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JEE Main & Advanced JEE Main Paper (Held On 25 April 2013)

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                    If an equation of a tangent to the curve,\[y=\cos (x+y),-1\le x\le 1+\pi \], is \[x+2y=k\]then \[k\]is equal to     JEE Main Online Paper ( Held On 25  April 2013 )

    A)                 1                                             

    B)                                        2

    C)                                        \[\frac{\pi }{4}\]                                              

    D)                                        \[\frac{\pi }{2}\]              

    Correct Answer: D

    Solution :

                    Let\[y=\cos (x+y)\] \[\Rightarrow \]\[\frac{dy}{dx}=-\sin (x+y)\left( 1+\frac{dy}{dx} \right)\]                             ?.(1) Now, given equation of tangent is\[x+2y=k\] \[\Rightarrow \]Slope\[=\frac{-1}{2}\] So, \[\frac{dy}{dx}=\frac{-1}{2}\]put this value in (1), we get \[\frac{-1}{2}=-\sin (x+y)\left( 1-\frac{1}{2} \right)\] \[\Rightarrow \]\[\sin (x+y)=1\]\[\Rightarrow \]\[x+y=\frac{\pi }{2}\]\[\Rightarrow \]\[y=\frac{\pi }{2}-x\] Now, \[\frac{\pi }{2}-x=\cos (x+y)\]\[\Rightarrow \]\[x=\frac{\pi }{2}\]and\[y=0\] Thus\[x+2y=k\Rightarrow \frac{\pi }{2}=k\]


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