Question
If the equation $${a_n}{x^n} + {a_{n - 1}}{x^{n - 1}} + ......+ {a_1}x = 0$$
$${a_1} \ne 0,\,n \geqslant 2,$$ has a positive root $$x = \alpha ,$$ then the equation $$n{a_n}{x^{n - 1}} + \left( {n - 1} \right){a_{n - 1}}{x^{n - 2}} + ....... + {a_1} = 0$$ has a positive root, which is
A.
greater than $$\alpha $$
B.
smaller than $$\alpha $$
C.
greater than or equal to $$\alpha $$
D.
equal to $$\alpha $$
Answer :
smaller than $$\alpha $$
Solution :
$${\text{Let}}\,f\left( x \right) = {a_n}{x^n} + {a_{n - 1}}{x^{n - 1}} + ........ + {a_1}x = 0$$
The other given equation,
$$\eqalign{
& n{a_n}{x^{n - 1}} + \left( {n - 1} \right){a_{n - 1}}{x^{n - 2}} + ....... + {a_1} = 0 = f'\left( x \right) \cr
& {\text{Given}}\,{a_1} \ne 0 \Rightarrow f\left( 0 \right) = 0 \cr
& {\text{Again}}\,f\left( x \right)\,{\text{has}}\,{\text{root}}\,\alpha , \Rightarrow f\left( \alpha \right) = 0 \cr
& \therefore f\left( 0 \right) = f\left( \alpha \right) \cr} $$
$$\therefore $$ By Roll’s theorem $$f'\left( x \right) = 0$$ has root between $$\left( {0,\alpha } \right)$$
Hence $$f'\left( x \right)$$ has a positive root smaller than $${\alpha .}$$