Differentiation Question 383
Question: $ f(x) $ and $ g(x) $ are two differentiable function on $ [0,2] $ such that $ f’’(x)-g’’(x)=0,f’(1)=2,g’(1)=4 $ , $ f(2)=3 $ , $ g(2)=9, $ then $ f(x)-g(x) $ at $ x=3/2 $ is
[AIEEE 2002]
Options:
A) 0
B) 2
C) 10
D) - 5
Show Answer
Answer:
Correct Answer: D
Solution:
$ {}^{( x\frac{dy}{dx}-y )}/ _{x^{2}=\frac{1}{x}-\frac{1}{a+bx}b=\frac{a}{x(a+bx)}} $
Integrate w.r.t. x, $ {f}’(x)-{g}’(x)=c $
At $ x=1 $ , $ {f}’(1)-{g}’(1)=c $
Therefore $ 2-4=c $
Therefore $ c=-2 $
Hence, $ {f}’(x)-{g}’(x)=-2 $ . Again integrate w.r.t. x, $ f(x)-g(x)=-2x+c_1 $ . At $ x=2 $ , $ f(2)-g(2)=-2\times 2+c_1 $
Therefore $ 3-9+4=c_1 $
Therefore $ c_1=-2 $
Then $ f(x)-g(x)=-2x-2=-(2x+2) $
$ f(3/2)-g(3/2)=-( 2\times \frac{3}{2}+2 )=-5 $ .
BETA
