If the equation $ a_{n} x^{n}+a_{n-1} x^{n-1}+\ldots+a_{1} x=0, a_{1} \neq 0, n \geq 2 $, has a positive root $ x=\alpha $, then the equation $ n a_{n} x^{n-1}+(n-1) a_{n-1} x^{n-2}+\ldots+a_{1}=0 $ has a positive root, which is

Question

If the equation $ a_{n} x^{n}+a_{n-1} x^{n-1}+\ldots+a_{1} x=0, a_{1} \neq 0, n \geq 2 $, has a positive root $ x=\alpha $, then the equation $ n a_{n} x^{n-1}+(n-1) a_{n-1} x^{n-2}+\ldots+a_{1}=0 $ has a positive root, which is, Smaller then $ \alpha $, Greater then $ \alpha $, Equal to $ \alpha $, Greater then or equal $ \alpha $

MARKS App · Accepted Answer

Let f x = a n x n + a n - 1 x n - 1 + . . . + a 1 x . . . 1 ∴ f ⁡ 0 = 0 Now f α = 0 Since, x = α is a root of differentiate equation 1 w.r.t. x , we get f ⁡ ' x = n a n x n - 1 + n - 1 a n - 1 x n - 2 + ... + a 1 = 0 Since, f α = f 0 = 0 . So, according to Rolle's theorem f ' x = 0 somewhere in between 0 & α . Hence, f ' x = 0 will have a root smaller than α .

Search any question & find its solution

Looking for more such questions to practice?