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KVPY Sample Paper KVPY Stream-SX Model Paper-9

  • question_answer
    \[\begin{align}   & if\,f(x)={{x}^{3}}+3{{x}^{2}}+6x+2\sin x,then\,\,the\, \\  & equation\frac{1}{x-f(1)}+\frac{2}{x-f(2)}+\frac{3}{x-f(3)}=0\,has \\ \end{align}\]

    A) No real roots     

    B) 1real root

    C) 2real roots

    D) more than 2real roots

    Correct Answer: C

    Solution :

    \[\begin{align}   & f'(x)=3{{x}^{2}}+6x+6+2\operatorname{cosx} \\  &  \\ \end{align}\],
    \[=3{{\left( x+1 \right)}^{2}}+3+\cos x>0\]for all x,
    So\[f\left( x \right)\]is an increasing function.
    Thus \[f(1)<f(2)<f(3)\]
    Let\[f(1)=a,f(2)=b\,and\,f(3)=c,\]
    Then \[a<b<c\]
    Give equation is \[\frac{1}{x-a}+\frac{2}{x-b}+\frac{3}{x-c}=0\]
    \[\begin{align}   & \Rightarrow \left( x-b \right)\left( x-c \right)+2\left( x-a \right)\left( x-c \right) \\  & +3\left( x-a \right)\left( x-b \right)=0 \\ \end{align}\]
    \[\text{g }\left( a \right)=\left( a-b \right)\left( a-c \right)+2.0+3.0>0;\]
    \[g(b)=2\left( b-a \right)\left( b-c \right)<0and\] And
    \[g(c)=3\left( c-a \right)\left( c-b \right)>0\]
    Hence, the equation g(x) =0 has a root in (a, b) and another in (b, c).


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