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JEE Main & Advanced JEE Main Paper (Held On 19 May 2012)

  • question_answer
    If\[f(x)=3{{x}^{10}}-7{{x}^{8}}+5{{x}^{6}}-21{{x}^{3}}+3{{x}^{2}}-7,\]then \[\underset{\alpha \to 0}{\mathop{\lim }}\,\frac{f\left( 1-\alpha  \right)-f\left( 1 \right)}{{{\alpha }^{3}}+3\alpha }\]is     JEE Main  Online Paper (Held On 19  May  2012)

    A) \[-\frac{53}{3}\]                                               

    B)                        \[\frac{53}{3}\]

    C)                        \[-\frac{55}{3}\]                                               

    D)                        \[\frac{55}{3}\]

    Correct Answer: B

    Solution :

                    Let\[f'(x)=3{{x}^{10}}-7{{x}^{8}}+5{{x}^{6}}-21{{x}^{3}}+3{{x}^{2}}-7\] \[f'(x)=30{{x}^{9}}-56{{x}^{7}}+30{{x}^{5}}-63{{x}^{5}}-63{{x}^{2}}+6x\] \[f'(1)=30-56+30-63+6\] \[=66-63-56=-53\] Consider\[\underset{\alpha \to 0}{\mathop{\lim }}\,\frac{f(1-\alpha )-f(1)}{{{\alpha }^{3}}+3\alpha }\] \[=\underset{\alpha \to 0}{\mathop{\lim }}\,\frac{f'(1-\alpha )(-1)-0}{3{{\alpha }^{2}}+3}\](By using L?hospital rule) \[=\frac{f'(1-0)(-1)}{3{{(0)}^{2}}+3}=\frac{-f'(1)}{3}=\frac{53}{3}\]


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