MCQ
If $F(x)=f(x)+f (\frac {1}{x})$, where $F(x)= \mathop \smallint \limits_{1 }^x \frac {log\, t}{1+t}$ then $F\left( e \right) = $ .
- A$1$
- B$2$
- ✓$0.5$
- D$0$
$f\left(\frac{1}{x}\right)=\int_{1}^{x} \frac{\log _{e}^{t}}{t+1} d t$
Let $t=\frac{1}{h}, d t=-\frac{1}{h^{2}} d t$
if $t=1, h-1$ or $t=\frac{1}{x}, 4-x$
$f\left(\frac{1}{x}\right)=\int_{1}^{x} \frac{\log _{e}\left(\frac{1}{h}\right)}{1+\frac{1}{h}}\left(-\frac{1}{h^{2}} d x\right)$
$f\left(\frac{1}{x}\right)=\int_{1}^{x} \frac{-\log h(-1)}{h+1} h d h$
$f\left(\frac{1}{x}\right)=\int_{1}^{x} 9 \frac{\log _{e}^{t}}{t(t+1)} d t ......(2)$
Adding 1 and 2
$f(x)+f\left(\frac{1}{x}\right)=\int_{1}^{x} \frac{\log _{e}^{t}}{t+1}\left(1+\frac{1}{t}\right) d t$
$\int_{1}^{x} \frac{\log _{e}^{t}}{t+1}\left(1+\frac{1}{t}\right) d t$
$\int_{1}^{x} \frac{\log _{e}^{t}}{t} d t$
$\log t=v$
$\frac{1}{t} d t=d v$
$f(x)+f\left(\frac{1}{x}\right)=\int_{0}^{\log x} v d v$
$=\left[\frac{v^{2}}{2}\right]_{0}^{\log x}$
$=\frac{\left(\log _{e}^{x}\right)^{2}}{2}$
$f(e)+f\left(\frac{1}{e}\right)=\frac{\left(\log _{e}^{e}\right)^{2}}{2}=\frac{1}{2}$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$1.$ If $1$ ball is drawn from each of the boxes $B_1, B_2$ and $B_3$, the probability that all $3$ drawn balls are of the same colour is
$(A)$ $\frac{82}{648}$ $(B)$ $\frac{90}{648}$ $(C)$ $\frac{558}{648}$ $(D)$ $\frac{566}{648}$
$2.$ If $2$ balls are drawn (without replacement) from a randomly selected box and one of the balls is white and the other ball is red, the probability that these $2$ balls are drawn from bo $B _2$ is
$(A)$ $\frac{116}{181}$ $(B)$ $\frac{126}{181}$ $(C)$ $\frac{65}{181}$ $(D)$ $\frac{55}{181}$
Give the answer question $1$ and $2.$