Interval or discrete values ??!!

Calculus Level 5

Let $x_1,x_2,x_2,.....,x_n$ be $n$ real values of $p$ for which the least value of $4x^2-4px+p^2-2p+2$ on the interval $0 \leq x \leq 2$ is equal to 3. Find the value of

$\dfrac{x_1+x_2+x_3+....+x_n}{n}.$

Note: Answer to 3 decimal points.

The answer is 3.874.

1 solution

Pranjal Jain
Oct 30, 2014

First find out the x-coordinate of minima by differentiating given equation wrt x which gives $x=\frac{p}{2}$

Now 3 cases will be there....

$\frac{p}{2}<0$ In this case minima will be at x=0 giving p= $1-\sqrt{2}$
$\frac{p}{2}>2$ In this case minima will be at x=2 giving p= $5+\sqrt{10}$
$0\leq\frac{p}{2}\leq 2$ In this case minima will be at $x=\frac{p}{2}$ giving no value of p

So $\frac{\displaystyle\sum_{i=1}^n x_{n}}{n}=\frac{1-\sqrt{2}+5+\sqrt{10}}{2}=\boxed{3.8740}$

Same method

U Z - 6 years, 7 months ago

Explained very nicely.!

Sandeep Bhardwaj - 6 years, 7 months ago

Did the same way!

Anurag Pandey - 4 years, 10 months ago

Interval or discrete values ??!!

The answer is 3.874.

1 solution

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