The function f (x) = 4 sin 3 x – 6 sin 2 x + 12 sin x + 100 is strictly

(A)Increasing in (Ο€, 3Ο€/2)Β  Β  Β  Β  Β  Β  Β  Β  Β  Β  Β  Β  Β  (B) decreasing in (Ο€/2,Ο€)

(C) decreasing in ((-Ο€)/2 ", "Β  Ο€/2)Β  `Β  Β  Β  Β  Β  (D) decreasingΒ  in ("0, "Β  Ο€/2)

The function f(x) = 4 sin3x – 6 sin2x + 12 sinx + 100 is strictly [MCQ - NCERT Exemplar - MCQs part 2 - Question 7 - NCERT Exemplar - MCQs - Serial order wise - Chapter 6 Class 12 Application of Derivatives part 3 - Question 7 - NCERT Exemplar - MCQs - Serial order wise - Chapter 6 Class 12 Application of Derivatives part 4 - Question 7 - NCERT Exemplar - MCQs - Serial order wise - Chapter 6 Class 12 Application of Derivatives part 5 - Question 7 - NCERT Exemplar - MCQs - Serial order wise - Chapter 6 Class 12 Application of Derivatives

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Question 7 The function f (x) = 4 sin3 x – 6 sin2 x + 12 sin x + 100 is strictly Increasing in (πœ‹, 3πœ‹/2) (B) decreasing in (πœ‹/2,πœ‹) (C) decreasing in ((βˆ’πœ‹)/2 ", " πœ‹/2) ` (D) decreasing in ("0, " πœ‹/2) f (π‘₯) = 4 sin^3⁑〖π‘₯βˆ’6 sin^2⁑〖π‘₯+12 sin⁑〖π‘₯+100γ€— γ€— γ€— Differentiating w.r.t 𝒙 f’ (𝒙) = 4 ×⁑〖3 〗⁑〖〖𝑠𝑖𝑛〗^2 π‘₯γ€— Γ— (sin⁑〖π‘₯)β€²γ€—βˆ’6 Γ— 2 sin⁑π‘₯Γ— (sin⁑π‘₯ )β€²+12 cos⁑π‘₯+0 = 12 sin^2 π‘₯ cos π‘₯ βˆ’ 12 sin π‘₯ cos π‘₯ + 12 cos π‘₯ = 12 cos π‘₯ (γ€–sin^2 π‘₯γ€—β‘γ€–βˆ’sin⁑〖π‘₯+1γ€— γ€— ) = 12 cos 𝒙 (γ€–γ€–π’”π’Šπ’γ€—^𝟐 𝒙〗⁑〖+ (πŸβˆ’π¬π’π§β‘π’™ )γ€— ) Now, we need to check sign of f’ (𝒙) Checking sign of γ€–π’”π’Šπ’γ€—^πŸβ‘γ€–π’™+(γ€–πŸβˆ’π’”π’Šπ’γ€—β‘π’™ ) γ€— Since, γ€–π’”π’Šπ’γ€—^𝟐 𝒙 β‰₯ 0 And βˆ’1 ≀ sin 𝒙 ≀ 1 1 β‰₯ βˆ’ sin π‘₯ β‰₯ βˆ’1 1 + 1 β‰₯ 1 βˆ’ sin π‘₯ β‰₯ 1 βˆ’ 1 2 β‰₯ 1 βˆ’ sin π‘₯ β‰₯ 0 0 ≀ 1 βˆ’ sin π‘₯ ≀ 2 So, (1 βˆ’ sin 𝒙) β‰₯𝟎 Therefore, γ€–π’”π’Šπ’γ€—^πŸβ‘γ€–π’™+(γ€–πŸβˆ’π’”π’Šπ’γ€—β‘π’™ ) γ€—β‰₯ 0 Either cos 𝒙β‰₯ 0 π‘₯ πœ–(0, πœ‹/2)βˆͺ(3πœ‹/2, 2πœ‹) i.e., 1st and 4th quadrant, which can be written as 𝒙 ∈ ((βˆ’π…)/𝟐,𝝅/𝟐) Or cos 𝒙 ≀ 0 π‘₯ πœ–(πœ‹/2,πœ‹)βˆͺ(πœ‹,3πœ‹/2) i.e., 2nd and 3rd quadrant, which can be written as 𝒙 ∈(𝝅/𝟐,πŸ‘π…/𝟐) Thus, f is increasing in the interval ((βˆ’π…)/𝟐,𝝅/𝟐) f is decreasing in the interval (𝝅/𝟐,πŸ‘π…/𝟐) Our options are (A) Increasing in (πœ‹, 3πœ‹/2) (B) decreasing in (πœ‹/2,πœ‹) (C) decreasing in ((βˆ’πœ‹)/2 ", " πœ‹/2) ` (D) decreasing in ("0, " πœ‹/2) As (πœ‹/2,πœ‹) lies in the interval (πœ‹/2,3πœ‹/2) So, f is decreasing in this interval (𝝅/𝟐,𝝅) So, the correct answer is (B)

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo