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# Integer Answer Type Questions: Trigonometric Functions & Equations | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE

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Q. 1. The number of all possible values of q where 0 < θ < π, for which the system of equations (y + z) cos 3θ = (xyz) sin 3θ

(xyz) sin 3θ = (y + 2z) cos 3θ  + y sin 3θ

have a solution (x0, y0, z0) with y0 z0 ≠ 0, is (2010)

Ans. Sol.  (3) The given equations are
xyz sin 3θ = ( y+ z ) cos3θ — (1)
xyz sin 3θ = 2 z cos3θ+ 2 y sin3θ — (2)
xyz sin3θ = y + 2z cos3θ+ y sin3θ — (3)

Operating (1) – (2) and (3) – (1), we get
(cos 3θ – 2 sinq )y – (cos 3θ)z = 0

and sin 3θ y + ( cos 3θ)z=0

which is homogeneous system of linear equation. But

y 0,z ≠ 0

⇒ cos 3θ = sin3θ

⇒   =

For θ ∈ (0,π) ⇒

∴ Three such solutions are possible.

Q. 2. The number of values of q in the interval, such that  for n = 0, ±1,±2 and tanθ = cot 5θ as well as sin 2θ = cos 4θ is (2010)

Ans. Sol.  (3)

⇒ cos θ cos 5θ – sin5θ sinθ= 0 ⇒ cos 6θ=0

Again sin 2θ= cos 4θ= 1 – 2 sin 2

⇒ 2sin2 2θ + sin 2θ –1= 0 ⇒ sin 2θ = –1

⇒

So common solutions are

∴ Number of solutions = 3.

Q. 3. The maximum value of the expression

Ans. Sol. Let

where g (θ) = sin 2θ+ 3 sinθ cos θ+ 5 cos2θ

Clearly f is maximum when g is minimum

Now (θ)  =

= 3 + 2cos 2θ +

Q. 4. The positive integer value of n > 3 satisfying the equation

(2011)

Ans. (7)

Sol.   We have,

= 2kπ where k ∈ Z

( n=not possible for any integral value of k)

As n > 3, for k = 0, we get n = 7.

Q. 5. The number of distinct solutions of the equation

in the interval [0, 2π] is  (JEE Adv. 2015)

Ans. (8)

Sol.   cos22x + cos4x + sin4x + cos6x + sin6x = 2

(cos22x – sin22x) = 0 ⇒ cos4x = 0

⇒ 4x = (2n + 1) or            x = (2n + 1)

For x∈[0, 2π], n can take values 0 to 7

∴ 8 solutions.

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## Maths 35 Years JEE Main & Advanced Past year Papers

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## Maths 35 Years JEE Main & Advanced Past year Papers

132 docs|70 tests

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