Question Bank [2022] — Maths (commerce) STD 12 Commerce / Arts — Question
Maharashtra BoardEnglish MediumSTD 12 Commerce / ArtsMaths (commerce)Question Bank [2022]3 Marks
Question
Find the values of x such that $f(x) = 2x^3 – 15x^2 + 36x + 1$ is increasing function
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Answer
$f(x)=2 x^3-15 x^2+36 x+1 \therefore f^{\prime}(x)=6 x^2-30 x+36$
$=6\left(x^2-5 x+6\right)$
$=6(x-3)(x-2)$
$f(x)$ is an increasing function, if $f^{\prime}(x)>0$
$\therefore 6(x-3)(x-2)>0$
$\therefore(x-3)(x-2)>0$
$a b>0 \Leftrightarrow a>0$ and $b>0$ or $a<0$ and $b<0$
$\therefore$ Either $(x-3)>0$ and $(x-2)>0$
or
$(x-3)<0 \text { and }(x-2)<0$
Case 1: $x-3>0$ and $x-2>0$
$\therefore x>3 \text { and } x>2$
$\therefore x>3$
Case 2: $x -3<0$ and $x -2<0$
$\therefore x<3 \text { and } x<2$
$\therefore x<2$
Thus, $f(x)$ is an increasing function for $x<2$ or $x>3$, i.e., $(-\infty, 2) \cup(3, \infty)$
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