Question

# (a) For real values of x, prove that the value of the expression 11x2+12x+6x2+4x+2 cannot lie between −5 and 3(b) x2+(a−b)x+(1−a−b)=0,a,bϵR. Find the condition on a, for which both roots of the equation are real and unequal.(c) Determine the values of x which satisfy the inequalities x2−3x+2>0 and x2−3x−4≤0.

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Solution

## (a) Let 11x2+12x+6x2+4x+2=y.∴ x2(11−y)+4x(3−y)+2(3−y)=0........(1)For real values of x, B2−4AC of (1) should be ≥0or 16(3−y)2−8(11−y)(3−y)≥0 i.e., +ive 8(3−y)[6−2y−11+y]≥0or 8(3−y)(−5−y)≥0or 8(y−3)(y+5)≥0 i.e. +iveor 8[y−(−5)](y−3) is +iveHence arguing as in part (a), y should not lie between −5 and 3.(b) For roots to be real and unequal D>0 i.e., +ive∴ (a−b)2−4(1−a−b)>0∀bϵRArranging as a quadratic in b,b2−2(a−2)b+(a2+4a−4) is +ive∀bϵR.Its sign will be same as of its first term i.e., +ive provided its Δ is −ive.∴ 4(a2−4a+4)−4(a2+4a−4)<0or −8a+8<0or a−1>0 ∴ a>1.(c) (x−1)(x−2)>0⇒x<1,x>2 (x+1)(x−4)≤0⇒−1≤x≤4.∴ −1≤x<1 and 2<x≤4 for both ∴ xϵ[−1,1]∪[2,4]. Both are semi-open; 1 and 2 are not included.

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