Integral Calculus Question 196
Question: The tangent of the curve $ y=f(x) $ at the point with abscissa $ x=1 $ from an angle of $ \pi /6 $ and at the point $ x=2 $ an angle of $ \pi /3 $ and at the point $ x=3 $ an angle of $ \pi /4 $ . If $ f’’(x) $ is continuous, then the value of $ \int\limits_1^{3}{f’’(x)f’(x)dx+\int\limits_2^{3}{f’’(x)dx}} $ is
Options:
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
Show Answer
Answer:
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}} $ $ =[ \frac{{{{ f’(x) }}^{2}}}{2} ]_1^{3}+[ f’(x) ]_2^{3} $ $ =\frac{1}{2}[ 1-\frac{1}{3} ]+[ 1-\sqrt{3} ]=\frac{4}{3}-\sqrt{3} $
BETA
