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JEE Main & Advanced Mathematics Indefinite Integrals Question Bank Critical Thinking

  • question_answer
    If \[\int_{{}}^{{}}{f(x)\sin x\cos x\ dx=\frac{1}{2({{b}^{2}}-{{a}^{2}})}\log (f(x))}+c\], then \[f(x)=\]

    A) \[\frac{1}{{{a}^{2}}{{\sin }^{2}}x+{{b}^{2}}{{\cos }^{2}}x}\]

    B) \[\frac{1}{{{a}^{2}}{{\sin }^{2}}x-{{b}^{2}}{{\cos }^{2}}x}\]

    C) \[\frac{1}{{{a}^{2}}{{\cos }^{2}}x+{{b}^{2}}{{\sin }^{2}}x}\]

    D) \[\frac{1}{{{a}^{2}}{{\cos }^{2}}x-{{b}^{2}}{{\sin }^{2}}x}\]

    Correct Answer: A

    Solution :

    • Since  \[\int_{{}}^{{}}{f(x)\,\sin x\cos x\,dx}=\frac{1}{2({{b}^{2}}-{{a}^{2}})}\log \left( f(x) \right)+c\]                   
    • Therefore \[f(x)\sin x\cos x\,=\frac{1}{2({{b}^{2}}-{{a}^{2}})}.\frac{1}{f(x)}{f}'(x)\]                   
    • Differentiating both sides \[w.r.t.\,\,x\]                   
    • \[\Rightarrow 2({{b}^{2}}-{{a}^{2}})\sin x\cos x=\frac{{f}'(x)}{f{{(x)}^{2}}}\]                   
    • \[\Rightarrow \int_{{}}^{{}}{(2{{b}^{2}}\sin x\cos x-2{{a}^{2}}\sin x\cos x)\,dx=}\int_{{}}^{{}}{\frac{{f}'(x)}{{{\{f(x)\}}^{2}}}}\,dx\]                   
    • \[\Rightarrow \pm \,(-{{b}^{2}}{{\cos }^{2}}x-{{a}^{2}}{{\sin }^{2}}x)=-\frac{1}{f(x)}\]                
    • \[\Rightarrow f(x)=\pm \frac{1}{({{a}^{2}}{{\sin }^{2}}x+{{b}^{2}}{{\cos }^{2}}x)}\].


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