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12th Class Mathematics Definite Integrals Question Bank Critical Thinking

  • question_answer
    If \[f(x)=\int_{-1}^{x}{|t|\,dt,}\] \[x\ge -1,\] then [MNR 1994]

    A) \[f\] and \[{f}'\] are continous for \[x+1>0\]

    B) \[f\] is continous but \[{f}'\] is not continous for \[x+1>0\]

    C) \[f\] and \[{f}'\] are not continous at \[x=0\]

    D) \[f\] is continous at \[x=0\] but \[{f}'\] is not so

    Correct Answer: A

    Solution :

    • Let us divide the interval into two sub-intervals\[{{I}_{1}}\], \[-1\le x<0\] so that x is ?ve and \[{{I}_{2}},x\ge 0\]so that x is +ve.           
    • For \[{{I}_{1}},f(x)=\int_{-1}^{x}{(-t)dt=-\frac{1}{2}({{x}^{2}}-1)}\]   .....(i)           
    • For \[{{I}_{1}},f(x)=\int_{-1}^{x}{(-t)dt=-\frac{1}{2}({{x}^{2}}-1)}\]           
    • \[=-\frac{1}{2}[{{t}^{2}}]_{-1}^{0}+\frac{1}{2}[{{t}^{2}}]_{0}^{x}=\frac{1}{2}(1+{{x}^{2}})\]     .....(ii)           
    • Hence the function can be defined as the following           
    • \[f(x)=\left\{ \begin{align}   & -\frac{1}{2}({{x}^{2}}-1),\text{If}-1\le x0 \\ \end{align} \right.\]           
    • For \[f,\,\,L=R=V=\frac{1}{2}\]at \[x=0\], so \[f\] is continuous at \[x=0\]. For \[f',\,\,L=R=V=0\]at \[x=0\], so \[f'\]is also continuous at \[x=0\]. Thus both \[f\]and \[f'\]are continuous at \[x=0\] and hence both are continuous for \[x>-1\]i.e., \[x+1>0\].


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