| Directions: (1 - 2) |
| The following questions consist of two statements, one labelled as "Assertion [A] and the other labelled as Reason [R]". You are to examine these two statements carefully and decide if Assertion [A] and Reason [R] are individually true and if so, whether the Reason [R] is the correct explanation for the given Assertion [A]. Select your answer from following options. |
Assertion (A):  is continuous at x = 0. |
Reason (R): Both \[h\left( x \right)={{x}^{2}}\],  are continuous at x = 0. |
A) Both A and R are individually true and R is the correct explanation of A.
B) Both A and R are individually true and R is not the correct explanation of A.
C) 'A' is true but 'R' is false
D) 'A' is false but 'R' is true
E) Both A and R are false.
Correct Answer: C
Solution :
Given Assertion (A) : \[f\left( x \right)=\left\{ \begin{matrix} {{x}^{2}}\sin \left( \frac{1}{x} \right),\,\,\,\,\,\,x\ne 0 \\ 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x=0 \\ \end{matrix} \right.\] \[\underset{x\to 0}{\mathop{Lt}}\,\,\,f\left( x \right)=f\left( 0 \right)\,\Rightarrow f\left( x \right)\](A finite Value) = 0 Also \[f\left( 0 \right)=0\] \[\therefore \,\,\,\underset{x\to 0}{\mathop{Lt}}\,\,f\left( x \right)=f\left( 0 \right)\Rightarrow f\left( x \right)\] is continuous at x = 0 Also given Reason (R) : \[h\left( x \right)={{x}^{2}}\], which is polynomial and hence continuous But g(x) is not continuous as \[\underset{x\to 0}{\mathop{Lt}}\,g\left( x \right)=\underset{x\to 0}{\mathop{Lt}}\,\sin \left( \frac{1}{x} \right)=\] = Not defined (Value oscillates between -1 to 1) \[\therefore \]Reason R is false Hence Option [C is the correct answer.
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