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UPSC NDA Math Model Paper 4

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Question : 39 of 120

Marks: +1, -0

The local maximum value of the function

f(x)=3 x^{4}+4 x^{3}-12 x^{2}+12

x=:

1
2
-2
0

Solution:

For the given function

f(x)=3 x^{4}+4 x^{3}-12 x^{2}+12

, first let's find the points of local maxima or minima:

f'(x)=12 x^{3}+12 x^{2}-24 x=0

\Rightarrow 12 x(x^{2}+x-2)=0

\Rightarrow x(x+2)(x-1)=0

\Rightarrow x=0

OR

x=-2

OR

x=1 .

f''(x)=36 x^{2}+24 x-24

f''(0)=36(0)^{2}+24(0)-24=-24

f''(-2)=36(-2)^{2}+24(-2)-24

=144-48-24=72

f''(1)=36(1)^{2}+24(1)-24

=36+24-24=36

Since, at x = 0 the value

f''(0) = -24 < 0

, the local maximum value of the function occurs at x = 0

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