JEE Main & Advanced AIEEE Paper (Held On 11 May 2011)

  • question_answer
    If function f(x) is differentiate at x = a, then \[\underset{x\to a}{\mathop{\lim }}\,\frac{{{x}^{2}}f(a)-{{a}^{2}}f(x)}{x-a}\]is:     AIEEE  Solved  Paper (Held On 11 May  2011)

    A) \[-{{a}^{2}}f'(a)\]

    B) \[af(a)-{{a}^{2}}f\,'(a)\]

    C) \[2af(a)-{{a}^{2}}f\,'(a)0\]

    D) \[2af(a)+{{a}^{2}}f'(a)\]

    Correct Answer: C

    Solution :

                    \[\underset{x\to a}{\mathop{\ell im}}\,\frac{{{x}^{2}}f(a)-{{a}^{2}}f(x)}{x-a}\]                 \[=\underset{x\to a}{\mathop{\ell im}}\,\frac{2xf(a)-{{a}^{2}}f'(x)}{1}\] \[=2af(a)-{{a}^{2}}f'(a)\] Alter\[\underset{x\to a}{\mathop{\ell im}}\,\frac{{{x}^{2}}f(a)-{{a}^{2}}f(x)}{x-a}\] \[=\underset{x\to a}{\mathop{\ell im}}\,\frac{{{x}^{2}}f(a)-{{a}^{2}}f(a)+{{a}^{2}}f(a)-{{a}^{2}}f(x)}{x-a}\] \[=\underset{x\to a}{\mathop{\ell im}}\,\frac{({{x}^{2}}-{{a}^{2}})f(a)-{{a}^{2}}f(x)-f((a))}{x-a}\] \[=\underset{x\to a}{\mathop{\ell im}}\,(x+a)f(a)-{{a}^{2}}\left\{ \frac{f(x)-f(a)}{(x-a)} \right\}\] \[=2af(a)-{{a}^{2}}f'(a)\]


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