Suppose $$f\left( x \right)$$ is differentiable at $$x = 1$$ and $$\mathop {\lim }\limits_{h \to 0} \frac{1}{h}f\left( {1 + h} \right) = 5,$$ then $$f'\left( 1 \right)$$ equals :
A.
$$3$$
B.
$$4$$
C.
$$5$$
D.
$$6$$
Answer :
$$5$$
Solution :
$$f'\left( 1 \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {1 + h} \right) - f\left( 1 \right)}}{h}\,;$$
As function is differentiable so it is continuous as it is given that $$\mathop {\lim }\limits_{h \to 0} \frac{{f\left( {1 + h} \right)}}{h} = 5$$ and hence $$f\left( 1 \right) = 0$$
Hence, $$f'\left( 1 \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {1 + h} \right)}}{h} = 5$$
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-