If a, b, p, q, r, s are constants and p(x – a) 2 + q(x – b) 2 = 5x 2 + 8x + 14 r(x – a) 2 + s(x – b) 2 = x 2 + 10x + 7. Then
(i) a + b =
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Ans.
(i)
Sol. Substituting successively x = b, x = a in the given
equations, we get
p(b – a) 2 = 5b 2 + 8b + 14 .....(1)
q(a – b) 2 = 5a 2 + 8a + 14 .....(2)
r(b – a) 2 = b 2 + 10b + 7 ....(3)
s(a – b) 2 = a 2 + 10a + 7 .....(4)
(1), (2) → p + q = 5
(3), (4) → r + s = 1
(1) + (2) → 4(a + b) + 5ab = –14
(3) + (4) → 5(a + b) + ab = –7
Solving, (a, b) = (1, –2) or (–2, 1),
(p, q, r, s) = (2, 3, –1, 2)
a + b = –1
(ii)
Sol. p + r = 2 – 1 = 1
(iii)
Sol. q + s = 3 + 2 = 5
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