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Question

(I) If x2+xa=0 has integral roots(P)2and aN,then a can be equal to(II) If the equation ax2+2bx+4c=16(Q)12has no real roots and a+c>b+4,then the integral value of c can be(III) If equation x2+2bx+9b14=0(R)1has only negative roots, then the integralvalues of b can be(IV) If N be the number of solutions of(S)30the equation |x|4x||2x=4, thenthe value of N is

Which of the following is the only CORRECT combination?

A
(I)(P),(Q),(S)
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B
(I)(P),(Q),(R)
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C
(II)(P),(Q)
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D
(III)(R)
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Solution

The correct option is A (I)(P),(Q),(S)
(I) D=1+4a
1+4a=(2λ+1)2
a=λ(λ+1),λI+

(II) Let f(x)=ax2+2bx+4c16
f(2)=4a4b+4c16=4(ab+c4)>0
As the given equation has no real roots.
f(x)>0 xR
f(0)>04c16>0c>4

(III) both roots are negative if, D0, the products of the roots must be positive & the sum of the roots must be negative
2b<0b>04b24(9b14)09b14>0
we get b(149,2][7,)

(IV) |x|x4||=2x+4
(i) x>4
4=2x+4x=0, which is not possible
(ii)x4
|x+x4|=2x+42|x2|=2x+4
(a)2x4
2x4=2x+4, which is not possible
(b)x2
2x+4=2x+4x=0
So, the equation has only one solution.

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