$$f\left( x \right)$$ and $$g\left( x \right)$$ are two differentiable functions on $$\left[ {0,\,2} \right]$$ such that $$f''\left( x \right) - g''\left( x \right) = 0,\,\,f'\left( 1 \right) = 2g'\left( 1 \right) = 4f\left( 2 \right) = 3g\left( 2 \right) = 9$$ then $$f\left( x \right) - g\left( x \right)$$ at $$x = \frac{3}{2}$$ is-
A.
$$0$$
B.
$$2$$
C.
$$10$$
D.
$$5$$
Answer :
$$5$$
Solution :
$$\eqalign{
& \because f''\left( x \right) - g''\left( x \right) = 0 \cr
& {\text{Integrating, }}f'\left( x \right) - g'\left( x \right) = c\,; \cr
& \Rightarrow f'\left( 1 \right) - g'\left( 1 \right) = c \cr
& \Rightarrow 4 - 2 = c \cr
& \Rightarrow c = 2 \cr
& \therefore f'\left( x \right) - g'\left( x \right) = 2\,; \cr
& {\text{Integrating, }}f\left( x \right) - g\left( x \right) = 2x + {c_1} \cr
& \Rightarrow f\left( 2 \right) - g\left( 2 \right) = 4 + {c_1} \cr
& \Rightarrow 9 - 3 = 4 + {c_1}\,; \cr
& \Rightarrow {c_1} = 2 \cr
& \therefore f\left( x \right) - g\left( x \right) = 2x + 2 \cr
& {\text{At }}x = \frac{3}{2},\,f\left( x \right) - g\left( x \right) = 3 + 2 = 5 \cr} $$
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-