JEE Main & Advanced Mathematics Differential Equations Question Bank Self Evaluation Test - Differential Equations

  • question_answer
    A function \[y=f(x)\] satisfies the condition \[f'(x)\sin x+f(x)\cos x=1\] being bounded when\[x\to 0\]. If\[l=\int_{0}^{\pi /2}{f(x)dx}\], then

    A) \[\frac{\pi }{2}<l<\frac{{{\pi }^{2}}}{4}\]

    B) \[\frac{\pi }{4}<l<\frac{{{\pi }^{2}}}{2}\]

    C) \[1<l<\frac{\pi }{2}\]

    D) \[0<l<1\]

    Correct Answer: A

    Solution :

    [a] \[\sin x\frac{dy}{dx}+y\cos x=1\] \[\frac{dy}{dx}+y\cot x=\cos ecx\] IF \[={{e}^{\int{\cot xdx}}}={{e}^{ln(sinx)}}=\sin x\] \[y\sin x=\int{\cos ecx.\sin xdx=x+C}\] IF \[x=0,y\] is finite \[\therefore C=0\] \[y=x(cosec\,x)=\frac{x}{\sin \,\,x}\] Now, \[l<\frac{{{\pi }^{2}}}{4}andl>\frac{\pi }{2}\] Hence, \[\frac{\pi }{2}<l<\frac{{{\pi }^{2}}}{4}\]

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