If p x - a 0 - a 1 x- a n x n and - p x - - e x-1 -1 - for all x- 0- then - a 1 -2 a 2 -3 a 3 -n a n -

The correct option is B ≤ 1 We have, #160;p'( 1 ) =lim _ h→ 0 p( 1+h ) p( 1 ) #160;/ h #160;⇒| p'( 1 ) #160;| =| lim _ h→ 0 p( 1+h ) p( ...

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Algebra of Limits

If p x = a ...

Question

If $p (x) = a_{0} + a_{1} x + . . . + a_{n} x^{n}$ and $| p (x) | \leq ∣ ∣ e^{x - 1} - 1 ∣ ∣$ for all $x \geq 0,$ then $| a_{1} + 2 a_{2} + 3 a_{3} + . . . + n a_{n} |$

\leq 1

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\geq 1

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\geq 0

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\leq 0

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Similar questions

p (x) = a_{0} + a_{1} x + \dots + a_{n} x_{n}

| p (x) | \leq | e^{x - 1} - 1 |

x \geq 0

| a_{1} + 2 a_{2} + \dots + n a_{n} |

{(1 + x + x^{2} + \cdot \cdot + x^{n})}^{n}

= a_{0} + a_{1} x + a_{2} x^{2} + \cdot \cdot \cdot + a_{m p} x^{m p}

a_{1} + 2 a_{2} + 3 a_{3} + \cdot \cdot \cdot + n p a_{n p}

a_{n} x^{n} + a_{n - 1} x^{n - 1} + \dots + a_{1} x = 0

x = α

n a_{n} x^{n - 1} + (n - 1) a_{n - 1} x^{n - 2} + \dots + a_{1} = 0

(x + 1) (x + 2) . . . (x + n) = A_{0} + A_{1} x + . . . . A_{n} x^{n}

A_{1} + 2 A_{2} . . . n A_{n}

n > 2

p (x) = x^{n} + a_{n - 1} x^{n - 1} + . . . . . + a_{1} x + a_{0}

a_{0}, a_{1}, . . . . . a_{n - 1}

n p (x) = (1 + x) p^{'} (x)

b = p (1)

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