Match the column :
Column-I | Column-II |
(i) Number of points of type ( cos θ, sin θ ) satisfying x2 + y2 – 2x – 2y + 1 = 0 is/are | [A] 4 |
(ii) Minimum value of cos4x – 6cos2x + 5 is | [B] – 2 |
(iii) No. of solution of tan 2x = tan6x in (0, 2π] is | [C] 2 |
(iv) The value of k for which the equation 2 cos4x – sin4x + k = 0 has at least one solution can be | [D] 0 |
Text Solution
Verified by Experts(i) [C]; (ii) [D]; (iii) [A]; (iv) [B]
Ans.
(i) [C]
(ii) [D]
(iii) [A]
(iv) [B], [D]
Sol. (i)
∴ cos 2 θ + sin 2 θ – 2cos θ – 2sin θ + 1 = 0
⇒ sin θ + cos θ = 1
Hence sin θ = 0, cos θ = 1& sin θ = 1, cos θ = 0 only satisfy the equation .
(ii)
∴ (cos 2 x – 3) 2 – 4 has minimum value if cos 2 x = 1
(iii)
∴ 6x = n π + 2x
x = 
where 2x, 6x ≠ ≠ k π + 
∴ x = 
, 2 π are solutions
(iv)
cos 4 x + cos 4 x – sin 4 x + k = 0
+ cos2x + k = 0
Let cos 2x = t
⇒ t 2 + 6 t + 1 + 4k = 0
⇒ (t + 3) 2 = 8 – 4k
∴ 4 ≤ ≤ 8 – 4k ≤ ≤ 16
– 4 ≤ ≤ – 4 k ≤ 8
–2 ≤ k ≤ 1
Prepare Smarter with CGP Edu
Get practice questions, solutions, and test series in one place.
Write a Review
Share your experience with this question and solution.
Commentary
Send your comment, doubt, correction, or feedback to admin.
Similar Questions
Explore conceptually related problems