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BITSAT2013MathematicsDeterminantsActual

The determinant | array ccc 1 & (x-3) & (x-3)² 1 & (x-4) & (x-4)² 1 & (x-5) & (x-5)² array |

Options

  1. A3 values of x
  2. B2 values of x
  3. C1 values of x
  4. DNo value of x

Correct answer

D. No value of x

Step-by-step solution

The given determinant vanishes, i.e., | array ccc 1 & x-3 & (x-3)² 1 & x-4 & (x-4)² 1 & x-5 & (x-5)² array |=0 Expanding along C ₁, we get array l (x-4)(x-5)²-(x-5)(x-4)²- (x-3)(x-5)² . .-(x-5)(x-3)² +(x-3)(x-4)² -(x-4)(x-3)²=0 (x-4)(x-5)(x-5-x+4) -(x-3)(x-5)(x-5-x+3) +(x-3)(x-4)(x-4-x+3)=0 array -(x-4)(x-5)+2(x-3)(x-5)-(x-3) (x-4)=0 -x²+9 x-20+2 x²-16 x+30-x²+7 x-12=0 -32+30=0 -2=0 Which is not possible, hence no value of x satisfies the given condition.

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