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in Continuity and Differentiability by (28.7k points)
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If the polynomial equation a0xn + an–1xn–1 + an–2xn–2 + …. a2x2 + a1x + a0 = 0

n being a positive integer, has two different real roots α and β, then between α and β, the equation n anxn–1 + (n–1) an–1xn–2 + … + a1 = 0 has

A. Exactly one root

B. Almost one root

C. At least one root

D. No root

1 Answer

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Best answer

Correct answer is C.

As the polynomial, nanxn–1 + (n–1) an–1xn–2 + … + a1 = 0 is a derivative of the polynomial a0xn + an–1xn–1 + an–2xn–2 +..............a2x2 + a1x + a0 = 0 --------- (i)

Putting x = 0 in equation (i),

f(0) = a0 < 0, {Y – Intercept of the graph is negative}

On the other hand,

\(\because\) an > 0 and ‘n’ is even, the leading term on xn, is positive for only x.

For |x| to be large, the term anxn will dominate, so

lim x→-∞ f (x) = +∞

lim x→+∞ f (x) = +∞

If lim x→+∞ f (x) = +∞, there must exist

Same number α < 0, where f(α > 0)

\(\because\) f(0) = a0 < 0,

\(\therefore\) α < β < 0, such that f(β) = 0

Also, there is some value 0 < α,

Where f(a) & so there exists,

0 < b < a with f(b) = 0

Additionally, the polynomial function (equation (i)) is continuous everywhere in R and consequently derivative in R.

\(\therefore\) a0xn + an–1xn–1 + an–2xn–2 +..............a2x2 + a1x + a0 = 0 is continuous on α, β and derivative on α, β.

Thus, it satisfies both the conditions of Rolle's Theorem.

As per the Rolle’s Theorem, between any two roots of a function f(x), there exists at least one root of its derivative.

Thus, the equation nanxn–1 + (n–1) an–1xn–2 + … + a1 = 0 will have at least one root between α & β.

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