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α,β,γ are roots of 2x3+x2-7=0 then the value of ∑αβ+βα is -3

For all real x,y the minimum value of the expression x2+2xy+3y2-6x-2y cannot be less than -11

If ab.c are real and satisfy 2a+3b+6c=0 then the equation ax2+bx+c=0 has real roots

There exists two distinct real values of ‘m’ for which the expression 2x2+mxy+3y2-5y-2 can be expressed as product of linear factors

answer is A.

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Detailed Solution

A) α+β+γ=-12αβ+βγ+γα=0αβγ=72αβγ1α+1β+1γ=0⇒1α+1β+1γ=0 Now (α+β+γ)1α+1β+1γ=1+αβ+αγ+βα+1+βγ+γα+γβ+1-12(0)=3+αβ+βα+αγ+γα+γβ+βγ∑αβ+βα=-3 B) x2+2xy+3y2-6x-2y≥kx2+x(2y-6)+3y2-2y-k≥0Δ≤0⇒(2y-6)2-4(1)3y2-2y-k≤0⇒4(y-3)2-43y2-2y-k≤0⇒y2-6y+9-3y2+2y+k≤0-2y2-4y+(9+k)=02y2+4y-(9+k)=0y∈R⇒Δ≤0⇒42-4(2)(-9(9+k))≤0⇒8k+88≤0⇒k≤-11 C) Consider f(x)=2ax3+3bx2+6cxf(0)=0, f(1)=0f(x)=6ax2+6bx+6c=6ax2+bx+c By Rolle's theorem f'(x) has a root between (0,1) D) 2x2+mxy+3y2-5y-2=(dx+by+c)(px+2y+r)⇒m2=49⇒m=7,-7

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