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12th Class Mathematics Definite Integrals Question Bank Critical Thinking

  • question_answer
    For which of the following values of m, the area of the region bounded by the curve \[y=x-{{x}^{2}}\] and the line \[y=mx\] equals\[\frac{9}{2}\] [IIT 1999]

    A) \[-4\]                                       

    B) \[-2\]

    C) 2    

    D) 4

    Correct Answer: B

    Solution :

    • The equation of curve is \[y=x-{{x}^{2}}\] \[\Rightarrow {{x}^{2}}-x=-y\] Þ \[{{\left( x-\frac{1}{2} \right)}^{2}}=-\left( y-\frac{1}{4} \right)\]                   
    • This is a parabola whose vertex is \[\left( \frac{1}{2},\frac{1}{4} \right)\]
    • Hence point of intersection of the curve and the line \[x-{{x}^{2}}=mx\Rightarrow x(1-x-m)=0\]i.e.,\[x=0\]or\[x=1-m\]                   
    • \[\therefore \,\,\frac{9}{2}=\int_{0}^{1-m}{(x-{{x}^{2}}-mx)dx=\left( \frac{{{x}^{2}}}{2}-\frac{{{x}^{3}}}{3}-m\frac{{{x}^{2}}}{2} \right)_{0}^{1-m}}\]                         
    • \[=(1-m)\frac{{{(1-m)}^{2}}}{2}-\frac{{{(1-m)}^{3}}}{3}=\frac{{{(1-m)}^{3}}}{6}\]                   
    • \ \[{{(1-m)}^{3}}=\frac{6\times 9}{2}=27\Rightarrow 1-m={{27}^{1/3}}=3\Rightarrow m=-2\]                   
    • Also, \[{{(1-m)}^{3}}-{{3}^{3}}=0\]                   
    • Þ \[(1-m-3)[{{(1-m)}^{2}}+9+(1-m)3]=0\]           
    • Þ \[{{(1-m)}^{2}}+3(1-m)+9=0\]           
    • Þ \[{{m}^{2}}-2m+1-3m+3+9=0\]Þ \[{{m}^{2}}-5m+13=0\]                   
    • \[m=\frac{5\pm \sqrt{25-52}}{2}\]i.e., m is imaginary                   
    • Hence, \[m=-2\].


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