Let $f(x)=2e^x + (a^2-5a+6)e^{-x} + (10a-2a^2-11)x - 3$ is increasing on values of $x$, then find number of integral values of $a$

Asked Sep 08 '21 at 18:50

Active Sep 08 '21 at 18:50

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The given solutions involves a kinda hard to notice manipulation shown as follows

$$f'(x)=2e^x-(a^2-5a+6)e^{-x} + (10a-2a^2 -11)$$ $$=2e^{-x}[e^x + (5a-a^2-6)][e^x + \frac 12]$$ $$\implies 2\le a \le 3$$

Now A different method i came across was to set $a^2-5a+6<0$ and $-2a^2+10a-11>0$ ie. turning all terms in the expression positive. This gives $1.633 < a < 3.366$ ie. $2 \le a \le 3$

This gives the same answer but i am worried about its validity, ie will it give extraneous roots or omit some important roots in other case, and was this just a happy accident ?

asked Sep 08 '21 at 18:50

Aditya

6,191

1

In the second approach, the function is increasing ... but it is not shown that it is NOT increasing for a different choice of $a.$ – user376343 Sep 08 '21 at 19:25
@user376343 so what does that imply? – Aditya Sep 09 '21 at 02:14
It implies this: It could occur that certain solution(s) are not found by the second method. – user376343 Sep 09 '21 at 18:56
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@user376343 so just got lucky in this car – Aditya Sep 10 '21 at 08:11

Let $f(x)=2e^x + (a^2-5a+6)e^{-x} + (10a-2a^2-11)x - 3$ is increasing on values of $x$, then find number of integral values of $a$

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