Question

For what value of \[\lambda \] is the function defined by \[f\left( x \right)=\left\{ \begin{matrix}
   \lambda \left( {{x}^{2}}-2x \right),if\ x\le 0 \\
   4x+1,\ if\ x>0 \\
\end{matrix} \right.\]
Continuous at x = 0? What about the continuity at x = 1?

Answer 1

Hint: In the given question, we have been given a function and asked to check the continuity of the function at x = 0 and at x = 1. At x = 0 if the given function is continuous, it will satisfy the given condition i.e. \[\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f\left( x \right)=f\left( 0 \right)\] . Then in the same way we will be checking the continuity of function at x = 1.

Complete step by step solution:
We have given that,
 \[f\left( x \right)=\left\{ \begin{matrix}
   \lambda \left( {{x}^{2}}-2x \right),if\ x\le 0 \\
   4x+1,\ if\ x>0 \\
\end{matrix} \right.\]
If the given function f(x) is continuous at x = 0, then
 \[\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f\left( x \right)=f\left( 0 \right)\]
Therefore,
 \[\Rightarrow \underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\lambda \left( {{x}^{2}}-2x \right)=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,4x+1=\lambda \left( {{0}^{2}}-2\left( 0 \right) \right)\]
Solving the above, we obtained
 \[\Rightarrow \underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\lambda \left( {{0}^{2}}-2\left( 0 \right) \right)=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,4\left( 0 \right)+1=\lambda \left( {{0}^{2}}-2\left( 0 \right) \right)\]
Simplifying the above by solving the numbers, we get
 \[\Rightarrow 0=1=0\]
Therefore, it can never be possible as 0 is not equal to 1.
Thus,
There will be no value of \[\lambda \] for which the given function f(x) is continuous at x = 0.
Now,
At x = 1,
If the given function f(x) is continuous at x = 1, then
 \[\underset{x\to 1}{\mathop{\lim }}\,f\left( x \right)=f\left( 1 \right)\]
Therefore,
 \[\Rightarrow \underset{x\to {{0}^{1}}}{\mathop{\lim }}\,4x+1=4\left( 1 \right)+1\]
Solving the above, we obtained
 \[\Rightarrow \underset{x\to 1}{\mathop{\lim }}\,4\left( 1 \right)+1=4\left( 1 \right)+1\]
Simplifying the above by solving the numbers, we get
 \[\Rightarrow 5=5\]
 \[\therefore \underset{x\to 1}{\mathop{\lim }}\,f\left( x \right)=f\left( 1 \right)\]
Thus,
For any values of \[\lambda \] , the given function f(x) is continuous at x = 1.
Hence this is the required answer.

Note: The key note while solving the question is that when we are checking the continuity a x = 1, then we need to take the only function where the value of ‘x’ defined is greater than 0. Most of the students make mistakes taking the other function for checking the continuity at x = 1, but that function is defined for the values of ‘x’ that are less than zero. Students should be very careful while doing the calculation part to avoid making any error.