Q.

Let f(x)=x2+b1x+c1,g(x)=x2+b2x+c2. Let the real roots of f(x)=0 be α,β and real roots of g(x)=0 be α+h,β+h. The least value of f(x) is -1/4. The least value of g(x) occurs at x = 7/2.The least value of g(x) isThe value of b2 isThe roots of f(x) = 0 are

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a

-14

b

-1

c

-13

d

-12

e

-5

f

9

g

-8

h

-7

i

3, -4

j

-3, 4

k

3, 4

l

-3, -4

answer is , , .

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Detailed Solution

(β−α)=((β+h)−(α+h))(β+α)2−4αβ=[(β+h)+(α+h)]2−4(β+h)(α+h)−b12−4c1=−b22−4c2D1=D2The least value of f(x) is−D14=−14⇒D1=1 and D2=1Therefore, the least value ofg(x) is −D24=−14The least value of g(x) occurs at−b22=72 or b2=−7Now, b22−4c2=D2or 49−4c2=1 or 484=c2 or c2=12⇒ x2−7x+12=0 or x=3,4(β−α)=((β+h)−(α+h))(β+α)2−4αβ=[(β+h)+(α+h)]2−4(β+h)(α+h)−b12−4c1=−b22−4c2D1=D2The least value of f(x) is−D14=−14⇒D1=1 and D2=1Therefore, the least value ofg(x) is −D24=−14The least value of g(r) occurs at−b22=72 or b2=−7Now, b22−4c2=D2or 49−4c2=1 or 484=c2 or c2=12⇒ x2−7x+12=0 or x=3,4(β−α)=((β+h)−(α+h))(β+α)2−4αβ=[(β+h)+(α+h)]2−4(β+h)(α+h)−b12−4c1=−b22−4c2D1=D2The least value of f(x) is−D14=−14⇒D1=1 and D2=1Therefore, the least value ofg(x) is −D24=−14The least value of g(r) occurs at−b22=72 or b2=−7Now, b22−4c2=D2or 49−4c2=1 or 484=c2 or c2=12⇒ x2−7x+12=0 or x=3,4
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