2. Questions Bank
  3. JEE Main & Advanced
  4. Mathematics
  5. Functions

JEE Main & Advanced Mathematics Functions Question Bank Critical Thinking

  • question_answer
    Let \[g(x)=x.\,f(x),\]where \[f(x)=\left\{ \begin{align}   & x\sin \frac{1}{x},\,x\ne 0 \\  & \,\,\,\,\,\,\,\,\,\,\,\,0,\,x=0 \\ \end{align} \right.\] at \[x=0\] [IIT Screening 1994; UPSEAT 2004]

    A) g is differentiable but g' is not continuous

    B) g is differentiable while f is not      

    C) Both f and g are differentiable           

    D) g is differentiable and g' is continuous

    Correct Answer: A

    Solution :

    •   \[f(x)=\left\{ \begin{array}{*{35}{r}}    x\,\sin \frac{1}{x}\,\,, & x\ne 0  \\    0\,\,\,\,, & x=0  \\ \end{array} \right.\], \[g(x)=\left\{ \begin{array}{*{35}{r}}    {{x}^{2}}\,\sin \frac{1}{x}\,\,, & x\ne 0  \\    0\,\,\,\,, & x=0  \\ \end{array} \right.\]           
    • \[L\,{f}'(0)=\underset{h\to 0}{\mathop{\lim }}\,\,\,\frac{f(0-h)-f(0)}{-h}\]                   
    • \[=\underset{h\to 0}{\mathop{\lim }}\,\,\frac{(0-h)\,\sin \,(-\frac{1}{h})-(0)}{-h}=\underset{h\to 0}{\mathop{\lim }}\,\,\,-\sin \,\left( \frac{1}{h} \right)\]
    • = a quantity which lies between ? 1 and 1           
    • \[R\,{f}'(0)=\underset{h\to 0}{\mathop{\lim }}\,\,\,\frac{f(0+h)-f(0)}{h}\]                   
    • \[=\underset{h\to 0}{\mathop{\lim }}\,\,\frac{(0+h)\,\,\sin \frac{1}{h}-0}{h}=\underset{h\to 0}{\mathop{\lim }}\,\,\,\sin \frac{1}{h}\]                   
    • = a quantity which lies between ? 1 and 1           
    • Hence \[L\,{f}'(0)\ne R\,{f}'(0)\]  \\[f(x)\] is not differentiable at \[x=0\]           
    • Now \[L\,{g}'(0)=\underset{h\to 0}{\mathop{\lim }}\,\,\,\frac{f(0-h)-f(0)}{0-h}\]           
    • \[L\,{g}'(0)=\underset{h\to 0}{\mathop{\lim }}\,\,\frac{{{(0-h)}^{2}}\sin \,(-\frac{1}{h})-0}{-h}=\underset{h\to 0}{\mathop{\lim }}\,h\,\sin \,\left( \frac{1}{h} \right)\]           
    • \[L\,{g}'(0)=0\times \left( -1\le \sin \frac{1}{h}\le 1 \right)\] Þ \[L\,{g}'(0)=0\]           
    • and \[R{g}'(0)=\underset{h\to 0}{\mathop{\lim }}\,\,\,\frac{f\,(0+h)-f(0)}{h}=\underset{h\to 0}{\mathop{\lim }}\,\frac{{{(0+h)}^{2}}\sin \left( \frac{1}{h} \right)-0}{h}\]                        
    • \[=\underset{h\to 0}{\mathop{\lim }}\,h\,\sin \left( \frac{1}{h} \right)=0\times \left( -1\le \sin \left( \frac{1}{h} \right)\le 1 \right)=0\]           
    • \[\because \,\,L\,{g}'(0)=R\,{g}'(0)\] then \[g(x)\] is differentiable at \[x=0\]           
    • Now \[g(x)={{x}^{2}}\sin \frac{1}{x}\] \[{g}'(x)=2x\,\sin \frac{1}{x}+{{x}^{2}}\cos \frac{1}{x}\times -\frac{1}{{{x}^{2}}}\]           
    • \[{g}'(x)=2x\,\sin \frac{1}{x}-\cos \frac{1}{x}\] Þ \[{g}'(x)=2\,f(x)-\cos \frac{1}{x}\]           
    • So, \[{g}'(x)\] is not differentiable at \[x=0\].


You need to login to perform this action.
You will be redirected in 3 sec spinner