A) \[\frac{-\pi }{4}\]
B) \[\frac{-\pi }{2}\]
C) \[\frac{3\pi }{2}\]
D) \[\pi \]
Correct Answer: A
Solution :
[a] \[f(t)=\int\limits_{2t}^{{{t}^{2}}}{{{\tan }^{-1}}\left| \frac{{{(1+t)}^{2}}-x}{1+x} \right|dx}\] ...(1) \[=\int\limits_{2t}^{{{t}^{2}}}{{{\tan }^{-1}}\left| \frac{1+x}{{{(1+t)}^{2}}-x} \right|dx}\] \[\left( \because \,\,\,\int\limits_{a}^{b}{f(x)\,\,dx=\int\limits_{a}^{b}{f(a+b-x)dx}} \right)\] ?.(2) Adding (1) and (2), we get \[2f(t)=\int\limits_{2t}^{{{t}^{2}}}{\frac{\pi }{2}.\,\,dx}=\frac{\pi }{2}({{t}^{2}}-2t)\] \[\therefore \,\,\,f(t)=\frac{\pi }{4}\,\left( {{(t-1)}^{2}}-1 \right)\] \[\therefore \] Minimum value of \[f(t)=-\frac{\pi }{4}\]
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