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If y(x) = vmatrix x & x & x+ x+1 \ 27 & 28 & 27 \ 1 & 1 & 1 vmatrix, x R, then d2ydx2 + y is equal to
(3) -1
Applying C3 C3 - C1 - C2:
y = vmatrix x & x & 1 \ 27 & 28 & -28 \ 1 & 1 & -1 vmatrix
Expanding along R1:
y = x(-28 + 28) - x(-27 + 28) + 1(27 - 28)
y = - x - 1
d2ydx2 = x
d2ydx2 + y = x - x - 1 = -1
Answer: (3)
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