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Byju's Answer

Standard VI

Mathematics

Degree of a Polynomial

Let fx=x3+2...

Question

Let $f (x) = (x^{3} + 2)^{30}$ . If $f^{n} (x)$ is a polynomial of degree $20$ , where $f^{n} (x)$ denotes the nth order derivative of $f (x)$ with respect to x, then the value of n is?

60

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40

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70

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50

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Solution

The correct option is C $70$
We have, $f (x) = (x^{3} + 2)^{30}$
Clearly, it is a polynomial of degree $90$ . It is given that $f^{n} (x)$ is a polynomial of degree $20$ .
$∴ 90 - n = 20$
$\Rightarrow n = 70$ .

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Similar questions

Q. Let

f (x) = (x^{3} + 2)^{30}

. If

f^{n} (x)

is a polynomial of degree

20

, where

f^{n} (x)

denotes the

n t h

derivative of

f (x)

w.r.t.

x

, then the value of

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is-

Q. lf

f (x) = {tan}^{- 1} x

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(1 + x^{2}) f^{(n + 2)} (x) + 2 x (n + 1) f^{(n + 1)} (x) + n (n + 1) f^{(n)} (x) =

(n \geq 1, f^{(n)} (x)

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, then show that

f^{'} (x) . f (n x) = f (x) . f^{'} (n x)

f (x) = (x^{2} + 8 x + 20) e^{x}

. Find

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and

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Let f(x)=(x3+2)30. If fn(x) is a polynomial of degree 20, where fn(x) denotes the nth order derivative of f(x) with respect to x, then the value of n is?

The correct option is C 70We have, f(x)=(x3+2)30Clearly, it is a polynomial of degree 90. It is given that fn(x) is a polynomial of degree 20.∴90−n=20⇒n=70.

Let $f (x) = (x^{3} + 2)^{30}$ . If $f^{n} (x)$ is a polynomial of degree $20$ , where $f^{n} (x)$ denotes the nth order derivative of $f (x)$ with respect to x, then the value of n is?

The correct option is C $70$
We have, $f (x) = (x^{3} + 2)^{30}$
Clearly, it is a polynomial of degree $90$ . It is given that $f^{n} (x)$ is a polynomial of degree $20$ .
$∴ 90 - n = 20$
$\Rightarrow n = 70$ .