Q89 / 150 · MHT CET 2007 · Mathematics

On the interval $[0,1]$ the function $x^{25}(1-x)^{75}$ takes its maximum value at the point

  • A
    0
  • Correct answer
    B
    $1 / 4$
  • C
    $1 / 2$
  • D
    $1 / 3$

Solution

Given, $f(x)=x^{25}(1-x)^{75}$ $$ \begin{aligned} \Rightarrow f^{\prime}(x) &=25 x^{24}(1-x)^{75}-75 x^{25}(1-x)^{74} \\ &=25 x^{24}(1-x)^{74}(1-4 x) \\ \therefore \quad f^{\prime}(x) &=0 \\ \Rightarrow \quad x &=0,1,1 / 4 \end{aligned} $$ If $x < 1 / 4$, then $$ f^{\prime}(x)=25 x^{24}(1-x)^{74}(1-4 x)>0 . $$ and if $x>1 / 4$, then $$ f^{\prime}(x)=25 x^{24}(1-x)^{74}(1-4 x) < 0 . $$ Thus, $f^{\prime}(x)$ changes its sign from positive to negative as $x$ passes through $1 / 4$ from left to right. Hence, $f(x)$ attains its maximum at $x=1 / 4$.

Hence, the correct answer is option: 'B'.

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