TIFR CSE 2010 | Part A | Question: 3

Question

The function $f (x) = 2.5 \log_e  \left( 2 + \exp \left( x^2  - 4x + 5 \right)\right)$ attains a minimum at $x = $?

• A. $0$
• B. $1$
• C. $2$
• D. $3$
• E. $4$

Answer 1

$f(x)$ will be minimum at $x = 2,$
Here is another approach.
Since log and exponent are monotonically increasing functions, the problem of minimizing $f(x)$ can be reduced to just minimizing the quadratic expression $x^{2} - 4x + 5,$
This quadratic expression can be written as $(x^{2} - 4x + 4) + 1$ which is equal to $(x - 2)^{2} + 1.$
Now since $(x - 2)^{2}$ can not be less than $0$, so $(x - 2)^{2} + 1$ can not be less than $1.$
Also $(x - 2)^{2} + 1$ will be at its minimum value $(= 1)$, when $x = 2.$

So, value of $f(x)$ will be minimum at $x = 2.$

Comments

Nice one :)

Answer 2

Just differentiate whole function and put value =0, x comes out to be 2.
1nd derivative +ve here, so minima

Comments

"1nd derivative +ve here, so minima"  how?

I know when 2nd derivative is +ve then that point is minima.

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Did you get this resolved?

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Can we decide if the roots are maxima or minima on the basis of the first derivative?

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I guess by “1nd” he meant “2nd”.

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after finding critical pt.s you can try first derivative test easily.

here f’’(x) is difficult to calc. then we can see f’(x) is +ve in x<2 and -ve after 2.

so function is decreasing till 2 then increasing. so x = 2 is pt. of minima.

Answer 3

In this question it is not necessary to either use differentiation or anything much different.

Just see that how this expression simplifies =>

After cancelling out log & Exponent we get something like =>

2.5ln2 + 2.5 ( x²-4x+5 )

Now 2.5ln2 will be there same in all x values.

Just put in all values one by one, it is not hard to see that , At x = 2 we get minimum value !

Answer c

Comments

I don't understood how you cancelled log.

After cancelling out log & Exponent we get something like =>

2.5ln2 + 2.5 ( x2-4x+5 )

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^
This solution is wrong, He assumed $log(a+b) = log(a)+log(b)$ which is clerly wrong.