JEE Main202325 Jan 2023Evening ShiftMathematicsContinuity and DifferentiabilityActual
If the function f x = 1 + | cos x | λ | cos x | , 0 < x < π 2 μ , x = π 2 e cot 6 x cot 4 x , π 2 < x < π is continuous at x = π 2 , then 9 λ + 6 log c μ + μ 6 - e 6 λ is equal to
Options
- A11
- B8
- C2 e 4 + 8
- D10
Correct answer
D. 10
Step-by-step solution
Given, f x is continuous at x = π 2 Now, solving L . H . L . at x = π 2 we get, lim x → π + 2 e cot   6 x cot   4 x = lim x → π + 2 e sin   4 x · cos   6 x sin   6 x · cos   4 x = e 2 / 3 Similarly, on simplification for R . H . L . we get lim x → π 2 - 1 + | cos x | λ | cos x | = e lim x → π 2 - λ = e λ ∵   f π 2 = μ For continuous function, f π 2 - = f π 2 = f π 2 + V