A function given by f(x) =ax^2+2bx+c/Ax^2+Bx+Chas points of extrema at

1.	A function given by f(x) =ax^2+2bx+c/Ax^2+Bx+Chas points of extrema at x = 1 and x = - 1, such that f(1) = 2,f(-1) = 3 and f(0) = 2.5. Then
Answer» Given: $\displaystyle f(x) = \frac{ax^2+2bx + c}{Ax^2 + Bx + C}$ ...(1) f(1) = 2 f(-1) = 3 f(0) = 2.5 To FIND: a =? Step-by-step explanation: Since f(x) has extreme value x = 1 and x = - 1 & f(1) = 2 f(-1) = 3 i.e. 2 ≤ f(x) ≤ 3 For points x = 1 & x = - 1, denominator must be MINIMUM at both these points & f(x) remains positive and quadratic. ∴ $\displaystyle D = k\ [ (x -1)^2 + (x +1)^2]$ $\displaystyle D = 2k \ (x-1)^2$ But D = $Ax^2 + Bx +C$ $Ax^2 + Bx +C = 2k \ (x-1)^2$ By comparing, A=2k B=0 C=2k i.e. A = C = 2k B = 0 Given that f(0) = 2.5 Put his in eq(1) $2.5 = \displaystyle \frac{ax^2+2bx + c}{Ax^2 + Bx + C}$ $\displaystyle \frac{c}{A} = 2.5$ c = 2.5A ...(2) Given that f(1) = 2 Put his in eq(1) $\displaystyle 2 = \frac{a+2b+c}{A+A}$ a + 2b + c = 4A ...(3) Given that f(-1) = 3 Put his in eq(1) $\displaystyle 3 = \frac{a+2b+c}{A+A}$ a - 2b + c = 6A ...(4) adding eq (3) & (4) we get 2a + 2c = 10 2a + 5A = 10A ∵ from eq (2) 2a = 5A a = 2.5A Correct option is b.

A function given by f(x) =ax^2+2bx+c/Ax^2+Bx+Chas points of extrema at x = 1 and x = - 1, such that f(1) = 2,f(-1) = 3 and f(0) = 2.5. Then

Answer»

Given:

$\displaystyle f(x) = \frac{ax^2+2bx + c}{Ax^2 + Bx + C}$ ...(1)

f(1) = 2

f(-1) = 3

f(0) = 2.5

To FIND:

a =?

Step-by-step explanation:

Since f(x) has extreme value x = 1 and x = - 1 &

f(1) = 2

f(-1) = 3

i.e. 2 ≤ f(x) ≤ 3

For points x = 1 & x = - 1, denominator must be MINIMUM at both these points & f(x) remains positive and quadratic.

∴ $\displaystyle D = k\ [ (x -1)^2 + (x +1)^2]$

$\displaystyle D = 2k \ (x-1)^2$

But D = $Ax^2 + Bx +C$

$Ax^2 + Bx +C = 2k \ (x-1)^2$

By comparing,

A=2k B=0 C=2k

i.e. A = C = 2k B = 0

Given that f(0) = 2.5

Put his in eq(1)

$2.5 = \displaystyle \frac{ax^2+2bx + c}{Ax^2 + Bx + C}$

$\displaystyle \frac{c}{A} = 2.5$

c = 2.5A ...(2)

Given that f(1) = 2

Put his in eq(1)

$\displaystyle 2 = \frac{a+2b+c}{A+A}$

a + 2b + c = 4A ...(3)

Given that f(-1) = 3

Put his in eq(1)

$\displaystyle 3 = \frac{a+2b+c}{A+A}$

a - 2b + c = 6A ...(4)

adding eq (3) & (4) we get

2a + 2c = 10

2a + 5A = 10A ∵ from eq (2)

2a = 5A

a = 2.5A

Correct option is b.