A) \[4\]
B) \[5-\frac{\pi }{4}\]
C) \[4+\frac{3\pi }{2}\]
D) None of these
Correct Answer: C
Solution :
[c] \[f(x)={{\sin }^{-1}}x-{{\cot }^{-1}}x+{{x}^{2}}+2x+6\] Domain of \[f(x)\] is \[[-1,1]\]. \[f'(x)=\frac{1}{\sqrt{1-{{x}^{2}}}}+\frac{1}{1+{{x}^{2}}}+2x+2>0\,\forall x\in [-1,1]\] So, \[f(x)\]is increasing in its domain. \[\therefore \,\,\,{{f}_{\min }}(at\,\,x=-1)=\frac{\pi }{2}-\frac{3\pi }{4}+5=5-\frac{5\pi }{4}\] \[{{f}_{\max }}(at\,\,x=1)=\frac{\pi }{2}-\frac{\pi }{4}+9=9+\frac{\pi }{4}\] \[\therefore \,\,\,{{f}_{\max .}}-{{f}_{\min ,}}=4+\frac{3\pi }{2}\]
You need to login to perform this action.
You will be redirected in
3 sec
