A) \[\frac{-\pi -2}{2}\]
B) \[-1-{{\cos }^{-1}}(2)\]
C) \[\frac{\pi +2}{2}\]
D) \[\frac{\pi -2}{2}\]
Correct Answer: C
Solution :
L.H.L. at x = 1 is . 1 R.H.L at x = 1 is \[a+{{\cos }^{-1}}(1+b)\] \[\Rightarrow \]\[-1=a+{{\cos }^{-1}}(1+b)\] \[{{\cos }^{-1}}(1+b)=-1-a\] ?(i) Now L.H.d. at \[x=1\] is - 1 R.H.D at x = 1 is \[\frac{-1}{\sqrt{1-{{(1+b)}^{2}}}}\] \[\Rightarrow \]\[{{(1+b)}^{2}}=0\]\[\Rightarrow \]\[b=-1\] \[a=-1-\frac{\pi }{2}\] \[\frac{a}{b}=\frac{-(2+\pi )}{2(-1)}=\frac{2+\pi }{2}\]You need to login to perform this action.
You will be redirected in
3 sec