If f x = a + tan - 1 ⁡ x - b ; x ≥ 1 x 2 ; x < 1 is differentiable at x = 1, then 4 a - b can be

Step-by-step solution

Also f x must be continuous at x = 1 &#8756; f 1 - = f 1 + = f 1 1 2 = a + tan - 1 &#8289; 1 - b &#8230; i f &#8242; x = 1 1 + x &#8722; b 2 ; x &#8805; 1 1 2 ; x &#60; 1 Now for f x to be differentiable at x = 1 f &#8242; 1 &#8722; = f &#8242; 1 + 1 1 + 1 - b 2 = 1 2 &#8658; 1 + 1 - b 2 = 2 &#8658; b = 2 or 0 &#8230; i i From equation i and i i , we get 1 2 = a &#177; &#960; 4 &#8658; a = 1 2 - &#960; 4 or 1 2 + &#960; 4 a , b &#8801; 1 2 - &#960; 4 , 0 or 1 2 + &#960; 4 , 2 Hence, 4 a - b = 2 - &#960; , &#960;