If x is real,the function f(x) = frac{(x - a) | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Let $y = \frac{(x - a)(x - b)}{(x - c)}$.Then $y(x - c) = x^2 - (a + b)x + ab$,which rearranges to $x^2 - (a + b + y)x + (ab + cy) = 0$.For $x$ to

Vedclass · Accepted Answer

$a < c < b$

Vedclass · Answer

(D) Let $y = \frac{(x - a)(x - b)}{(x - c)}$.Then $y(x - c) = x^2 - (a + b)x + ab$,which rearranges to $x^2 - (a + b + y)x + (ab + cy) = 0$.For $x$ to be real,the discriminant $D$ must be $\ge 0$:$D = (a + b + y)^2 - 4(ab + cy) \ge 0$.Expanding this,we get $y^2 + 2y(a + b - 2c) + (a - b)^2 \ge 0$.For the function to assume all real values $y$,this quadratic in $y$ must be non-negative for all $y$. However,since the coefficient of $y^2$ is positive,this quadratic will take all real values if its discriminant $D_y $D_y = [2(a + b - 2c)]^2 - 4(a - b)^2 $4(a + b - 2c)^2 - 4(a - b)^2 $(a + b - 2c)^2 - (a - b)^2 Using $A^2 - B^2 = (A-B)(A+B)$:$(a + b - 2c - a + b)(a + b - 2c + a - b) $(2b - 2c)(2a - 2c) $4(b - c)(a - c) $(c - b)(c - a) This inequality holds if $c$ lies between $a$ and $b$,i.e.,$a < c < b$ or $b < c < a$.

If $x$ is real,the function $f(x) = \frac{(x - a)(x - b)}{(x - c)}$ will assume all real values,provided

Similar Questions

If $\frac{5}{2}$ is the sum of two roots of the equation $6x^6-25x^5+31x^4-31x^2+25x-6=0$,then the sum of all non-real roots of the equation is

Let $a, b, c, d$ be real numbers between $-5$ and $5$ such that $|a|=\sqrt{4-\sqrt{5-a}}$,$|b|=\sqrt{4+\sqrt{5-b}}$,$|c|=\sqrt{4-\sqrt{5+c}}$,and $|d|=\sqrt{4+\sqrt{5+d}}$. Then,the product $abcd$ is

Let $a \in \mathbb{R}$ and let $\alpha, \beta$ be the roots of the equation $x^2+60^{\frac{1}{4}} x+a=0$. If $\alpha^4+\beta^4=-30$,then the product of all possible values of $a$ is $......$

If the quadratic equation formed by eliminating $x$ from $x^2+\alpha x+\beta=0$ and $xy+l(x+y)+m=0$ has the same roots as that of the given quadratic equation,then the set of values of $\beta$ is

If $\tan \alpha$ equals the integral solution of the inequality $4x^2 - 16x + 15 < 0$ and $\cos \beta$ equals the slope of the bisector of the first quadrant,then $\sin(\alpha + \beta)\sin(\alpha - \beta)$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module