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12th Class Mathematics Definite Integrals Question Bank Critical Thinking

  • question_answer
    If \[f(x)=A\sin \left( \frac{\pi x}{2} \right)+B,\] \[{f}'\left( \frac{1}{2} \right)=\sqrt{2}\] and \[\int_{0}^{1}{f(x)\,dx=\frac{2A}{\pi },}\] then the constants \[A\] and \[B\] are respectively [IIT 1995]

    A) \[\frac{\pi }{2}\] and \[\frac{\pi }{2}\]                           

    B) \[\frac{2}{\pi }\] and \[\frac{3}{\pi }\]

    C) \[\frac{4}{\pi }\] and 0     

    D) 0 and \[-\frac{4}{\pi }\]

    Correct Answer: C

    Solution :

    • \[f(x)=A\sin \left( \frac{\pi x}{2} \right)+B,\,f'\left( \frac{1}{2} \right)=\sqrt{2}\],                  
    • \[\int_{0}^{1}{f(x)dx=\frac{2A}{\pi }}\],                              
    • (given)                   
    • Þ \[\int_{0}^{1}{\left\{ A\sin \left( \frac{\pi x}{2} \right)+B \right\}dx=\frac{2A}{\pi }}\]                   
    • Þ \[\left| -\frac{2A}{\pi }\cos \frac{\pi x}{2}+Bx \right|_{0}^{1}=\frac{2A}{\pi }\]                   
    • Þ \[B-\left( \frac{-2A}{\pi } \right)=\frac{2A}{\pi }\Rightarrow B=0\]                   
    • \[\therefore \,\,f(x)=A\sin \frac{\pi x}{2}\Rightarrow f'(x)=\frac{\pi A}{2}\cos \frac{\pi x}{2}\]                   
    • \[\therefore \,f'\left( \frac{1}{2} \right)=\frac{\pi A}{2}\left( \frac{1}{\sqrt{2}} \right)=\sqrt{2}\Rightarrow \pi A=4\Rightarrow A=\frac{4}{\pi }\]                    Hence \[A=\frac{4}{\pi },B=0\].


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