The global maximum and minimum values of f(x) = 3x^4 - 8x^3 + 12x^2 - 48x + 25, x are M and m respectively then value of 2M + m is equal to.

JEE MainMathematicsEasyMonotonocity

The global maximum and minimum values of f(x) = 3x^4 - 8x^3 +
12x^2 - 48x + 25, x $\in$ are M and m respectively then value
of 2M + m is equal to.

Show answer and solution

f(x) = 3x^4 - 8x^3 + 12x^2 - 48x + 25

\therefore

f' (x) = 12x^3 - 24x^2 + 24x - 48 = 0

\Rightarrow

x^3 - 2x^2 + 2x - 4 = 0

\Rightarrow

(x^2 + 2) (x - 2) = 0

\Rightarrow

x = 2

\in

\therefore

f(0) = 25
f(2) = 48 - 64 + 48 - 96 + 25 = - 39
f(3) = 243 - 216 + 108 - 144 + 25 = 16

Source: Grass-G / M-12 Mathematics Paper 2 Nov.docx

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