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BCECE Engineering BCECE Engineering Solved Paper-2014

  • question_answer
     If \[f'(2)=6,f'(1)=4,\], then \[\underset{h\to 0}{\mathop{\lim }}\,\frac{f(2h+2+{{h}^{2}})-f(2)}{f(h-{{h}^{2}}+1)-f(1)}\] is equal to

    A) 3                                             

    B) \[-\frac{3}{2}\]

    C) \[\frac{3}{2}\]                                   

    D) Does not exist

    Correct Answer: A

    Solution :

     \[\underset{h\to 0}{\mathop{\lim }}\,\frac{f(2h+2+{{h}^{2}})-f(2)}{f(h-{{h}^{2}}+1)-f(1)}\] \[\underset{h\to 0}{\mathop{\lim }}\,\frac{\{f'(2h+2+{{h}^{2}})\}(2+2h)-0}{\{f'(h-{{h}^{2}}+1)\}(1-2h)-0}\]                 (using L Hospital?s rule) \[=\frac{f'(2).2}{f'(1).1}=\frac{6\times 2}{4\times 1}\] \[=\frac{12}{4}=3\]


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