Question

# List IList II (A)If the quadratic equation x2−2αx+2α=0 has two real(P)0and distinct roots x1 and x2 such that |x1–x2|≤2√3, then α can be(B)If y=x2+4xx2+4x+6, then the integers in range of y is/are(Q)1(C)If (2−n)x2−n<8x+4 ∀ x∈R, then n4 can be(R)2(D)If the equation cos2x+2acosx+6a=17 has a solution,(S)3then a can be(T)−1 Which of the following is the only CORRECT combination? (A)→(R),(S)(A)→(S),(T)(B)→(P),(Q)(B)→(Q),(T)

Solution

## The correct option is B (A)→(S),(T)(A)  If the quadratic equation x2−2αx+2α=0 has two real and distinct roots, then  D>0⇒4α2−8α>0⇒4α(α−2)>0⇒α∈(−∞,0)∪(2,∞)   …(i) Also, given that |x1−x2|≤2√3⇒(x1+x2)2−4x1x2≤12⇒4α2−8α≤12⇒(α−3)(α+1)≤0⇒α∈[−1,3]   …(ii) From (i) and (ii), α∈[−1,0)∪(2,3] ∴α can be −1,3 (A)→(S),(T) (B) y=x2+4xx2+4x+6⇒y=1−6(x+2)2+2⇒y∈[−2,1) ∴y=−2,−1,0 (B)→(P),(T) (C) (2−n)x2−n<8x+4 ∀ x∈R ⇒(n−2)x2+8x+(n+4)>0 ⇒n−2>0⇒n>2   …(i) and D<0 ⇒64−4(n−2)(n+4)<0⇒n2+2n−24>0⇒(n+6)(n−4)>0 ⇒n∈(−∞,−6)∪(4,∞)   …(ii) From (i) and (ii), we get ⇒n>4  ⇒n4>1 (C)→(R),(S) (D) cos2x+2acosx+6a=17 ⇒2cos2x+2acosx+6a−18=0⇒2(cosx−3)(cosx+3)+2a(cosx+3)=0 ⇒(cosx+3)[2(cosx−3)+2a]=0 ⇒cosx+3=0⇒cosx=−3 (not possible) or 2(cosx−3)+2a=0 ⇒cosx=3−a We know that, ⇒−1≤cosx≤1⇒−1≤3−a≤1⇒−4≤−a≤−2⇒2≤a≤4 (D)→(R),(S)

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