Let f : R → R be defined as f x = λ x 2 - 5 x + 6 μ 5 x - x 2 - 6 x < 2 e tan ( x - 2 ) x - [ x ] x > 2 μ x = 2 where x is the greatest integer less than or equal to x . If f is continuous at x = 2 , then λ + μ is equal to :

Step-by-step solution

If f x is continuous at x = 2 , then lim x → 2 + f x = lim x → 2 - f x = lim x → 2 f x Here, lim x → 2 + f x = lim x → 2 + e tan ( x - 2 ) x - 2 = e 1 = e lim x → 2 - f x = lim x → 2 - λ x 2 - 5 x + 6 μ 5 x - x 2 - 6 = lim x → 2 - - λ ( x - 2 ) ( x - 3 ) μ ( x - 2 ) ( x - 3 ) = - λ μ ∵ x 2 - 5 x + 6 = x - 2 x - 3   if x < 2 or x > 3 - x - 2 x - 3   if 2 < x&#160