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3) One maximum and one minimumThe function \( f(x)=(x-1)^{2}+2(x-2)^{2}+3(x-3)^{2}+\ldots \ldots+n(x-n)^{2} \) has minimum at \( x= \)1) \( \frac{n+1}{3} \)2) \( \frac{2 n+1}{3} \)3) \( \frac{n-1}{3} \)4) \( \frac{2 n-1}{3} \) |
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Answer» f(x) = (x-1)2 + 2(x-2)2 + 3(x-3)2 + ......+n(x-n)2 ∴f'(x) = 2(x-1) + 4(x-2)+6(x-3)+ ....+ 2n (x-n) = 2x(1 + 2 + 3 + ....+ n) -2 (12 + 22 + 32 + .... + n2) = \(\frac{2x\times n(n+1)}{2}-2\times n\frac{n(n+1)(2n+1)}{6}\) = n(n + 1)x - \(\frac{n(n+1)(2n+1)}{3}\) = \(\frac{n(n+1)}{3}\) (3x -(2n + 1)), n>1 ∴ f''(x) = n(n+1)>0 ∴ point where f'(x) = 0 is point of minima. f'(x) = 0 gives x = \(\frac{2n+1}{3}\) Hence, x = \(\frac{2n+1}{3}\) is point of minima of given function f(x), Hence, option (2) is correct. |
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