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JEE Main & Advanced Mathematics Functions Question Bank Critical Thinking

  • question_answer
    The values of a and b such that \[\underset{x\to 0}{\mathop{\lim }}\,\frac{x(1+a\cos x)-b\sin x}{{{x}^{3}}}=1\], are               [Roorkee 1996]

    A) \[\frac{5}{2},\ \frac{3}{2}\]

    B) \[\frac{5}{2},\ -\frac{3}{2}\]

    C) \[-\frac{5}{2},\ -\frac{3}{2}\]

    D) None of these

    Correct Answer: C

    Solution :

    • \[\underset{x\to 0}{\mathop{\lim }}\,\,\,\frac{x\,(1+a\cos x)-b\,\sin x}{{{x}^{3}}}=1\]           
    • Þ \[\underset{x\to 0}{\mathop{\lim }}\,\,\frac{x\left\{ 1+a\,\left( 1-\frac{{{x}^{2}}}{2\,\,!}+\frac{{{x}^{4}}}{4\,\,!}-\frac{{{x}^{6}}}{6\,\,!}+... \right) \right\}-b\,\left\{ x-\frac{{{x}^{3}}}{3\,\,!}+\frac{{{x}^{5}}}{5\,\,!}-... \right\}}{{{x}^{3}}}=1\]           
    • Þ\[\underset{x\to 0}{\mathop{\lim }}\,\,\frac{(1+a-b)+{{x}^{2}}\,\left( \frac{b}{3\,\,!}-\frac{a}{2\,\,!} \right)+{{x}^{4}}\left( \frac{a}{4\,\,!}-\frac{b}{5\,\,!} \right)+...}{{{x}^{2}}}=1\]  .....(i)           
    • If \[1+a-b\ne 0,\]then L.H.S. \[\to \infty \] as \[x\to 0\] while  R.H.S. =1, therefore \[1+a-b=0.\]           
    • Now from (i), \[\underset{x\to 0}{\mathop{\lim }}\,\,\frac{{{x}^{2}}\left( \frac{b}{3\,\,!}-\frac{a}{2\,\,!} \right)+{{x}^{4}}\,\left( \frac{a}{4\,\,!}-\frac{b}{5\,\,!} \right)+...}{{{x}^{2}}}=1\]           
    • \[\Rightarrow \,\,\frac{b}{3\,!}-\frac{a}{2\,\,!}=1\,\Rightarrow \,\,b-3a=6\]. Solving \[1+a-b=0\] and \[b-3a=6,\] we get \[a=-5/2,\,\,b=-3/2\].


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