Question

If $f(x)={ }^{40} C_{1} \cdot x(1-x)^{39}+2 \cdot{ }^{40} C_{2} x^{2}(1-x)^{38}+3^{40} C_{3} \times$ $x^{3}(1-x)^{37}+\ldots+40 \cdot{ }^{40} C_{40} x^{40},$ then the value of $f(3)$ is not divisible by

• A. 5
• B. 6
• C. 8
• D. None of these

Answer 1

Answer: (D)

$f(x)={ }^{40} C_{1} \cdot x(1-x)^{39}+2 \cdot{ }^{40} C_{2} x^{2}(1-x)^{38}$
$+3{ }^{40} C_{3} x^{3}(1-x)^{37}+\ldots+40 \cdot{ }^{40} C_{40} x^{40}$
$T_{r}=r \cdot{ }^{40} C_{r} x^{r}(1-x)^{40-r}=40 x \cdot{ }^{39} C_{r-1} x^{r-1}(1-x)^{40-r}$
$\therefore f(x)=40 x(x+1-x)^{39}=40 x$