If the minimum value of f(x) = x^2 + 2bx + 2c^2 is greater than the ma

Question

If the minimum value of \(f(x) = x^2 + 2bx + 2c^2\) is greater than the maximum value of \(g(x) = -x^2 - 2cx + b^2\), then which of the following must be true?

A. \(|c| > √2|b|\)

B. \(√2|c| > b\)

C. \(0 < c < √2b\)

D. \(c < b\)

E. \(c = b\)

freedom128 · Accepted Answer

f(x) = x^2 + 2bx + 2c^2 is minimum when f'(x) = 0 --> 2x+2b=0 --> x_min = -b. 
** Minimum value: f(x=-b)= -b^2 + 2c^2

g(x) = -x^2 -2cx + b^2 is maximum when g'(x) = 0 --> -2x-2c=0 --> x_max = -c.
** Maximum value: g(x=-c)= c^2 + b^2

f(x=-b) > g(x=-c) 
-b^2 + 2c^2 > c^2 + b^2
c^2 > 2b^2
|c|> √2|b|

FINAL ANSWER IS (A)

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ArunSharma12 · Answer

f(x)=x^2+2bx+2c^2 : minimum value is obtained at -b/2a = -2b/2 = -b
 f(-b) = b^2 -2b^2 + 2c^2 = 2c^2 -b^2

g(x)=−x^2−2cx+b^2 : maximum value is obtained at -b/2a = 2c/-2 = -c
 g(-c) = -c^2 + 2c^2 + b^2 = c^2 + b^2

2c^2 -b^2 > c^2 + b^2 
 c^2 > 2b^2 
 |c| > √2|b| 
Ans: A

onoviosu · Answer

To find the min/max of an algebraic equation, use the formula

C - (b^2/4a)

Applying this to the first equation you get 
2C^2 - b^2

On the second you get
b^2 + c^2

Solving the inequality we get A.

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debjit1990 · Answer

Ans:B
f(x)=x2+2bx+2c2
f(0)=2c2(Taking 0 to make the value as low as possible)
g(x)=−x2−2cx+b2
g(0)=b2(Taking 0 to make the value as high as possible, as it's making all the negative values as 0)

2c2>b2
c>(b/root2)..√2|c|>b

HoneyLemon · Answer

Tough one for me . 
Is the answer A ?

Below is my reasoning . 
The minimum value of f(x)=x^2+2bx+2c^2 is 
greater than 
 the maximum value of g(x)=−x^2−2cx+b^2,

By looking at this quadratic equation I could think of equation of parabola . Now the corresponding co-ordinates of vertex will provide us the info about the maximum and minimum value. 
 To find the co-ordinates of vertex let’s write the parabola in its vertex form.

f(x)=x^2+2bx+2c^2 
= > 
 y =x^2+2bx+2c^2 
 = x^2+2bx+ (b^2 – b^2 )+ 2c^2 (Added and subtracted b^2 )
 = (X+b)^2 +( 2c^2 – b^2)
= > 
 Y – ( 2c^2 – b^2) = (X+b)^2 
 Now the parabola is in Y – k = (X -h)^2 form with vertex at (h , k)
 
So vertex of this parabola is {-b , ( 2c^2 – b^2) } at which the value is minimum.

To find minimum value of x^2+2bx+2c^2 , lets put x-coordinate in the equation .

Min (F(X)) = (-b)^2+2b(-b)+2c^2 
 = 2c^2 – b^2

Similarly we can find the vertex of g(x)=−x^2−2cx+b^2
= > 
Y =−x^2−2cx+b^2 
 = −(x^2+2cx +c^2 -c^2) +b^2
 = - (X + C)^2 + (b^2+c^2)

SO vertex of g(x)=−x^2−2cx+b^2 is (- c , (b^2+c^2) ) at which the value of the curve will be maximum .

To find Max value of −x^2−2cx+b^2, lets put x-coordinate in the equation . 
Max (G(X)) = − (-c)^2−2c(-c)+b^2 = b^2 +c^2

Now 
 2c^2 – b^2 > b^2 +c^2 
= > c^2 > 2 b^2 
= > |c| > Sqrt(2) |b|
 Hence A , IMO.

ScottTargetTestPrep · Answer

We see that the two functions are parabolas of the form f(x) = ax^2 + bx + c. We need to find the vertex of each parabola, and we first determine the x-coordinate of the vertex by using the formula x = -b/2a. We then determine the y-value (minimum or maximum value of the parabola) by substituting the value of the x-coordinate into the original quadratic equation.

For the first parabola, the x-coordinate of the vertex is -(2b)/ = -b, and so the minimum value of the function f(x) is f(-b) = (-b)^2 + 2b(-b) + 2c^2 = -b^2 + 2c^2.

Likewise, the x-coordinate of the vertex of the second parabola is -(-2c)/  = -c, so the maximum value of the function g(x) is g(-c) = -(-c)^2 - 2c(-c) + b^2 = c^2 + b^2.

Since we are given that the former is greater than the latter, we have:

-b^2 + 2c^2 > c^2 + b^2

c^2 > 2b^2

Taking the square root of both sides, we have:

√(c^2) > √(2b^2)

√(c^2) > √2 * √(b^2)

|c| > √2 * |b|

Answer: A

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If the minimum value of f(x) = x^2 + 2bx + 2c^2 is greater than the ma

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