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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 35

  • question_answer
    Let \[Y=SX\]and \[Z=tX\] such that\[\left| \begin{matrix}    X & Y & Z  \\    {{X}_{1}} & {{Y}_{1}} & {{Z}_{1}}  \\    {{X}_{2}} & {{Y}_{2}} & {{Z}_{2}}  \\ \end{matrix} \right|+\left| \begin{matrix}    {{S}_{1}} & {{t}_{1}}  \\    {{S}_{2}} & {{t}_{2}}  \\ \end{matrix} \right|={{X}^{n}},\]all the variables being differentiable functions of x and lower suffices denote the derivative with respect to x. Then n =

    A) \[1\]                      

    B)        \[2\]                      

    C) \[3\]       

    D)        \[4\]

    Correct Answer: C

    Solution :

    [c] \[\Delta =\left| \begin{matrix}    X & SX & tX  \\    {{X}_{1}} & S{{X}_{1}}+{{S}_{1}}X & t{{X}_{1}}+{{t}_{1}}X  \\    {{X}_{2}} & S{{X}_{2}}+2{{S}_{1}}{{X}_{1}}+{{S}_{2}}X & t{{X}_{2}}+2{{t}_{1}}{{X}_{1}}+{{t}_{2}}X  \\ \end{matrix} \right|\] \[=\left| \begin{matrix}    X & 0 & 0  \\    {{X}_{1}} & {{S}_{1}}X & {{t}_{1}}X  \\    {{X}_{2}} & 2{{S}_{1}}{{X}_{1}}+{{S}_{2}}X & 2{{t}_{1}}{{X}_{1}}+{{t}_{2}}X  \\ \end{matrix} \right|\] (Applying \[{{C}_{2}}\to {{C}_{2}}-S{{C}_{1}}\] and \[{{C}_{3}}\to {{C}_{3}}-t{{C}_{1}}\]) \[={{X}^{2}}\left| \begin{matrix}    {{S}_{1}} & {{t}_{1}}  \\    2{{S}_{1}}{{X}_{1}}+{{S}_{2}}X & 2{{t}_{1}}{{X}_{1}}+{{t}_{2}}X  \\ \end{matrix} \right|\] \[={{X}^{3}}\left| \begin{matrix}    {{S}_{1}} & {{t}_{1}}  \\    {{S}_{2}} & {{t}_{2}}  \\ \end{matrix} \right|\] (Applying \[{{R}_{2}}\to {{R}_{2}}-2{{X}_{1}}{{R}_{1}}\] )              \[\therefore \,\,\,\,\,\,n=3\]              


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