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JEE Main & Advanced Mathematics Applications of Derivatives Question Bank Critical Thinking

  • question_answer
    At what points of the curve \[y=\frac{2}{3}{{x}^{3}}+\frac{1}{2}{{x}^{2}},\]tangent makes the equal angle with axis [UPSEAT 1999]

    A) \[\left( \frac{1}{2},\,\frac{5}{24} \right)\] and \[\left( -1,\,-\frac{1}{6} \right)\]

    B) \[\left( \frac{1}{2},\,\frac{4}{9} \right)\] and \[\left( -1,\,0 \right)\]

    C) \[\left( \frac{1}{3},\,\frac{1}{7} \right)\] and \[\left( -3,\,\frac{1}{2} \right)\]

    D) \[\left( \frac{1}{3},\,\frac{4}{47} \right)\] and \[\left( -1,\,-\frac{1}{3} \right)\]

    Correct Answer: A

    Solution :

    • \[y=\frac{2}{3}{{x}^{3}}+\frac{1}{2}{{x}^{2}}\] Þ \[\frac{dy}{dx}=2{{x}^{2}}+x\]   ?..(i)           
    • Now tangent makes equal angle with axis           
    • \ \[y={{45}^{o}}\] or \[-{{45}^{o}}\]           
    • \ \[\frac{dy}{dx}=\tan (\pm {{45}^{o}})=\pm \tan ({{45}^{o}})=\pm 1\]           
    • \ From equation (i), \[2{{x}^{2}}+x=1\] (taking +ve sign)           
    • Þ \[2{{x}^{2}}+x-1=0\] Þ \[(2x-1)\,(x+1)=0\]           
    • \\[x=\frac{1}{2},-1\]           
    • From the given curve, when \[x=\frac{1}{2}\], \[y=\frac{2}{3}.\frac{1}{8}+\frac{1}{2}.\frac{1}{4}=\frac{1}{12}+\frac{1}{8}=\frac{5}{24}\] and when \[x=-1\],   \[y=\frac{2}{3}(-1)+\frac{1}{2}.1\]= \[-\frac{2}{3}+\frac{1}{2}=-\frac{1}{6}\]           
    • Therefore, required points are \[\left( \frac{1}{2},\,\frac{5}{24} \right)\] and \[\left( -1,\,-\frac{1}{6} \right)\].


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