If $$f\left( {x + y} \right) = f\left( x \right).f\left( y \right)\forall x.y$$ and $$f\left( 5 \right) = 2,\,f'\left( 0 \right) = 3,$$ then $$f'\left( 5 \right)$$ is-
A.
$$0$$
B.
$$1$$
C.
$$6$$
D.
$$2$$
Answer :
$$6$$
Solution :
$$f\left( {x + y} \right) = f\left( x \right) \times f\left( y \right)$$
Differentiate 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 $$f'\left( x \right) = f'\left( 0 \right)f\left( x \right)\,;$$
$$ \Rightarrow f'\left( 5 \right) = 3f\left( 5 \right){\text{ }} = 3 \times 2{\text{ }} = 6$$
Releted MCQ Question on Calculus >> Differentiability and Differentiation
Releted Question 1
There exist a function $$f\left( x \right),$$ satisfying $$f\left( 0 \right) = 1,\,f'\left( 0 \right) = - 1,\,f\left( x \right) > 0$$ for all $$x,$$ and-
A.
$$f''\left( x \right) > 0$$ for all $$x$$
B.
$$ - 1 < f''\left( x \right) < 0$$ for all $$x$$
C.
$$ - 2 \leqslant f''\left( x \right) \leqslant - 1$$ for all $$x$$
If $$f\left( a \right) = 2,\,f'\left( a \right) = 1,\,g\left( a \right) = - 1,\,g'\left( a \right) = 2,$$ then the value of $$\mathop {\lim }\limits_{x \to a} \frac{{g\left( x \right)f\left( a \right) - g\left( a \right)f\left( x \right)}}{{x - a}}$$ is-
Let $$f:R \to R$$ be a differentiable function and $$f\left( 1 \right) = 4.$$ Then the value of $$\mathop {\lim }\limits_{x \to 1} \int\limits_4^{f\left( x \right)} {\frac{{2t}}{{x - 1}}} dt$$ is-