JEE MAIN - Mathematics (2020 - 4th September Morning Slot - No. 5)
Suppose a differentiable function f(x) satisfies the identity
f(x+y) = f(x) + f(y) + xy2 + x2y, for all real x and y.
$$\mathop {\lim }\limits_{x \to 0} {{f\left( x \right)} \over x} = 1$$, then f'(3) is equal to ______.
f(x+y) = f(x) + f(y) + xy2 + x2y, for all real x and y.
$$\mathop {\lim }\limits_{x \to 0} {{f\left( x \right)} \over x} = 1$$, then f'(3) is equal to ______.
Answer
10
Explanation
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, $$\mathop {\lim }\limits_{x \to 0} {{f(x)} \over x} = 1$$
This is in $$ \frac{0}{0} $$ form, so we can apply L' hospital rule.
$$\mathop {\lim }\limits_{x \to 0} {{f'(x)} \over 1} = 1$$
$$ \Rightarrow f'(0) = 1$$
Putting value of f'(0) at equation (2), we get
f'(x) = 1 + x2
$$ \therefore $$ f'(3) = 1 + 32 = 10
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, $$\mathop {\lim }\limits_{x \to 0} {{f(x)} \over x} = 1$$
This is in $$ \frac{0}{0} $$ form, so we can apply L' hospital rule.
$$\mathop {\lim }\limits_{x \to 0} {{f'(x)} \over 1} = 1$$
$$ \Rightarrow f'(0) = 1$$
Putting value of f'(0) at equation (2), we get
f'(x) = 1 + x2
$$ \therefore $$ f'(3) = 1 + 32 = 10
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