MCQ
The value of the function $(x - 1){(x - 2)^2}$ at its maxima is
  • A
    $1$
  • B
    $2$
  • C
    $0$
  • ${4 \over {27}}$

Answer

Correct option: D.
${4 \over {27}}$
d
(d) Given $f(x) = (x - 1){(x - 2)^2}$

$f(x) = (x - 1)({x^2} + 4 - 4x)$; $f(x) = ({x^3} - 5{x^2} + 8x - 4)$

Now $f'(x) = 3{x^2} - 10x + 8$, $f'(x) = 0$

==> $3{x^2} - 10x + 8 = 0$ ==> $(3x - 4)(x - 2) = 0$==> $x = \frac{4}{3}$, 2

Now $f''(x) = 6x - 10$

$f''(4/3) = 6 \times 4/3 - 10 < 0$

$f''(2) = 12 - 10 > 0$

Hence at $x = \frac{4}{3}$ the function will occupy maximum value.

$\therefore $ Maximum value $=f(4/3) = 4/27$.

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