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L.C.M. of x2 – 1 and x2 – x – 2 will be(A) (x2 – 1)(x – 2)(B) (x2 – 1)(x + 2)(C) (x – 1)2 (x + 2)(D) (x + 1)2 (x – 2) |
Answer» Answer is (A) (x2 – 1)(x – 2) x2 – 1 = (x – 1)(x + 1) …(i) x2 – x – 2 = x2 – 2x + x – 2 = x (x – 2) + 1 (x – 2) = (x + 1)(x – 2) …(ii) from eqn (i) and (ii) (x + 1)(x – 1 )(x – 2) L.C.M. = (x2 – 1)(x – 2) |
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