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Question

List IList II(A)If x2+xa=0 has integral roots(P)2and aN,than a can be equal to(B)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(C)If the equation x2+2bx+9b14=0(R)1has only negative roots, then the integralvalues of b can be(D)If n is 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
(A)(P), (Q), (S)
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B
(A)(P), (Q), (R)
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C
(B)(P), (Q)
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D
(C)(R)
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Solution

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

(B)
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)>0
4c16>0c>4

(C)
Both roots are negative if, D0, the product of the roots must be positive and the sum of the roots must be negative.
2b<0b>04b24(9b14)09b14>0

We get b(149,2][7,)

(D)
|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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