Question

The number of critical points of the function\[f\left( x \right) = \left| {x - 1} \right|\left| {x - 2} \right|\]is
A.1
B.2
C.3
D.None of these

Answer 1

Hint: First we define the function in interval \[x < 1\], \[1 \leqslant x < 2\]and \[x \geqslant 2\]after that differentiate the function with respect to x in given interval. If the value of x lies in the interval then value of x is our critical point if not then it will not be a critical point. After that we calculate the left and right hand derivative of a function at branch point if the derivative does not exist at branch point then the point is also known as critical point.

Complete step-by-step answer:
We have given \[f\left( x \right) = \left| {x - 1} \right|\left| {x - 2} \right|\]
Define \[f\left( x \right) = \left\{
  \left( {x - 1} \right)\left( {x - 2} \right);x < 1 \\
   - \left( {x - 1} \right)\left( {x - 2} \right);1 \leqslant x < 2 \\
  \left( {x - 1} \right)\left( {x - 2} \right);x \geqslant 2 \\
   \right.\]
\[ \Rightarrow \]\[f\left( x \right) = \left\{
  {x^2} - 3x + 2;x < 1 \\
   - \left( {{x^2} - 3x + 2} \right);1 \leqslant x < 2 \\
  {x^2} - 3x + 2;x \geqslant 2 \\
   \right.\]
For \[x < 1\]we have,
\[f\left( x \right) = {x^2} - 3x + 2\]
Differentiating with respect to x we get,
\[f'\left( x \right) = 2x - 3\]
As we know to find critical points we have to equate derivatives of \[f\left( x \right)\] with zero.
So, \[f'\left( x \right) = 0\]
\[ \Rightarrow 2x - 3 = 0\]
\[ \Rightarrow x = \dfrac{3}{2}\]
As \[x = \dfrac{3}{2} = 1.5\]does not lies in \[x < 1\]
so \[x = \dfrac{3}{2}\] is not a critical point of f(x) for \[x < 1\].
Which implies that when\[x < 1\]the function \[f\left( x \right)\]does not have any critical point.
For \[1 \leqslant x < 2\]we have,
\[f\left( x \right) = - ({x^2} - 3x + 2)\]
Differentiating with respect to x we get,
\[f'\left( x \right) = - \left( {2x - 3} \right)\]
\[ \Rightarrow f'\left( x \right) = - 2x + 3\]
As we know to find critical points we have to equate derivatives of \[f\left( x \right)\] with zero.
So,\[f'\left( x \right) = 0\]
\[ \Rightarrow - 2x + 3 = 0\]
\[ \Rightarrow 2x = 3\]
\[ \Rightarrow x = \dfrac{3}{2}\]
As \[x = \dfrac{3}{2}\]lies between\[1 \leqslant x < 2\].
So, \[x = \dfrac{3}{2}\]is a critical point of f(x) in the interval \[1 \leqslant x < 2\].
For \[x \geqslant 2\]we have,
\[f\left( x \right) = {x^2} - 3x + 2\]
Differentiating with respect to x we get,
\[f'\left( x \right) = 2x - 3\]
As we know to find critical points we have to equate derivatives of \[f\left( x \right)\] with zero.
So, \[f'\left( x \right) = 0\]
\[ \Rightarrow 2x - 3 = 0\]
\[ \Rightarrow x = \dfrac{3}{2}\]=1.5
As \[1.5 < 2\]it will not lie in the interval \[x \geqslant 2\]
So, \[x = \dfrac{3}{2}\] is not a critical point of f(x) for\[x \geqslant 2\].
Which implies that when \[x \geqslant 2\] the function \[f\left( x \right)\]does not have any critical point.
Now, we calculate the left and right hand derivative of a function at branch point if the derivative does not exist then the point is known as critical point.
\[f'\left( x \right) = \left\{
  2x - 3;x < 1 \\
   - \left( {2x - 3} \right);1 \leqslant x < 2 \\
  2x - 3;x \geqslant 2 \\
   \right.\]
For \[x = 1\]
Consider,
\[Lf'\left( x \right) = \]\[2x - 3\] \[ \ldots \left( 1 \right)\]
Now, put x=1 in equation (1)
 \[ \Rightarrow Lf'\left( 1 \right) = 2\left( 1 \right) - 3\]
\[ \Rightarrow Lf'\left( 1 \right) = - 1\]
\[Rf'\left( x \right) = - \left( {2x - 3} \right)\] \[ \ldots \left( 2 \right)\]
Now, put x=1 in equation (2)
\[ \Rightarrow Rf'\left( 1 \right) = - \left( {2\left( 1 \right) - 3} \right)\]
\[ \Rightarrow Rf'\left( 1 \right) = - \left( {2 - 3} \right)\]
\[ \Rightarrow Rf'\left( 1 \right) = - \left( { - 1} \right)\]
\[ \Rightarrow Rf'\left( 1 \right) = 1\]
\[ \Rightarrow Lf'\left( 1 \right) \ne Rf'\left( 1 \right)\]
\[ \Rightarrow \]\[f'\left( 1 \right)\] does not exist.
Similarly,
 Now we calculate for\[x = 2\]
Consider,
\[Lf'\left( x \right) = - \left( {2x - 3} \right)\] \[ \ldots \left( 3 \right)\]
Put \[x = 2\]in equation (3)
\[ \Rightarrow Lf'\left( 2 \right) = - \left( {2\left( 2 \right) - 3} \right)\]
\[ \Rightarrow Lf'\left( 2 \right) = - \left( {4 - 3} \right)\]
\[ \Rightarrow Lf'\left( 2 \right) = - 1\]
\[Rf'\left( x \right) = 2x - 3\] \[ \ldots \left( 4 \right)\]
Put \[x = 2\]in equation (4)
\[ \Rightarrow Rf'\left( 2 \right) = 2\left( 2 \right) - 3\]
\[ \Rightarrow Rf'\left( x \right) = 4 - 3\]
\[ \Rightarrow Rf'\left( x \right) = 1\]
\[ \Rightarrow Lf'\left( 2 \right) \ne Rf'\left( 2 \right)\]
\[ \Rightarrow \]\[f'\left( 2 \right)\] does not exist.
Therefore, the total number of critical points is 3.
Critical points of \[f\left( x \right) = \left| {x - 1} \right|\left| {x - 2} \right|\] are \[\left\{ {1,2,\dfrac{3}{2}} \right\}\]
Hence, option C. 3 is the correct answer.


Note: Critical point of a function: ‘When dealing with function of a real variable a critical point is a point in the domain of the function where the function is either not differentiable or the derivative is equal to zero.’