Application of DerivativeHard
Question
Let p(x) = a0 + a1 x2 + a2x4 + ..... + anx2n be a polynomial ina real variable xwith 0 < a0 < a1 < a1 < .....n The function P(x) has
Options
A.neither a maximum nor a minimum
B.only one maximum
C.only one minimum
D.only one maximum and only one minimum
Solution
Given P(x) = a0 + a1x2 + a2x4 + ...... + anx2n
where, an > an-1 > an-2 ....> a2 > a1 > a0 > 0
⇒ P′(x) = 2a1x + 4a2x3 + .... + 2nanx2n-1
= 2x(a1 + 2a1x2 + .... + nanx2n-2 ..... (i)
where, (a1 + 2a2x2 + 3a3x4 + ..... + nanx2n-2) > 0
for all x ∈ R
Thus,
ie P′(x) changes sign from (-ve) to (+ ve) at x = 0.
∴ P(x) attains minimun at x = 0.
Hence, it has only one minimun at x = 0.
where, an > an-1 > an-2 ....> a2 > a1 > a0 > 0
⇒ P′(x) = 2a1x + 4a2x3 + .... + 2nanx2n-1
= 2x(a1 + 2a1x2 + .... + nanx2n-2 ..... (i)
where, (a1 + 2a2x2 + 3a3x4 + ..... + nanx2n-2) > 0
for all x ∈ R
Thus,
ie P′(x) changes sign from (-ve) to (+ ve) at x = 0.
∴ P(x) attains minimun at x = 0.
Hence, it has only one minimun at x = 0.
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