A) \[\frac{4\sqrt{3}-1}{3\sqrt{3}}\]
B) \[\frac{3\sqrt{3}-1}{2}\]
C) \[\frac{4-3\sqrt{3}}{3}\]
D) None of these
Correct Answer: C
Solution :
[c] According to question \[f'(1)=tan\frac{\pi }{6}=\frac{1}{\sqrt{3}}\] \[f'(2)=tan\frac{\pi }{3}=\sqrt{3}\] and \[f'(3)=tan\frac{\pi }{4}=1\] so, \[\int\limits_{1}^{3}{f''(x)f'(x)dx+\int\limits_{1}^{3}{f''(x)dx}}\] \[=\left[ \frac{{{\left\{ f'(x) \right\}}^{2}}}{2} \right]_{1}^{3}+\left[ f'(x) \right]_{2}^{3}\] \[=\frac{1}{2}\left[ 1-\frac{1}{3} \right]+\left[ 1-\sqrt{3} \right]=\frac{4}{3}-\sqrt{3}\]
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