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Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives
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The smallest value of the polynomial $x^3-18x^2+96x$ in [0,9] is

\[(A)\;126\quad(B)\;0\quad(C)\;135\quad(D)\;160.\]

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1 Answer

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Toolbox:
  • To obtain the absolute maxima or minima for the function $f(x)$
  • (i) Find $f'(x)$ and put $f'(x)=0$
  • (ii) Obtain the points from $f'(x)=0$
  • (iii) By phthogoras theorem, find $f''(x)$ and check the value of $f''(x)$ for each of the points obtained if $f''(x)>0$ it has a minimum at that point. If $f''(x)<0$ then it has a maximum at that point.
Step 1
$ y=x^3-18x^2+96x$
On differentiating w.r.t $x$ we get,
$ \large\frac{dy}{dx}=3x^2-36x+96$
when $ \large\frac{dy}{dx}=0$
$ 3x^2-36x+96=0$
$ \Rightarrow 3(x^2-12x+32)=0$
$ \Rightarrow 3(x-8)(x-4)=0$
$ \Rightarrow x=8, 4$
Let us consider the points when $x=0, 4, 8, 9$
Step 2
when $x=0$
$y=0$
Step 3
when $x=4$
$y=4^3-18(4)^2+96(4)$
$ = 260$
Step 4
when $x=8$
$y=8^3-18(8)^2+96(8)$
$ = 120$
Step 5
when $ x = 9$
$y=9^3-18(9)^2+96(9)$
$ = 135$
Hence the smallest value of the polynomial is 0
Hence B is the correct option.
answered Aug 11, 2013 by thanvigandhi_1
 
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