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Zero of 3y3 + 8y2 – 1 is(A) 1(B) (-1)(C) 0(D) None of these |
Answer» Answer is (D) None of these Let p(y) = 3y3 + 8y2 – 1 p(1) = 3 × (1)3 + 8(1)2 – 1 = 3 + 8 – 1 = 10 ≠ 0 p(-1) = 3(-1)3 + 8(-1)2 – 1 = -3 + 8 – 1 = 4 ≠ 0 and p(0) = 3 × (0)3 + 8(0)2 – 1 = 0 + 0 – 3 = -3 ≠ 0 Thus by putting y = 1, -1 the expression obtained is not equal to zero. |
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