JEE Advance - Mathematics (2000 - No. 4)
Suppose $$p\left( x \right) = {a_0} + {a_1}x + {a_2}{x^2} + .......... + {a_n}{x^n}.$$ If
$$\left| {p\left( x \right)} \right| \le \left| {{e^{x - 1}} - 1} \right|$$ for all $$x \ge 0$$, prove that
$$\left| {{a_1} + 2{a_2} + ........ + n{a_n}} \right| \le 1$$.
$$\left| {p\left( x \right)} \right| \le \left| {{e^{x - 1}} - 1} \right|$$ for all $$x \ge 0$$, prove that
$$\left| {{a_1} + 2{a_2} + ........ + n{a_n}} \right| \le 1$$.
Consider the limit as x approaches 1 and apply L'Hopital's rule.
Consider the limit as x approaches 0 and apply L'Hopital's rule.
Consider the derivative of p(x) and e^(x-1) - 1 at x = 1.
Consider the integral of p(x) from 0 to 1 and compare it with the integral of e^(x-1) - 1 from 0 to 1.
Consider the second derivative of p(x) and e^(x-1) - 1 at x=0.
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