Suppose p x - a 0- a 1 x - a n x n- If - p x - - e x -1-1- for all x - 0- then the maximum value of - a 1-2 a 2- na n - is equal to

We know that, p(x) = a_0 + a_1x +⋯ + a_nx_n ⇒ p'(x) = a_1 +2a_2 x +⋯ na_n x^n 1 ⇒ p'(1) = nbsp;a_1 + 2a_2 + ⋯ + na_n As, |p(x)| ≤ |e^x 1−1| ⇒ |p(1)| ≤ 0 ⇒ p(1 ...

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Byju's Answer

Standard XII

Mathematics

First Principle of Differentiation

Suppose p x =...

Question

Suppose $p (x) = a_{0} + a_{1} x + \dots + a_{n} x_{n}$ . If $| p (x) | \leq | e^{x - 1} - 1 |$ for all $x \geq 0$ , then the maximum value of $| a_{1} + 2 a_{2} + \dots + n a_{n} |$ is equal to

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Solution

We know that,
$p (x) = a_{0} + a_{1} x + \dots + a_{n} x_{n} \Rightarrow p^{'} (x) = a_{1} + 2 a_{2} x + \dots n a_{n} x^{n - 1} \Rightarrow p^{'} (1) = a_{1} + 2 a_{2} + \dots + n a_{n}$
As,
$| p (x) | \leq | e^{x - 1} - 1 | \Rightarrow | p (1) | \leq 0 \Rightarrow p (1) = 0 | p (h + 1) | \leq | e^{h} - 1 | \forall h \geq - 1 \Rightarrow | p (h + 1) - p (1) | \leq | e^{h} - 1 | \Rightarrow ∣ ∣ ∣ \frac{p (h + 1) - p (1)}{h} ∣ ∣ ∣ \leq ∣ ∣ ∣ \frac{e^{h} - 1}{h} ∣ ∣ ∣$
Taking limit $h \to 0$
$lim h \to 0 ∣ ∣ ∣ \frac{p (h + 1) - p (1)}{h} ∣ ∣ ∣ \leq lim h \to 0 ∣ ∣ ∣ \frac{e^{h} - 1}{h} ∣ ∣ ∣ \Rightarrow | p^{'} (1) | \leq 1$

Hence
$| a_{1} + 2 a_{2} + \dots + n a_{n} | \leq 1$

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Similar questions

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Suppose p(x)=a0+a1x+⋯+anxn. If |p(x)|≤|ex−1−1| for all x≥0, then the maximum value of |a1+2a2+⋯+nan| is equal to

Suppose $p (x) = a_{0} + a_{1} x + \dots + a_{n} x_{n}$ . If $| p (x) | \leq | e^{x - 1} - 1 |$ for all $x \geq 0$ , then the maximum value of $| a_{1} + 2 a_{2} + \dots + n a_{n} |$ is equal to