question_answer

Assertion [A]: If then \[f'\left( x \right)\]does not exist

Reason [R]: \[f'\left( x \right)\]is not continuous at x =2.

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 L.H.D. \[\underset{x\to {{2}^{2-}}}{\mathop{Lt}}\,\frac{f(x)-f(2)}{x-2}=\underset{h\to 0}{\mathop{Lt}}\,\frac{f(2-h)-f(2)}{-h}\] \[=\underset{h\to 0}{\mathop{Lt}}\,\frac{(2-h+1)-3}{-h}=\underset{h\to 0}{\mathop{Lt}}\,\frac{3-h-3}{-h}=\underset{h\to 0}{\mathop{Lt}}\,\frac{-h}{-h}=1\] R.H.D. \[\underset{x\to {{2}^{+}}}{\mathop{Lt}}\,\frac{f(x)-f(2)}{x-2}=\underset{h\to 0}{\mathop{Lt}}\,\frac{f(2+h)-f(2)}{h}\] \[=\underset{h\to 0}{\mathop{Lt}}\,\frac{2(2+h)-1-3}{h}\] \[=\underset{h\to 0}{\mathop{Lt}}\,\frac{4+2h-1-3}{h}=\underset{h\to 0}{\mathop{Lt}}\,\frac{2h}{h}=2\] Clearly \[LHD\ne RHD\] \[\Rightarrow \,f\left( x \right)\]is not differentiable at x = 2 \[\Rightarrow \,f'\left( x \right)\] does not exist \[\therefore \] Assertion [A] is true Also LH \[Lt=\underset{x\to {{2}^{-}}}{\mathop{Lt}}\,\,f\left( x \right)=\underset{x\to {{2}^{-}}}{\mathop{Lt}}\,\left( x+1 \right)=3\] RH \[Lt=\underset{x\to {{2}^{+}}}{\mathop{Lt}}\,\,f\left( x \right)=\underset{x\to {{2}^{+}}}{\mathop{Lt}}\,\,\left( 2x-1 \right)=2\times 2-1=3\] \[\therefore \] LH Lt = RH Lt \[\Rightarrow f\left( x \right)\] is continuous at \[x=2\] \[\therefore \] Reason (R) is false Hence option [C] is the correct answer.