Let $f (x) = (x - 2) (x^{4} - 4 x^{3} + 6 x^{2} - 4 x + 1)$ then value of local minimum of f is

- 2 / 3

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- {(4 / 5)}^{4}

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- 4^{4} / 5^{5}

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- {(4 / 5)}^{5}

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Solution

The correct option is C $- 4^{4} / 5^{5}$
$f (x) = (x - 2) (x^{4} - 4 x^{3} + 6 x^{2} - 4 x + 1) = (x - 2) {(x - 1)}^{4}$
for local maximum $f^{'} (x) = 4 (x - 2) {(x - 1)}^{3} + {(x - 1)}^{4} = {(x - 1)}^{3} (5 x - 9) = 0$
$\Rightarrow x = 1, \frac{9}{5}$
$f^{''} (x) = 5 {(x - 1)}^{3} + 3 (5 x - 9) {(x - 1)}^{2}$
at $x = \frac{9}{5}$ , $f^{''} (x) > 0$
Therefore, $f (x)$ has local minimum at $x = \frac{9}{5}$
Thus, $f (\frac{9}{5}) = - {(\frac{4}{5})}^{5}$
Ans: C

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Let f(x)=(x−2)(x4−4x3+6x2−4x+1) then value of local minimum of f is

The correct option is C −44/55f(x)=(x−2)(x4−4x3+6x2−4x+1)=(x−2)(x−1)4for local maximum f′(x)=4(x−2)(x−1)3+(x−1)4=(x−1)3(5x−9)=0⇒x=1,95f′′(x)=5(x−1)3+3(5x−9)(x−1)2at x=95, f′′(x)>0Therefore, f(x) has local minimum at x=95Thus, f(95)=−(45)5Ans: C

Let $f (x) = (x - 2) (x^{4} - 4 x^{3} + 6 x^{2} - 4 x + 1)$ then value of local minimum of f is