BITSAT 2013 Paper

The given determinant vanishes, i.e.,
|

1	(x−3)	(x−3)2
1	(x−4)	(x−4)2
1	(x−5)	(x−5)2

| = 0
Expanding along C1, we get
(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
⇒ – (x – 4)(x – 5) + 2(x – 3)(x – 5) – (x – 3) (x – 4) = 0
⇒ – x2 + 9x – 20+ 2x2 – 16x + 30 – x2 + 7x – 12 = 0
⇒ – 32 + 30 = 0 ⇒ –2 = 0
Which is not possible, hence no value of x satisfies the given condition