Let f : - π 4 , π 4 → R be defined as, f x = 1 + sin x 3 a sin x , - π 4 < x < 0 b , x = 0 e cot 4 x / cot 2 x , 0 < x < π 4 If f is continuous at x = 0 then the value of 6 a + b 2 is equal to:

Step-by-step solution

If f x is continuous at point x = a , then lim x → 0 f x = lim x → 0 + f x = lim x → 0 - f x From the question, we can say that lim x → 0 f x = b And lim x → 0 + e cot 4 x cot 2 x = lim x → 0 + e tan 2 x tan 4 x = lim x → 0 + e tan 2 x 2 × 2 x tan 4 x 4 x = e 1 2 = b And lim x → 0 - 1 + sin x 3 a sin x = e lim x → 0 - sin x 3 a sin x = e 3 a From the above we can say that e 3 a = e 1 2 ⇒ a = 1 6 ⇒ 6 a = 1 So, 6 a + b 2 = 1 + e