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Question
If Rolle's theorem holds for the function f(x) = x3 + px2 + qx + 5, x ∈ [1, 3] with c = `2 + (1)/sqrt(3)`, find the values of p and q.
Solution
The Rol's theorem hold for the function f(x) = x3 + px2 + qx + 5, x ∈ [1, 3]
∴ f(1) = f(3)
∴ 13 + p(1)2 + q(1) + 5 = 33 + p(3)2 + q(3) + 5
∴ 1 + p + q + 5 = 27 + 9p + 3q + 5
∴ 8p + 2q = – 26
∴ 4p + q = – 13 ...(1)
Also, there exists at least one point c ∈ (1, 3) such that (c) = 0.
Now, f'(x) = `d/dx(x^3 + px^2 + qx + 5)`
= 3x2 + p x 2x + q x 1 + 0
= 3x2 + 2px + q
∴ f'(c) = 3c2 + 2pc + q, where c = `2 + (1)/sqrt(3)`
∴ f'(c) = `3(2 + 1/sqrt(3))^2 + 2p(2 + 1/sqrt(3)) + q`
= `3(4 + 4/sqrt(3) + 1/3) + 4p + (2p)/sqrt(3) + q`
= `12 + 12/sqrt(3) + 1 + 4p + (2p)/sqrt(3) + q`
= `4p + (2p)/sqrt(3) + q + 13 + (12)/sqrt(3)`
But f'(c) = 0
∴ `4p + (2p)/sqrt(3) + q + 13 + (12)/sqrt(3)` = 0
∴ `(4sqrt(3) + 2)p + sqrt(3)q + (13sqrt(3) + 12)` = 0
∴ `(4sqrt(3) + 2)p + sqrt(3)q = -13sqrt(3) - 12` ...(2)
Multiplying equation (1) by `sqrt(3)`, we get
`4sqrt(3)p + sqrt(3)q = -13sqrt(3)`
Subtracting this equation from (2), we get
2p = – 12
∴ p = – 6
∴ from (1), 4(– 6) + q = – 13
∴ q = 11
Hence, p = – 6 and q = 11.
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