 ) > 0
⇒ 2 sin x – 1 > 0 (∵ – 1 ≤ sin x ≤ 1)
⇒ sin x > 1/2 ⇒ x ∈ (π/6, 5π/6) ....(1)
Also x^{2} – x – 2 < 0
⇒ (x – 2) (x + 1) < 0 ⇒ – 1 < x < 2 ....(2)
Combining (1) and (2) x ∈ (π/6, 2).
The minimum value of the expression sin α + sin β + sin γ , where α, β, γ are real numbers satisfying α + β + γ = π is
sin α + sin β + sin γ
∴ Each lies between –1
and 1, therefore
∴ min value = – 2.
The number of values of x in the interval [0, 5π] satisfying the equation 3 sin^{2} x – 7 sin x + 2 = 0 is (1998  2 Marks)
3 sin^{2} x – 7 sin x + 2 = 0, put sin x = s
⇒ (s – 2) (3s – 1) = 0 ⇒ s = 1/3
(s = 2 is not possible)
Number of solutions of sin x = from the following graph is 6 between [0, 5π]
Which of the following number(s) is/are rational? (1998  2 Marks)
We know that sin (irrational)
(irrational)
sin 15° . cos 15° =(2 sin 15° cos 15°)
sin 15° cos 75° = sin 15° cos (90 – 15°)
= sin 15° sin 15° = sin2 15°
(irrational)
For a positive integer n, let (1999  3 Marks) Then
Similarly,
If then (2009)
Given that
⇒
⇒
⇒
⇒
⇒
⇒
⇒
⇒
Also
For the solution (s) of is (are)
(2009)
We have
⇒
⇒
⇒
⇒
⇒
⇒
⇒
Let be such that then φ cannot satisfy (2012)
As tan
∴ θ ∈ II or IV quadrant ...(1)
And
⇒ θ ∈ III or IV quadrant ...(2)
Also θ ∈ [0, 2π] ...(3)
Combining above three equations (1), (2) and (3); θ ∈ IV
quadrant and more precisely
Now,
But
The number of points in for which x^{2} – x sin x – cos x = 0, is (JEE Adv. 2013)
(c) Let f(x) = x^{2} – x sin x – cos x
∴ f' (x) = 2x – x cos x = x(2 – cos x)
∴ f is increasing on and decreasing on
∴ y = f(x) meets xaxis twice.
i.e., f(x) = 0 has two points in
Let f(x) = x sin πx, x > 0. Then for all natural numbers n, f'(x) vanishes at (JEE Adv. 2013)
We have f(x) = x sin πx, x > 0
We observe, from graph of y = tanπx and y = –πx that they intersect at unique point in the intervals
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