SATHEE: Application of Derivatives 4 Question 26

Application of Derivatives 4 Question 26

####27. If $P (x) = a_{0} + a_{1} x^{2} + a_{2} x^{4} + \dots + a_{n} x^{2 n}$ is a polynomial in a real variable $x$ with $0 < a_{0} < a_{1} < a_{2} < \dots < a_{n}$ . Then, the function $P (x)$ has

$(1986, 2 M)$

(a) neither a maximum nor a minimum

(b) only one maximum

(d) only one maximum and only one minimum

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Answer:

Correct Answer: 27. $(a, c)$

Solution:

Given, $P (x) = a_{0} + a_{1} x^{2} + a_{2} x^{4} + \dots + a_{n} x^{2 n}$ where, $a_{n} > a_{n - 1} > a_{n - 2} > \dots > a_{2} > a_{1} > a_{0} > 0$

$\Rightarrow P^{'} (x) = 2 a_{1} x + 4 a_{2} x^{3} + \dots + 2 n a_{n} x^{2 n - 1}$

$= 2 x (a_{1} + 2 a_{2} x^{2} + \dots + n a_{n} x^{2 n - 2})$

where, $(a_{1} + 2 a_{2} x^{2} + 3 a_{3} x^{4} + \dots + n a_{n} x^{2 n - 2}) > 0, \forall x \in R$ .

Thus,

$P^{'} (x) > 0, when x > 0$

$P^{'} (x) < 0, when x < 0$

i.e. $P^{'} (x)$ changes sign from $(- ve)$ to (+ve) at $x = 0$ .

$∴ P (x)$ attains minimum at $x = 0$ .

Hence, it has only one minimum at $x = 0$ .

Application of Derivatives 4 Question 26

Answer:

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Application of Derivatives 4 Question 26

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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