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JEE Main & Advanced JEE Main Paper Phase-I (Held on 08-1-2020 Morning)

  • question_answer
    For a>0, let the curves \[{{C}_{1}}:{{y}^{2}}=ax\] and \[{{C}_{2}}:{{x}^{2}}=ay\] intersect at origin 0 and a point P. let the line \[x=b(0<b<a)\]intersect the chord OP and the x-axis at points Q and R, respectively. If the line x = b bisects the area bounded by the curves, \[{{C}_{1}}\] and \[{{C}_{2}}\], and the area of \[\Delta OQR=\frac{1}{2},\] then 'a' satisfies the equation [JEE MAIN Held On 08-01-2020 Morning]

    A) \[{{x}^{6}}+6{{x}^{3}}-4=0\]

    B) \[{{x}^{6}}-12{{x}^{3}}-4=0\]

    C) \[{{x}^{6}}-6{{x}^{3}}+4=0\]

    D) \[{{x}^{6}}-12{{x}^{3}}+4=0\]

    Correct Answer: D

    Solution :

    Area between \[{{y}^{2}}=ax\] and \[{{x}^{2}}=ay\]is \[\frac{16\left( \frac{a}{4} \right)\left( \frac{a}{4} \right)}{3}=\frac{{{a}^{2}}}{3}\] \[\therefore \int\limits_{0}^{b}{\left( \sqrt{ax}-\frac{{{x}^{2}}}{a} \right)dx=\frac{{{a}^{2}}}{6}}\]                 ...(i) Equation of AB is y = x \[\therefore \frac{1}{2}\cdot b\cdot b=\frac{1}{2}\]          \[\Rightarrow b=1\]         .....(ii) By (i) and (ii) \[\int\limits_{0}^{1}{\left( \sqrt{a}\sqrt{x}-\frac{{{x}^{2}}}{a} \right)dx=\frac{{{a}^{2}}}{6}}\] \[\Rightarrow \frac{\sqrt{a}{{x}^{3/2}}}{3/2}-{{\left. \frac{{{x}^{3}}}{3a} \right|}^{1}}_{0}=\frac{{{a}^{2}}}{6}\] \[\Rightarrow \frac{2}{3}\sqrt{a}-\frac{1}{3a}=\frac{{{a}^{2}}}{6}\] \[\Rightarrow 4{{a}^{3/2}}-2={{a}^{3}}\] \[\Rightarrow 4{{a}^{3/2}}={{a}^{3}}+2\] \[\Rightarrow 16{{a}^{2}}={{a}^{6}}+4{{a}^{3}}+4\] \[\Rightarrow {{a}^{6}}-12{{a}^{3}}+4=0\] Hence a satisfy \[{{x}^{6}}-12{{x}^{3}}+4=0\]


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