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
- A3 values of x
- B2 values of x
- C1 values of x
- 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.