Question

(a) Show that if x is real, the expression $\frac{x^2-bc}{2x-b-c}$ has no real values between b and c.
(b) For real x, the function $\frac{(x-a)(x-b)}{x-c}$ will assume all real values provided
(i) $a>b>c$
(ii) $a<b<c$
(iii) $a>c>b$
(iv) $a<c<b$
(c) If the roots of the equation $a{x}^2+bx+c=0$ are real and of opposite sign then the roots of the equation $\alpha {(x-\beta )}^2+\beta {(x-\alpha )}^2=0$ are

• A. Positive
• B. Negative
• C. Neal and of opposite sign
• D. Imaginary

Answer 1

The correct option is C Neal and of opposite sign

(a) Proceeding as usual, $y^2-y(b+c)+bc\ge 0$ or $(y-b)(y-c)\ge 0$ i.e., $+ive$

Above is possible only when y does not lie between b and c.

(b) Ans (iii) and (iv) are both correct.

Let $y=\frac{(x-4)(x-b)}{x-c}$

or $y(x-c)=x^2+(a+b)x+ab$

or $x^2-(a+b+y)x+ab+cy=0$

$\Delta ={(a+b+y)}^2-4(ab+cy)$

$=y^2+2y(a+b-2c)+{(a-b)}^2$

Since x is real and y assumes all real values, we must have $\Delta \ge 0$ for all real values of y. The sign of a quadratic in y is same as of first term provided its discriminant $B^2-4AC<0$

This will be so if $4{(a+b-2c)}^2-4{(a-b)}^2<0$

or $44(a+b-2c+a-b)(a+b-2c-a+b)<0[P^2-Q^2]$

or $16(a-c)(b-c)<0$

or $16(c-a)(c-b)=-ive$

$\therefore$ c lies between a and b, i.e., $a<c<b$.........(1)

Where $a<b$, but if $b<a$ then the above condition will be

$b<c<a$ or $a>c>b$.........(2)

Hence from (1) and (2) we observe that both (c) and (d) are correct answers.

(c) Ans. (c).

Given $b^2=4ac\ge 0$ and product $=\frac{c}{a}=-ive$

The given equation is

$(\alpha +\beta )x^2-4\alpha \beta x+\alpha \beta (\alpha +\beta )=0$

or $x^2-\frac{4\alpha \beta }{\alpha +\beta }x+\alpha \beta =0$

or $x^2+4\frac{c}{b}x+\frac{c}{a}=0$

$\Delta =16\frac{c^2}{b^2}-4\frac{c}{z}$

$\Delta$ is clearly $+ive$ as $\frac{c}{a}$ is $-ive$. Hence the roots are real and their product being $\frac{c}{a}=-ive$ so that they are of opposite sign. Hence (c) is correct.