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# If x is real, find the maximum and minimum values of x2+14x+9x2+2x+3.(b) Prove that for any real values of x the expression x+ax2+bx+c2 will have any value if b2>4c2 and a2+c2<ab.

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Solution

## (a) Let x2+14x+9x2+2x+3=y∴ x2(1−y)+2x(7−y)+3(3−y)=0........(1)For real values of x, B2−4AC of (1) should be ≥0∴ 4(7−y)2−12(1−y)(3−y)≥0or (40−14y+y2)−3(3−4y+y2)≥0or −2y2−2y+40≥0or −2(y2+y−20)≥0or (y+5)(y−4) is ≤0or [y−(−5)](y−4) is −ive.By §5 y should lie between −5 and 4.Therefore the maximum value is 4 and minimum is −5.(b) y=x+ax2+bx+c2∴ yx2+x(by−1)+(c2y−a)=0Since x is real ∴Δ≥0∴ (by−1)2−4y(c2y−a)≥0or y2(b2−4c2)+2y(2a−b)+1=+ive.........(1)The sign of a quadratic expression is same as of its first terms i.e. of b2−4c2 provided its discriminant D is −iveor 4(2a−b)2−4(b2−4c2)=−iveor 16(a2+c2−ab)=−ive or a2+c2<abHence the sign of expression (1) is same as of b2−4c2 i.e. +ive. Hence y will have any value.

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