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:

Calculation:

GIven, f(x) f”(x) – f'(x)2 = 0

Let h(x) = f(x)/f'(x)

⇒ h'(x) = 0

⇒ h(x) = k

⇒ f(x)/f'(x) = k

⇒ f(x) = k f’(x)

Putting k = 0, we get

⇒ f(0) = k f'(0)

⇒ k = 1/2

Now, f(x) = \(\frac{1}{2}\) f'(x)

⇒ 2 = f'(x)/f(x)

Integrating on both sides, we get

⇒ ∫ 2dx = ∫ f'(x)/f(x) dx

⇒ 2x = ln |f(x)| + C

Now, f(0) = 1

⇒ C = 0

∴ 2x = ln |f(x)|

⇒ f(x) = ±e2x

As f(0) = 1

⇒ f(x) = e2x

∴ f(1) = e2 ≈ 7.38

∴ The value of f(1) lies in the interval (6, 9).

The correct answer is Option 2.