Applications Of Derivatives Question 229
Question: Let $ f(x)=\int\limits_0^{x}{\frac{\cos t}{t}dt,x>0} $ then $ f(x) $ has
[Kurukshetra CEE 2002]
Options:
A) Maxima when $ n=-2,,-4,,-6,,….. $
B) Maxima when $ n=-1,,-3,,-5,,…. $
C) Minima when $ n=0,,2,,4,…. $
D) Minima when $ n=1,3,5…. $
Show Answer
Answer:
Correct Answer: B
Solution:
$ f(x)=\int\limits_0^{x}{\frac{\cos t}{t}},dt $ , $ x>0 $
therefore $ {f}’(x)=\frac{\cos x}{x} $ , $ x>0 $
therefore $ {f}’(x)=0\Rightarrow \frac{\cos x}{x}=0 $
therefore $ x=(2n+1)\frac{\pi }{2} $ , for $ n\in z $ .
Now $ {f}’’(x)=\frac{-x\sin x-\cos x}{x^{2}} $
\ $ {f}’’[(2n+1)\pi /2]=\frac{-2}{(2n+1)\pi },{{(-1)}^{n}} $ = $ \frac{2{{(-1)}^{n+1}}}{(2n+1)\pi } $ .
Thus $ {f}’’(x)>0 $
$ n=-2,,-4,,-6,…….. $
$ {f}’’(x)<0 $
$ n=0,,2,,4,,…….. $
$ {f}’’(x)>0 $
$ n=1,,3,,5…….. $
$ {f}’’(x)<0 $
$ n=-1,,-3,,-5…….. $
Thus $ f(x) $ attain maximum for $ n=-1,,-3,,-5 $ ,…………. and minimum for $ n=1,,3,,5 $ ,…………..
BETA
