JEE MAIN - Mathematics (2002 - No. 10)
If f(x + y) = f(x).f(y) $$\forall $$ x, y and f(5) = 2, f'(0) = 3, then
f'(5) is
f'(5) is
0
1
6
2
Explanation
$$f\left( {x + y} \right) = f\left( x \right) \times f\left( y \right)$$
Differeniate with respect to $$x,$$ treating $$y$$ as constant
$$f'\left( {x + y} \right) = f'\left( x \right)f\left( y \right)$$
Putting $$x=0$$ and $$y=x$$, we get
$$f'\left( x \right) = f'\left( 0 \right)f\left( x \right);$$
$$ \Rightarrow f'\left( 5 \right) = 3f\left( 5 \right) = 3 \times 2 = 6.$$
Differeniate with respect to $$x,$$ treating $$y$$ as constant
$$f'\left( {x + y} \right) = f'\left( x \right)f\left( y \right)$$
Putting $$x=0$$ and $$y=x$$, we get
$$f'\left( x \right) = f'\left( 0 \right)f\left( x \right);$$
$$ \Rightarrow f'\left( 5 \right) = 3f\left( 5 \right) = 3 \times 2 = 6.$$
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