Let the numbers 2,b,c be in an A.P. and A=\begin{vmatrix} 1 &1 &1 \\ 2 &b &c \\ 4 &b^{2} &c^{2} \end{vmatrix}.. If \det (A)\epsilon \left [ 2,16 \right ], then c lies in the interval : 

  • Option 1)

    [2,3)

  • Option 2)

    \left ( 2+2^{\frac{3}{4}},4 \right )

  • Option 3)

    \left [ 4,6 \right ]

     

  • Option 4)

    \left [ 3,2+2^{\frac{3}{4}} \right ]

 

Answers (1)


A=\begin{vmatrix} 1 &1 &1 \\ 2 &b &c \\ 4 & b^{2} &c^{2} \end{vmatrix}

C_2\rightarrow C_2-C_1;C_3-C_1

A=\begin{vmatrix} 1 &0 &0 \\ 2 &b-2 &c-2 \\ 4 &b^{2}-4 & c^{2}-4 \end{vmatrix}=(b-2)(c^{2}-4)-(c-2)(b^{2}-4)

\left | A \right |=(b-2)(c-2)(c-b)

2,b,c are in A.P

2,2+d,2+2d\rightarrow AP.

\left | A \right |=(d)(2d)(d)=2d^{3}\epsilon \left [ 2,16 \right ]

d^{3}\equiv \left [ 1,8 \right ]

d\equiv \left [ 1,2 \right ]

C=2+2d

C\equiv \left [ 4,6 \right ]


Option 1)

[2,3)

Option 2)

\left ( 2+2^{\frac{3}{4}},4 \right )

Option 3)

\left [ 4,6 \right ]

 

Option 4)

\left [ 3,2+2^{\frac{3}{4}} \right ]

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