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Q. If $ y={{\cos }^{-1}}\cos (|x|-f(x)), $ where $ f(x)=\left\{ \begin{matrix} 1, & if & x>0 \\ -1, & if & x<0 \\ 0, & if & x=0 \\ \end{matrix} \right. $ Then, $ {{(dy/dx)}_{x=\frac{5\pi }{4}}} $ is equal to

J & K CETJ & K CET 2005

Solution:

$ \therefore $ $ y={{\cos }^{-1}}\cos (x-1),x > 0 $
$ \Rightarrow $ $ y=x-1,0\le x-1\le \pi $
$ \Rightarrow $ $ y=x-1,1\le x\le \pi +1 $
$ \therefore $ $ y=x-1,1\le x\le \pi +1 $
At $ x=\frac{5\pi }{4}\in [1,\,\pi +1] $
$ \Rightarrow $ $ \frac{dy}{dx}=1 $
$ \Rightarrow $ $ {{\left( \frac{dy}{dx} \right)}_{x=\frac{5\pi }{4}}}=1 $