JEE Advance - Mathematics (2018 - Paper 2 Offline - No. 11)
Let f : R $$ \to $$ R be a differentiable function with f(0) = 1 and satisfying the equation f(x + y) = f(x) f'(y) + f'(x) f(y) for all x, y$$ \in $$ R.
Then, the value of loge(f(4)) is ...........
Then, the value of loge(f(4)) is ...........
Answer
2
Explanation
Given,
$$f(x + y) = f(x)f'(y) + f'(x)f(y),\,\forall x,y \in R$$ and f(0) = 1
Put x = y = 0, we get
f(0) = f(0) f'(0) + f'(0) f(0)
$$ \Rightarrow 1 = 2f'(0) \Rightarrow f'(0) = {1 \over 2}$$
Put x = x and y = 0, we get
f(x) = f(x) f'(0) + f'(x) f(0)
$$ \Rightarrow f(x) = {1 \over 2}f(x) + f'(x)$$
$$ \Rightarrow f'(x) = {1 \over 2}f(x) \Rightarrow {{f'(x)} \over {f(x)}} = {1 \over 2}$$
On integrating, we get
$$\log f(x) = {1 \over 2}x + C$$
$$ \Rightarrow f(x) = A{e^{{1 \over 2}x}}$$, where eC = A
If f(0) = 1, then A = 1
Hence, $$f(x) = {e^{{1 \over 2}x}}$$
$$ \Rightarrow {\log _e}f(x) = {1 \over 2}x$$
$$ \Rightarrow {\log _e}f(4) = {1 \over 2} \times 4 = 2$$
$$f(x + y) = f(x)f'(y) + f'(x)f(y),\,\forall x,y \in R$$ and f(0) = 1
Put x = y = 0, we get
f(0) = f(0) f'(0) + f'(0) f(0)
$$ \Rightarrow 1 = 2f'(0) \Rightarrow f'(0) = {1 \over 2}$$
Put x = x and y = 0, we get
f(x) = f(x) f'(0) + f'(x) f(0)
$$ \Rightarrow f(x) = {1 \over 2}f(x) + f'(x)$$
$$ \Rightarrow f'(x) = {1 \over 2}f(x) \Rightarrow {{f'(x)} \over {f(x)}} = {1 \over 2}$$
On integrating, we get
$$\log f(x) = {1 \over 2}x + C$$
$$ \Rightarrow f(x) = A{e^{{1 \over 2}x}}$$, where eC = A
If f(0) = 1, then A = 1
Hence, $$f(x) = {e^{{1 \over 2}x}}$$
$$ \Rightarrow {\log _e}f(x) = {1 \over 2}x$$
$$ \Rightarrow {\log _e}f(4) = {1 \over 2} \times 4 = 2$$
Comments (0)
