Definite Integration Question 344
Question: If $ f(x)=\int _{-1}^{x}{|t|dt,} $
$ x\ge -1, $ then
[MNR 1994]
Options:
A) $ f $ and $ {f}’ $ are continous for $ x+1>0 $
B) $ f $ is continous but $ {f}’ $ is not continous for $ x+1>0 $
C) $ f $ and $ {f}’ $ are not continous at $ x=0 $
D) $ f $ is continous at $ x=0 $ but $ {f}’ $ is not so
Show Answer
Answer:
Correct Answer: A
Solution:
Let us divide the interval into two sub-intervals $ I_1 $ , $ -1\le x<0 $ so that x is -ve and $ I_2,x\ge 0 $ so that x is +ve.
For $ I_1,f(x)=\int _{-1}^{x}{(-t)dt=-\frac{1}{2}(x^{2}-1)} $ …..(i)
For $ I_1,f(x)=\int _{-1}^{x}{(-t)dt=-\frac{1}{2}(x^{2}-1)} $
$ =-\frac{1}{2}[t^{2}] _{-1}^{0}+\frac{1}{2}[t^{2}]_0^{x}=\frac{1}{2}(1+x^{2}) $ …..(ii)
Hence the function can be defined as the following
$ f(x)= \begin{cases} & -\frac{1}{2}(x^{2}-1),If-1\le x0 \\ \end{cases} . $
For $ f,L=R=V=\frac{1}{2} $ at $ x=0 $ , so $ f $ is continuous at $ x=0 $ . For $ f’,L=R=V=0 $ at $ x=0 $ , so $ f’ $ is also continuous at $ x=0 $ . Thus both $ f $ and $ f’ $ are continuous at $ x=0 $ and hence both are continuous for $ x>-1 $ i.e., $ x+1>0 $ .
BETA
