What is the NO-SHORTCUT approach for learning great Mathematics?

Learn MoreThis is a Test of Mathematics Solution Subjective 65 (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.

Show that for all real x, the expression $ {ax^2} $ + bx + C ( where a, b, c are real constants with a > 0), has the minimum value $ {\frac{4ac - b^2}{4a}} $ . Also find the value of x for which this minimum value is attained.

f (x) $ {ax^2} $ + bx + c

Now minimum derivative = 0 & 2nd order derivative > 0.

$ {\frac{df(x)}{dx}} $ = 2ax + b

Or $ {\frac{d^2f(x)}{dx^2}} $ =2a

Now 2a> so 2nd order derivative > 0 so $ {\frac{d^2f(x)}{dx^2}} $ = 2.

So minimum occurs when

${\frac{df(x)}{d(x)}} $ = 0 or 2ax + b = 0

or 2ax = -b

or x = $ {\frac{-b}{2a}} $ (ans)

At x = $ {\frac{-b}{2a}} $

$ {ax^2} $ + bx + c

= $ {a\times {\frac{b^4}{4a^2}}} $ + $ {b\times {\frac{-b}{2a}}} $ + c

= $ {\frac{4ac-b^2}{4a}} $ (proved)

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