QJEEM20Ma556e: JEE Main 2020 Mathematics - Class 12 JEEM20Ma556e Solution: Suppose a differentiable function f(x) satisfies the identity…

Given, f(x + y) = f(x) + f(y) + xy² + x²y ...(1)

differentiating partially with respect to x,

f'(x+y) = f'(x) + 0 + y² + y(2x) [y = constant]

Put x = 0 and y = x

$$\therefore$$ f'(x) = f'(0) + x² ....(2)

putting x = y = 0 at equation (1),

f(0) = 2f(0)

$$\Rightarrow$$ f(0) = 0

Given, $$\underset{x\to 0}{lim}\frac{f(x)}{x}=1$$

This is in $$\frac{0}{0}$$ form, so we can apply L' hospital rule.

$$\underset{x\to 0}{lim}\frac{f^{\prime }(x)}{1}=1$$

$$\Rightarrow f^{\prime }(0)=1$$

Putting value of f'(0) at equation (2), we get

f'(x) = 1 + x²

$$\therefore$$ f'(3) = 1 + 3² = 10