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MAXIMA AND MINIMA

Sujoy Das
22/02/2018 0 0

QUESTION

Q) Let ‘A’ = (x^2 - 3x + 4) / (x^2 + 3x + 4)
Then the minimum value that ‘A’ can undertake for all real values of x, lies in which of these intervals?
A) (-inf, -2)
B) (-2, -1)
C) (-1, 0)
D) ( 0, 1)

 

ANSWER

(A - 1)x2 + 3(A + 1)x + 4(A - 1) = 0

Since x is real, hence D >= 0

So, 9(A + 1)2 – 4.4.(A - 1)2 >= 0
or 7A2 - 50A + 7 <= 0

or (7A - 1).(A - 7) <= 0

so 1/7 <= A <=7
Hence the minimum value that ‘A’ can adopt is 1/7, which lies in the interval (0,1)
Option – (D)

 

 

 

 

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