Let $$f\left( a \right) = g\left( a \right) = k$$ and their nth derivatives $${f^n}\left( a \right),\,{g^n}\left( a \right)$$ exist and are not equal for some $$n.$$ Further if $$\mathop {\lim }\limits_{x \to a} \frac{{f\left( a \right)g\left( x \right) - f\left( a \right) - g\left( a \right)f\left( x \right) + f\left( a \right)}}{{g\left( x \right) - f\left( x \right)}} = 4$$ then the value of $$k$$ is-
A.
$$0$$
B.
$$4$$
C.
$$2$$
D.
$$1$$
Answer :
$$4$$
Solution :
$$\eqalign{
& \mathop {\lim }\limits_{x \to a} \frac{{f\left( a \right)g'\left( x \right) - g\left( a \right)f'\left( x \right)}}{{g'\left( x \right) - f'\left( x \right)}} = 4\,\,\left( {{\text{By L'Hospital rule}}} \right) \cr
& \mathop {\lim }\limits_{x \to a} \frac{{k\,g'\left( x \right) - k\,f'\left( x \right)}}{{g'\left( x \right) - f'\left( x \right)}} = 4 \cr
& \therefore k = 4 \cr} $$
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-