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Let a, b, c, p, q be the real numbers....

Let `a, b, c, p, q` be the real numbers. Suppose `alpha,beta` are the roots of the equation `x^2+2px+ q=0`. and `alpha,1/beta` are the roots of the equation `ax^2+2 bx+ c=0`, where `beta !in {-1,0,1}`. Statement I `(p^2-q) (b^2-ac)>=0` Statement 11 `b !in pa` or `c !in qa`.

A

Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-1.

B

Statement-1 is True, Statement-2 is True, Statement-2 is not a correct explanation for Statement-1.

C

Statement-1 is True, Statement-2 is False.

D

Statement-1 is False, Statement-2 is True.

Text Solution

Verified by Experts

The correct Answer is:
B
Since `alpha, beta` are the roots of the equation `x^(2) + 2px + q = 0 and alpha, (1)/(beta)` are the roots of the equation `ax^(2) + 2 bx + c = 0`.
`alpha + beta = - 2p, alpha beta = q, alpha+(1)/(beta) = -(2b)/(a) and (alpha)/(beta) = (c)/(a)`
`therefore" "(p^(2)-q)(b^(2)-ac)={((alpha+beta)/(-2))^(2)-alpha beta}" "{((a+(1)/(beta))/(-2))^(2)-(alpha)/(beta)}a^(2)`
`rArr" "(p^(2)-q)(b^(2)-ac)=((alpha-beta)^(2))/(4)xx(1)/(4)(alpha-(1)/(beta))^(2)a^(2) ge 0`
So, statement -1 is correct.
If `beta = 1`, then `alpha + beta = - 2 p rArr alpha + 1 = - 2p and, alpha+(1)/(beta) = - (2b)/(a) rArr alpha + 1 = - (2b)/(a)`
`therefore" "-2p = - (2b)/(a) rArr b = pa`
But, `beta ne 1` (given). Therefore, `b ne pa`.
Also, `beta = 1`
`rArr" "alpha = q and alpha alpha = (c)/(a)" "["Putting" beta=1 "in" alpha beta = q and (alpha)/(beta)=(c)/(a)]`
`rArr" "q = (c)/(a) or, c = aq`
But, `beta ne 1 "So", c ne qa`.
So, statement -2 is true. But, it is not a correct explanation for statement-1.
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