Question
The equation of a curve is $$y = f\left( x \right).$$ The tangents at $$\left( {1,\,f\left( 1 \right)} \right),\,\left( {2,\,f\left( 2 \right)} \right)$$ and $$\left( {3,\,f\left( 3 \right)} \right)$$ make angles $$\frac{\pi }{6},\,\frac{\pi }{3}$$ and $$\frac{\pi }{4}$$ respectively with the positive direction of the x-axis. Then the value of $$\int_2^3 {f'\left( x \right)f''\left( x \right)dx + \int_1^3 {f''\left( x \right)dx} } $$ is equal to :
A.
$$ - \frac{1}{{\sqrt 3 }}$$
B.
$$\frac{1}{{\sqrt 3 }}$$
C.
0
D.
none of these
Answer :
$$ - \frac{1}{{\sqrt 3 }}$$
Solution :
$$\eqalign{
& {\text{Here,}}\,\,f'\left( 1 \right) = \tan \frac{\pi }{6} = \frac{1}{{\sqrt 3 }} \cr
& f'\left( 2 \right) = \tan \frac{\pi }{3} = \sqrt 3 \cr
& f'\left( 3 \right) = \tan \frac{\pi }{4} = 1 \cr
& {\text{Now, }}\int_2^3 {f'\left( x \right).f''\left( x \right)dx = \left[ {\frac{1}{2}{{\left\{ {f'\left( x \right)} \right\}}^2}} \right]_2^3 = \frac{1}{2}\left[ {{{\left\{ {f'\left( 3 \right)} \right\}}^2} - {{\left\{ {f'\left( 2 \right)} \right\}}^2}} \right]} \cr
& \int_1^3 {f''\left( x \right)dx} = \left[ {f'\left( x \right)} \right]_1^3 = f'\left( 3 \right) - f'\left( 1 \right) \cr
& \therefore {\text{ value}}\, = \frac{1}{2}\left\{ {{1^2} - {{\left( {\sqrt 3 } \right)}^2}} \right\} + 1 - \frac{1}{{\sqrt 3 }} = - \frac{1}{{\sqrt 3 }} \cr
& \cr} $$