Korean College Entrance Exam Question

$m$ is a natural number.

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Let $n=2m$ .

And let's divide $\sum_{k=1}^{n} \left|f(x^k)\right|$ into two parts.

Like $\sum_{k=1}^{m} \left|f(x^{2k-1})\right|$ and $\sum_{k=1}^{m} \left|f(x^{2k})\right|$ .

Then let's take a closer look at $\sum_{k=1}^{m} \left|f(x^{2k})\right|$ . What's the minimum of $f(x^{2k})$ ?

According to the question, $f(x^{2k})=e^{x^{2k}+1}-1$ . And the smallest $x^{2k}+1$ can get is when $x=0$ , since $x^{2k}=(x^2)^k$ and $x^2$ is always greater than or equal to $0$ .

So naturally the minimum of $f(x^{2k})$ is $e-1$ , which is positive.

Then we can conclude that no matter the value of $x$ , $f(x^{2k})>0$ , and the function $\left|f(x^{2k})\right|$ is always differentiable, so we can eliminate that from our calculation.

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Now let $n=2m-1$ .

And let's divide $\sum_{k=1}^{n} \left|f(x^k)\right|$ into two parts.

Like $\sum_{k=1}^{m} \left|f(x^{2k-1})\right|$ and $\sum_{k=1}^{m} \left|f(x^{2k-2})\right|$ .

Then let's take a closer look at $\sum_{k=1}^{m-1} \left|f(x^{2k})\right|$ . What's the minimum of $f(x^{2k})$ ?

Wait, this is the same thing as before. I'll gladly skip doing the whole thing again.

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So whether $n=2m-1$ or $n=2m$ , you only get $\sum_{k=1}^{m} \left|f(x^{2k-1})\right|$ as influenceable over the differentiability of $g(x)$ .

Then let's define $p(x)=100\left|f(x)\right|-\sum_{k=1}^{m} \left|f(x^{2m-1})\right|$ .

Hmm. In order for g(x) to be differentiable, $\lim_{h\to+0} {\frac{p(x+h)-p(x)}{h}}=\lim_{h\to+0} {\frac{p(x-h)-p(x)}{-h}}$ has to be satisfied.

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According to the question, $f(x^{2k-1})=e^{x^{2k-1}+1}-1$ .

If $x\geq-1$ , you can see that $x^{2k-1}\geq-1$ , which means $x^{2k-1}+1\geq0$ . Therefore $e^{x^{2k-1}+1}\geq1$ and finally $e^{x^{2k-1}+1}-1\geq0$ . $f(x^{2k-1})\geq0$ .

Use a similar method to $x<-1$ , and you get $f(x^{2k-1})<0$ .

Substitute $k=1$ .

If $x\geq-1$ , $f(x)\geq0$ , and if $x<-1$ , $f(x)<0$ .

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So we can say that,

If $x\geq-1$ , $p(x)=100f(x)-\sum_{k=1}^{m} f(x^{2k-1})$ .

And if $x<-1$ , $p(x)=-100f(x)+\sum_{k=1}^{m} f(x^{2k-1})$

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$\lim_{h\to+0} \frac{p(-1+h)-p(-1)}{h}=\lim_{h\to+0} \left(100\times\frac{f(-1+h)-f(-1)}{h}-\frac{\sum_{k=1}^{m} \left(f\left((-1+h)^{2k-1}\right)-f\left((-1)^{2k-1}\right)\right)}{h}\right)$

$=\lim_{h\to+0} \left(100\times\frac{e^h-1}{h}-\frac{\sum_{k=1}^{m} \left(e^{(h-1)^{2k-1}+1}-1\right)}{h}\right)=\lim_{h\to+0} \left(100-\frac{\sum_{k=1}^{m} \left((h-1)^{2k-1}+1\right)}{h}\right)$

$=\lim_{h\to+0} \left(100-\frac{\sum_{k=1}^{m} \left((2k-1)h+h^2\times(\cdot\cdot\cdot)\right)}{h}\right)=100-\sum_{k=1}^{m}(2k-1)$

$=100-m^2$

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Apply the same method to $\lim_{h\to+0} \frac{p(-1-h)-p(-1)}{-h}$ and you'll get $\lim_{h\to+0} \frac{p(-1-h)-p(-1)}{-h}=-100+m^2$ .

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Nice, so we need to make the natural number $m$ satisfy $\lim_{h\to+0} {\frac{p(x+h)-p(x)}{h}}=\lim_{h\to+0} {\frac{p(x-h)-p(x)}{-h}}$ .

So the equation would be $100-m^2=-100+m^2$ . Simplify that and we'll get $m^2=100$ .

$\therefore m=10$ .

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$\therefore n=2m-1=19 \quad or \quad n=2m=20$ .

So the value we're trying to get is $19+20=39$ .

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Answer: 39

Can u post all the questions from that entrance paper Only of maths section