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List IList II (A)If the quadratic equation x22αx+2α=0 has two real(P)0and distinct roots x1 and x2 such that |x1x2|23, then α can be(B)If y=x2+4xx2+4x+6, then the integers in range of y is/are(Q)1(C)If (2n)x2n<8x+4  xR, 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? 
  1. (A)(R),(S)
  2. (A)(S),(T)
  3. (B)(P),(Q)
  4. (B)(Q),(T)


Solution

The correct option is B (A)(S),(T)
(A)  If the quadratic equation x22αx+2α=0 has two real and distinct roots, then 
D>04α28α>04α(α2)>0α(,0)(2,)   (i)

Also, given that
|x1x2|23(x1+x2)24x1x2124α28α12(α3)(α+1)0α[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+6y=16(x+2)2+2y[2,1)
y=2,1,0
(B)(P),(T)

(C) (2n)x2n<8x+4  xR
(n2)x2+8x+(n+4)>0
n2>0n>2   (i)
and D<0
644(n2)(n+4)<0n2+2n24>0(n+6)(n4)>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+6a18=02(cosx3)(cosx+3)+2a(cosx+3)=0
(cosx+3)[2(cosx3)+2a]=0
cosx+3=0cosx=3 (not possible)
or 2(cosx3)+2a=0
cosx=3a

We know that,
1cosx113a14a22a4
(D)(R),(S) 
 

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