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J & K CET Engineering J and K - CET Engineering Solved Paper-2009

  • question_answer
    If \[f\,(x)=\underset{y\to x}{\mathop{lim}}\,\,\frac{{{\sin }^{2}}y-{{\sin }^{2}}x}{{{y}^{2}}-{{x}^{2}}},\]then \[\int{4x\,\,f(x)\,\,dx}\]is equal to

    A)  \[\cos \,\,2x+c\]      

    B)  \[2\,\cos \,\,2x+c\]

    C)  \[-\,\cos \,\,2x+c\]    

    D)  \[-2\,\cos \,\,2x+c\]

    Correct Answer: C

    Solution :

    \[f(x)=\underset{y\to x}{\mathop{\lim }}\,\,\,\,\frac{{{\sin }^{2}}\,y-\,{{\sin }^{2}}x}{{{y}^{2}}-{{x}^{2}}}\] \[\left[ \frac{0}{0}\,form \right]\] \[=\underset{y\to x}{\mathop{\lim }}\,\,\frac{2\sin \,y\,\cos \,y-0}{2y-0}\] \[=\frac{\sin \,2x}{2x}\] \[\therefore \]\[\int{4x\,\,f(x)\,\,dx=\int{4x\,\,\left( \frac{\sin 2x}{2x} \right)}\,\,dx}\] \[=2\int{\sin \,\,2x\,\,dx}\] \[=-\cos \,2x+c\]


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