Find the maximum value of 1 1 1 1 1 + sin θ 1 1 1 1 + cos θ

Let Δ = 1 1 1 1 1 + sin θ 1 1 1 1 + cos θ Applying R 2 → R 2 - R 1 and R 3 → R 3 - R 1 , we get Δ = 1 1 1 0 sin θ 0 0 0 cos θ = sin θ cos θ = sin 2 θ 2 We know that, - 1 ≤ sin 2 θ ≤ 1 ⇒ - 1 2 ≤ sin 2 θ 2 ≤ 1 2 ∴ Maximum value of sin 2 θ 2 is 1 2

If x sin θ cos θ - sin θ - x 1 cos θ 1 x = 8 , write the value of x .

Given x sin θ cos θ - sin θ - x 1 cos θ 1 x = 8 Expanding along R 1 , we get x - x 2 - 1 - sin θ ( - x sin θ - cos θ ) + cos θ ( - sin θ + x cos θ ) = 8 ⇒ - x 3 - x + x sin 2 θ + sin θ cos θ - sin θ cos θ + x cos 2 θ = 8 ⇒ - x 3 - x + x sin 2 θ + cos 2 θ = 8 ⇒ - x 3 - x + x = 8 ⇒ x 3 + 8 = 0 ⇒ x 3 + 2 3 = 0 We know that, a 3 + b 3 = ( a + b ) a 2 − a b + b 2 Therefore, x 3 + 2 2 = x + 2 x 2 - 2 x + 4 ⇒ x + 2 = 0 or x 2 - 2 x + 4 = 0 ⇒ x = - 2 or x = 2 ± 4 - 16 2 Hence, x = - 2 , x = 1 ± i 3

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MATHEMATICS CLASS XII VOLUME-1>Chapter 6 - Determinants>MULTIPLE CHOICE QUESTIONS (MCQs)

R D Sharma Solutions for Chapter: Determinants, Exercise 1: MULTIPLE CHOICE QUESTIONS (MCQs)

Author:R D Sharma

R D Sharma Mathematics Solutions for Exercise - R D Sharma Solutions for Chapter: Determinants, Exercise 1: MULTIPLE CHOICE QUESTIONS (MCQs)

Attempt the free practice questions on Chapter 6: Determinants, Exercise 1: MULTIPLE CHOICE QUESTIONS (MCQs) with hints and solutions to strengthen your understanding. MATHEMATICS CLASS XII VOLUME-1 solutions are prepared by Experienced Embibe Experts.