RAJASTHAN ­ PET Rajasthan PET Solved Paper-2011

  • question_answer
    If \[y=sin(sinx)\]and \[\frac{{{d}^{2}}y}{d{{x}^{2}}}+\frac{dy}{dx}\tan x+f(x)=0,\]then\[f(x)\]equals to

    A)  \[si{{n}^{2}}x.sin(cos\text{ }x)\]

    B)  sin2 x cos(sin x)

    C)  \[co{{s}^{2}}x.sin(\cos \text{ }x)\]

    D)  \[co{{s}^{2}}x\text{ }sin\text{ }(sin\text{ }x)\]

    Correct Answer: D

    Solution :

     \[\frac{dy}{dx}=\cos (\sin x).\cos x\] \[\Rightarrow \] \[\frac{{{d}^{2}}y}{d{{x}^{2}}}=-\cos (\sin x).\sin x\] \[+\cos [-\sin (\sin x)\cos x]\] \[\Rightarrow \] \[\frac{{{d}^{2}}y}{d{{x}^{2}}}+\frac{dy}{dx}\tan x=-{{\cos }^{2}}x.\sin (\sin x)\]


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