Published by:
CGP EDU Academic Team
Published on: August 13, 2026
Consider a function f(x) = 
(4 – 3x 2 ) where ' α ' is a positive parameter.
(i) No. of points of extrema of f(x) for a given value of α is –
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Ans.
(i)
Sol. f(x) = 
(4 – 3x 2 )
f '(x) = – (4 – 3x 2 ) + 
(– 6x)
= ax 2 – 6 
x – 4
D = 36 
+ 144 > 0
two critical points given by
x = 
= 
= 
, 
x = 
is point of local max. and x = 
is local min.
(ii)
Sol. f 
= 
= 
= 4 
= 
( α 2 – 3) (3 – α 2 )
= 
f 
= 
= 4 
= 
∴ f 
– f 
= 
+ 
= 
= 
[9 α 4 – 6 α 2 + 1 + α 6 – 6 α 4 + 9 α 2 ]
= 
( α 6 + 3 α 4 + 3 α 2 + 1) = 
= 
(iii)
Sol. Least value = 32/9
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