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JEE Main & Advanced JEE Main Paper (Held On 9 April 2016)

  • question_answer
    If the function \[f(x)=\left\{ \begin{matrix}    -x, & x<1  \\    a+{{\cos }^{-1}} & (x+b),1\le x\le 2  \\ \end{matrix} \right.\] is differentiable at x = 1, then\[\frac{a}{b}\]is equal to :

    A) \[\frac{-\pi -2}{2}\]                         

    B) \[-1-{{\cos }^{-1}}(2)\]

    C) \[\frac{\pi +2}{2}\]         

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

    Correct Answer: C

    Solution :

                    L.H.L. at x = 1 is . 1 R.H.L at x = 1 is \[a+{{\cos }^{-1}}(1+b)\] \[\Rightarrow \]\[-1=a+{{\cos }^{-1}}(1+b)\] \[{{\cos }^{-1}}(1+b)=-1-a\]                                         ?(i) Now  L.H.d. at \[x=1\] is - 1 R.H.D at x = 1 is \[\frac{-1}{\sqrt{1-{{(1+b)}^{2}}}}\] \[\Rightarrow \]\[{{(1+b)}^{2}}=0\]\[\Rightarrow \]\[b=-1\] \[a=-1-\frac{\pi }{2}\] \[\frac{a}{b}=\frac{-(2+\pi )}{2(-1)}=\frac{2+\pi }{2}\]                


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