Let f be a twice differentiable function defined on R such that f 0 = 1 , f ' 0 = 2 and f ' x ≠ 0 for all x ∈ R . If f x f ' x f ' x f '' x = 0 , for all x ∈ R , then the value of f 1 lies in the interval

Step-by-step solution

We have, f x f ' x f ' x f ' ' x = 0 ⇒ f x f ' ' x - f ' x 2 = 0 ⇒ f ' ' x f ' x = f ' x f x On integrating both side, we get ln f ' x = ln f x + ln c ⇒ f ' x = cf x ⇒ f ' x f x = c Again integrating, we get ln   f x = c x + k 1 ⇒ f x = k e c x Since, f 0 = 1 = k Therefore, f ' 0 = c = 2 Now, f x = e 2 x Hence, f 1 = e 2 ∈ 6 , 9