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Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives
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Find the values of \(x\) for which \(y = [x(x – 2)]^2\) is an increasing function.

$\begin{array}{1 1} (A)\;\text{f decreasing for 0< x< 1 and x> 2} \\(B)\;\text{f increasing for 0< x< 1 and x> 2} \\ (C)\;\text{f is neither increasing nor decreasing for 0 < x < 1 and x> 2} \\ (D)\;\text{Cannot be determined.} \end{array} $

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Toolbox:
  • A function $f(x)$ is said to be a strictly increasing function on $(a,b)$ if $x_1 < x_2\Rightarrow f(x_1) < f(x_2)$ for all $x_1,x_2\in (a,b)$
  • If $x_1 < x_2\Rightarrow f(x_1) > f(x_2)$ for all $x_1,x_2\in (a,b)$ then $f(x)$ is said to be strictly decreasing on $(a,b)$
  • A function $f(x)$ is said to be increasing on $[a,b]$ if it is increasing (decreasing) on $(a,b)$ and it is increasing (decreasing) at $x=a$ and $x=b$.
  • The necessary sufficient condition for a differentiable function defined on $(a,b)$ to be strictly increasing on $(a,b)$ is that $f'(x) > 0$ for all $x\in (a,b)$
  • The necessary sufficient condition for a differentiable function defined on $(a,b)$ to be strictly decreasing on $(a,b)$ is that $f'(x) < 0$ for all $x\in (a,b)$
Step 1:
Let $f(x)=[x(x-2)]^2$
On expanding we get,
$f(x)=x^2(x^2-4x+4)$
$\qquad=x^4-4x^3+4x^2$
Differentiating w.r.t $x$
$f'(x)=4x^3-12x^2+8x$
$\qquad=4x(x^2-3x+2)$
Step 2:
For the function to be increasing,
$f'(x) > 0$
$\Rightarrow 4x(x^2-3x+2) > 0$
On factorizing we get,
$4x(x-1)(x-2)> 0$
Step 3:
Consider $0 < x < 1$
$f'(x) =(+)(-)(-)$=negative
For $x > 2$
$f'(x) =(+)(+)(+)$=positive
Hence the function is increasing for $0 < x < 1$ and $x > 2$
answered Jul 9, 2013 by sreemathi.v
 

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