Question

# Column IColumn II(A) If the roots of the equation(P)7x3−9x2+26x−k=0 are positiveand in A.P., then k is equal to(B) If the roots of the equation(Q)11x3−14x2+kx−64=0 are positiveand in G.P., then k is equal to(C) If the roots of the equation(R)246x3−kx2+6x−1=0 are positiveand in H.P., then k is equal to(D)The harmonic mean for the roots of(S)26equation x3−11x2+3x−26=0 is(T)56 Which of the following is the only CORRECT combination?

A
(A)(Q),(B)(T),(C)(S),(D)(R)
B
(A)(R),(B)(S),(C)(Q),(D)(Q)
C
(A)(R),(B)(T),(C)(Q),(D)(S)
D
(A)(Q),(B)(P),(C)(R),(D)(S)

Solution

## The correct option is C (A)→(R),(B)→(T),(C)→(Q),(D)→(S)(A) Let the roots be a−d,a,a+d ⇒a−d+a+a+d=9 ⇒a=3 So 3 is the root of the equation x3−9x2+26x−k=0 ⇒33−9(3)2+26(3)−k=0 ⇒k=24 (A)→(R) (B) Let the roots be ar,a,ar ⇒(ar)(a)(ar)=64 ⇒a=4 So 4 is the root of the equation x3−14x2+k−64=0 ⇒43−14(4)2+k(4)−64=0 ⇒k=56 (B)→(T) (C) f(x)=6x3−kx2+6x−1=0, here roots are in H.P. f(1x)=x3−6x2+kx−6=0, here the roots will be in A.P. Let the roots be a−d,a,a+d ⇒a−d+a+a+d=6 ⇒a=2 So 2 is the root of the equation x3−6x2+kx−6=0 ⇒k=11 ​​​​​​​(C)→(Q) (D) x3−11x2+3x−26=0 Let the roots be α,β,γ H.M.=31α+1β+1γ=3αβγαβ+βγ+γα=26 ​​​​​​​(D)→(S)

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