Question

Let f(x) be a function which satisfies f(x³)f’(x) = f’(x)f’(x³) + f”(x²). Given that f(1) = 1 and f”’(1) = `1/4`, then value of 4(f’(1) + f'”(1)) is ______.

विकल्प

• 0.00

• 1.00

• 2.00

• 3.00

Answer 1

Let f(x) be a function which satisfies f(x³)f’(x) = f’(x)f’(x³) + f”(x²). Given that f(1) = 1 and f”’(1) = `1/4`, then value of 4(f’(1) + f'”(1)) is 3.00.

Explanation:

Let f”(1) = b and f’(1) = a

Put x = 1 in given equation a² + b = a

b = a – a²

Differentiate the given equation

3x²f’(x³)f’(x) + f(x³)f”(x) = f”(x)f’(x³) + 3x²f’(x)f’’(x³) + 2xf’’’(x²)

Put x = 1

3a² + b = `ab + 3ab + 1/2`

3a² + a – a² = `4a(a - a^2) + 1/2`

`4a^3 - 2a^2 + a - 1/2` = 0

`4a^2(a - 1/2) + (a - 1/2)` = 0

a = `1/2` and b = `1/4`

4(a + b) = 3