Let g x be a linear function and f x = g x , x ≤ 0 1 + x 2 + x 1 x , x > 0 , is continuous at x = 0 . If f ' 1 = f − 1 , then the value of g 3 is

Step-by-step solution

Let, g x = a x + b Now function f x in continuous at x = 0 ⇒ lim x → 0 + f x = f 0 ⇒ lim x → 0 1 + x 2 + x 1 x = b ⇒ 0 = b ⇒ g x = a x Now, for x > 0 f x = 1 + x 2 + x 1 x ⇒ log f x = log 1 + x 2 + x 1 x ⇒ log f x = log 1 + x - log 2 + x x ⇒ 1 f x × f ' x = x 1 1 + x - 1 2 + x - log 1 + x - log 2 + x x 2 ⇒ f ' x = f x x 1 1 + x - 1 2 + x - log 1 + x - log 2 + x x 2 ⇒ f ' 1 = f 1 1 2 - 1 3 - log 2 - log 3 1 2 ⇒ f ' 1 = 2 3 1 6 - log 2 3 And, f − 1 = g − 1 = − a ⇒ a = 2 3 log 2 3 − 1 9 Now, finding g 3 = 3 a + 0 ⇒ g