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JEE Main & Advanced JEE Main Online Paper (Held On 23 April 2013)

  • question_answer
                    If \[f(x)=\sin (\sin x)\]and            \[f''(x)+\tan xf(x)+g(x)=0,\]then g(\[x\]) is :     JEE Main Online Paper ( Held On 23  April 2013 )

    A)                  \[{{\cos }^{2}}x\cos (\sin x)\]    

    B)                                         \[{{\sin }^{2}}x\cos (\cos x)\]

    C)                                         \[{{\sin }^{2}}x\sin (\cos x)\]     

    D)                                         \[{{\cos }^{2}}x\sin (\sin x)\]

    Correct Answer: D

    Solution :

                     \[f(x)=\sin (\sin x)\] \[\Rightarrow \]\[f'(x)=\cos (\sin x).\cos x\] \[\Rightarrow \]\[f''(x)=-\sin (\sin x).{{\cos }^{2}}x+\cos (\sin x).(-\sin x)\] \[\Rightarrow \]\[=-{{\cos }^{2}}x.\sin (\sin x)-\sin x.\cos (\sin x)\] Now\[f''(x)+\tan x.f'(x)+g(x)=0\] \[\Rightarrow \]\[g(x)={{\cos }^{2}}x.\sin (\sin x)+\sin x.\cos (\sin x)\] Now\[f''(x)+\tan x.f'(x)+g(x)=0\] \[\Rightarrow \]\[g(x)={{\cos }^{2}}x.\sin (\sin x)+\sin x.\cos (\sin x)\] \[-\tan x.\cos x.\cos (\sin x)\] \[\Rightarrow \]\[g(x)={{\cos }^{2}}x.\sin (\sin x).\]


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