Let $$F\left( x \right) = f\left( x \right) + f\left( {{1 \over x}} \right),$$ where $$f\left( x \right) = \int\limits_l^x {{{\log t} \over {1 + t}}dt,} $$ Then $$F(e)$$ equals
A
$$1$$
B
$$2$$
C
$$1/2$$
D
$$0$$
Explanation
Given $$f\left( x \right) = f\left( x \right) + f\left( {{1 \over x}} \right),$$
where $$f\left( x \right) = \int_1^x {{{\log \,t} \over {1 + t}}} \,dt$$
$$\therefore$$ $$F\left( e \right) = f\left( e \right) + f\left( {{1 \over e}} \right)$$
The value of $$\int\limits_1^a {\left[ x \right]} f'\left( x \right)dx,a > 1$$ where $${\left[ x \right]}$$ denotes the greatest integer not exceeding $$x$$ is
A
$$af\left( a \right) - \left\{ {f\left( 1 \right) + f\left( 2 \right) + .............f\left( {\left[ a \right]} \right)} \right\}$$
B
$$\left[ a \right]f\left( a \right) - \left\{ {f\left( 1 \right) + f\left( 2 \right) + ...........f\left( {\left[ a \right]} \right)} \right\}$$
C
$$\left[ a \right]f\left( {\left[ a \right]} \right) - \left\{ {f\left( 1 \right) + f\left( 2 \right) + ...........f\left( a \right)} \right\}$$
D
$$af\left( {\left[ a \right]} \right) - \left\{ {f\left( 1 \right) + f\left( 2 \right) + .............f\left( a \right)} \right\}$$
Explanation
Let $$a = k + h$$ where $$k$$ is an integer such that
$$\left[ a \right] = k$$ and $$0 \le h < 1$$
$$\therefore$$ $$\int\limits_1^a {\left[ x \right]f'\left( x \right)dx = \int\limits_1^2 {1f'\left( x \right)} } \,dx$$
$$\,\,\,\,\,\,\,\,\,\,\,\, + \int\limits_2^3 {2f'\left( x \right)dx + } ....$$