CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2009

  • question_answer
    If\[y={{\sec }^{-1}}[\cos ecx]+\cos e{{c}^{-1}}[\sec x]\] \[+{{\sin }^{-1}}[\cos x]+{{\cos }^{-1}}[\sin x],\]then\[\frac{dy}{dx}\]is equal to

    A)  0                                            

    B)  2

    C)  \[-2\]                   

    D)         \[-\,4\]

    E)  1

    Correct Answer: D

    Solution :

    Given, \[y={{\sec }^{-1}}(\cos ecx)+\cos {{\sec }^{-1}}(\sec x)\] \[+{{\sin }^{-1}}(\cos x)+{{\cos }^{-1}}(\sin x)\] \[={{\sec }^{-1}}\left[ \sec \left( \frac{\pi }{2}-x \right) \right]+\cos {{\sec }^{-1}}\left[ \cos ec\left( \frac{\pi }{2}-x \right) \right]\] \[+{{\sin }^{-1}}\left[ \sin \left( \frac{\pi }{2}-x \right) \right]+{{\cos }^{-1}}\left[ \cos \left( \frac{\pi }{2}-x \right) \right]\] \[=\frac{\pi }{2}-x+\frac{\pi }{2}-x+\frac{\pi }{2}-x+\frac{\pi }{2}-x\] \[\Rightarrow \]               \[y=2\pi -2x\] On differentiating w.r.t.\[x,\]we get \[\Rightarrow \]               \[\frac{dy}{dx}=-4\]


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