CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

Consider the functions $$f\left( x \right)=sin\left( x-1 \right) $$ and $$g\left( x \right)=\cot ^{ -1 }{ \left[ x-1 \right]  } $$

Assertion: The function $$\displaystyle F\left( x \right) =f\left( x \right).g\left( x \right)$$ is discontinuous at $$x=1$$

Reason: If $$f\left( x \right) $$ is discontinuous at $$x=a$$ and $$g\left( x \right) $$ is also discontinuous at $$x=a$$ then the product function $$f\left( x \right) .g\left( x \right) $$ is discontinuous at $$x=a$$.


A
Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
loader
B
Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
loader
C
Assertion is correct but Reason is incorrect
loader
D
Both Assertion and Reason are incorrect
loader

Solution

The correct option is C Assertion is correct but Reason is incorrect
$$f\left( x \right) =sin\left( x-1 \right) ;g\left( x \right) =\cot ^{ -1 }{ \left[ x-1 \right]  } $$
$$\displaystyle \therefore F\left( x \right) =f\left( x \right) .g\left( x \right) =\begin{cases} \begin{matrix} -\cot ^{ -1 }{ \left[ x-1 \right]  } ; & x<1 \end{matrix} \\ \begin{matrix} 0; & x=1 \end{matrix} \\ \begin{matrix} \cot ^{ -1 }{ \left[ x-1 \right]  } ; & x>1 \end{matrix} \end{cases}$$
$$\displaystyle =\begin{cases} \begin{matrix} -\cot ^{ -1 }{ \left( -1 \right) ; }  & x<1 \end{matrix} \\ \begin{matrix} 0; & x=1 \end{matrix} \\ \begin{matrix} \cot ^{ -1 }{ \left( 0 \right) ; }  & x>1 \end{matrix} \end{cases}=\begin{cases} \begin{matrix} -\left( \pi -\cot ^{ -1 }{ 1 }  \right) ; & x<1 \end{matrix} \\ \begin{matrix} 0; & x=1 \end{matrix} \\ \begin{matrix} \pi /2; & x>1 \end{matrix} \end{cases}$$
$$\displaystyle =\begin{cases} \begin{matrix} -3\pi /4; & x<1 \end{matrix} \\ \begin{matrix} 0; & x=1 \end{matrix} \\ \begin{matrix} \pi /2; & x>1 \end{matrix} \end{cases}$$
$$\displaystyle \Rightarrow F\left( { 1 }^{ - } \right) =-\frac { 3\pi  }{ 4 } ;F\left( { 1 }^{ + } \right) =\frac { \pi  }{ 2 } $$ and $$F\left( 1 \right) =0$$
$$\Rightarrow f\left( x \right) $$ is discontinuous at $$x=1$$
Therefore assertion is correct but product of two discontinuous functions may be a continuous function.
So reason is wrong.

Mathematics

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More



footer-image