Question

If g(x)=max|y^2-xy| then the minimum value of g(x) for real x is....
Here y can go from [0,1]

Answer 1

@mayankdevnani @parthkohli @priyar..help

Answer 2

is here absolute value? if so then minimum value is 0

Answer 3

Lol...bt not in the options....nvm

Answer 4

N yes it's definitely an absolute value...

Answer 5

@imqwerty WUd u help

Answer 6

okay

1st we gotta find the stationary point by taking \(f(y)=y^2-xy\) and solving \(\large \frac{df}{dy}=0\)
we will get \(y=\large \frac{x}{2}\)
now this can be local minimum or local maximum so we gotta compare it with extremities \(f(0)\) and \(f(1)\)

\(\begin{align}
g(x) &= \max\left(f(0)\;,f(1),f\left(\frac{x}{2}\right)\right)\\
&= \max\left(0\;,1-x,\frac{-x^2}{4}\right)\\
&= \max\left(0\;,1-x\right)\\
&= \begin{cases}
0 & x>1 \\
1-x & x\le 1
\end{cases}
\end{align}\)

Answer 7

Correct then....

Answer 8

As PR the graph we can get its minimum value as 0 from ur method....

Answer 9

N actually qwerty that was the function in x how could u differentiate it vid respect to x ?

Answer 10

Ayush btao Naa

Answer 11

sry sry i didn't see ur cmnts here

Answer 12

So see it now....lol

Answer 13

yes \:)
Note that \(g(x)\) is \(\ge 0\) and \(x \le 1\). so the minimum value of \(g(x)\) is clearly 0.

Answer 14

what are the options then ?

Answer 15

1/4
1/2
3+√8
3-√8

Answer 16

continuing @imqwerty response
\[g(x) = \max(0, |1-x|, |\frac{-x^2}{4}|)\]
If we graph these we can find the point where g(x) is smallest

\[1-x = \frac{x^2}{4} \rightarrow x^2 +4x - 4 = 0\]
\[x = 2 \sqrt{2} - 2\]
\[\min(g(x)) = 1-x = 1 - (2\sqrt{2}-2) = 3 - 2 \sqrt{2}\]