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  3. RAJASTHAN ­ PET

RAJASTHAN ­ PET Rajasthan PET Solved Paper-2002

  • question_answer
    If\[y=|x-1|+|x-3|,\]then the value of\[\frac{dy}{dx}\]at \[x=2\]is

    A)  1                   

    B)  0

    C)  \[-1\]               

    D)  does not exist

    Correct Answer: B

    Solution :

     \[y=|x-1|+|x-3|\] At  \[x=2,y=x-1-(x-3)\] \[\Rightarrow \] \[y=2\] \[\therefore \] \[\frac{dy}{dx}=0\]


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