GATE IT 2008 | Question: 31

Question

If $f(x)$ is defined as follows, what is the minimum value of $f(x)$ for $x \in (0, 2]$ ?

$$f(x) = \begin{cases}  \frac{25}{8x} &\text{ when } x \leq \frac{3}{2} \\ x+ \frac{1}{x} &\text { otherwise}\end{cases}$$

• A. $2$
• B. $2 \frac{1}{12}$
• C. $2\frac{1}{6}$
• D. $2\frac{1}{2}$

Comments

The given function is neither continuous nor differentiable at x=3/2. But still, the minima exists on 3/2. But I have a doubt. If the function is not continuous at 3.2, then how the minima at that point existing?

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Good Explanation 

https://gateoverflow.in/3341/gate-it-2008-question-31?show=362595#a362595

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https://gateoverflow.in/3341/gate-it-2008-question-31?show=461352#a461352
solution based on intution

Answer 1

$$f'(x) = \begin{cases}  \frac{-25x^{-2}}{8} \text{;  } x \leq \frac{3}{2} \text{,             decreasing in this interval}\\ 1- \frac{1}{x^2} \text {;  otherwise,     increasing in this interval}\end{cases}$$

$$f''(x) > 0, x \in \left(0, \frac{3}{2}\right]\text{;  }  \text{ $f(x)$ will be  minimum at $x = \frac{3}{2}$}\\ f''(x) < 0 \text {, otherwise; $\ \ \ $  $f(x)$ will  be maximum at $x = 2$}$$

$f(x)$ is decreasing in $(0, \frac{3}{2}]$ and in $\left(\frac{3}{2},  2\right]$ it is increasing.
So, minimum value of $f(x) $ must be at $x = \frac{3}{2}$
$f(\frac{3}{2}) = \frac{25}{12} = 2 \frac{1}{12}$

Comments

Best explanation

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The given function is neither continuous nor differentiable at x=3/2. But still, the minima exists on 3/2. But I have a doubt. If the function is not continuous at 3.2, then how the minima at that point existing?

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Can anyone explain this a little bit more

Answer 2

Answer is B. The solution is as follows-

Comments

If the function is not continuous at 3/2, then how the minima at that point existing?

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At x=2, f(x) will have maximum value for x+1/x

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excellent solution with next level intuitive solution.

Answer 3

Answer: B

at x = $\frac{3}{2},$

f(x) =$2\left(\frac{1}{12}\right).$

Comments

how f(3/2) = 2/12 ????
shouldn't be 25/12 ??

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$\frac{25}{12}$ = 2$\frac{1}{12}$

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can we say that function is decreasing in the interval x(,3/2] and then increasing after 3/2, so the possible minima is at 3/2???

Answer 4

$f(x)=\frac{25}{8x}$   when x<=$\frac{3}{2}$

        $=x+\frac{1}{x}$   otherwise

Now if we check continuity at x=3/2,

Left hand limit ,

$\lim_{x->\frac{3}{2}^{-}}f(x)$

=$\lim_{x->\frac{3}{2}^{-}}\frac{25}{8x}$

=$\frac{25}{8}*\frac{2}{3}$

=$\frac{25}{12}$

Right hand limit,

$\lim_{x->\frac{3}{2}^{+}}f(x)$

=$\lim_{x->\frac{3}{2}^{+}}x+\frac{1}{x}$

=$\frac{3}{2}+\frac{2}{3}$

=$\frac{13}{6}$

so, The function is not continuous at $x=\frac{3}{2}$ so it is not differentiable in the range $(0,2]$ .

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Now finding the minimum value using differentiation in the range is not correct as it is not differentiable in the range.

one important result want to point out

a function can have local maxima or minima at not critical point.For strictly monotonically increasing or decreasing function has the maximum or minimum value lies in the endpoints of the interval.

If we look carefully 

$f(x)=\frac{3.125}{x}$   when x<=$\frac{3}{2}$

        $=x+\frac{1}{x}$   otherwise

then f(x) is decreasing in the range $(0,\frac{3}{2}]$  and increasing in the range $(\frac{3}{2},2]$.

As when x is in the range  $(0,\frac{3}{2}]$ then ,as x>0 and increases, then function $\frac{3.185}{x}$ so , f(x) decreases till x=3/2 .

Now when x is in the range $(\frac{3}{2},2]$ then as x>3/2 and increases ,then function f(x)=$x+\frac{1}{x}$ is increases.

so the function has minimum point in x=3/2 as till x=3/2 the function decreases then increases and maximum point at x=2 in $(0,2]$ .

so , minimum value 

f(3/2)=$\frac{25}{12}$

the maximum value 

f(2)=$2+\frac{1}{2}=2.5$

So correct answer is (B).

Answer 5

I used this approach based on intution

Final answer-B

Comments

very intuitive buddy !  great job.....

Answer 6

YOU CAN USE GRAPH TO UNDERSTAND HOW IT WORKS !!