Question

Let $f\left(x\right)=(2x-a)(2x-c)+(2x-b)(2x-d)$, where $a<b<c<d$, then

• A. $f\left(x\right)=0$ has atleast one real solution in $(\frac{a}{2},\frac{b}{2})$
• B. $f\left(x\right)=0$ has no real solution in $(\frac{a}{2},\frac{b}{2})$
• C. $f\left(x\right)=0$ has atleast one real solution in $(\frac{b}{2},\frac{c}{2})$
• D. $f\left(x\right)=0$ has no real solution in $(\frac{c}{2},\frac{d}{2})$

Answer 1

The correct option is A $f\left(x\right)=0$ has atleast one real solution in $(\frac{a}{2},\frac{b}{2})$

Given : $f\left(x\right)=(2x-a)(2x-c)+(2x-b)(2x-d)$,where $a<b<c<d$
Since, $f\left(x\right)$ is a polynomial. So, it is continuous in $R$.
Now, $f\left(\frac{a}{2}\right)=(a-b)(a-d)>0,$
$f\left(\frac{b}{2}\right)=(b-a)(b-c)<0$,
$f\left(\frac{c}{2}\right)=(c-b)(c-d)<0$ and
$f\left(\frac{d}{2}\right)=(d-a)(d-c)>0$
From $I.V.T.$ if there is change in sign for continuous function $f\left(x\right)$ in $[a,b]$, then $f\left(x\right)=0$ have atleast one solution in $(a,b).$
$\therefore f\left(x\right)=0$ have atleast one solution in $(\frac{a}{2},\frac{b}{2})$ and in $(\frac{c}{2},\frac{d}{2}).$

Since, it is a quadratic equation, then it has only two roots.
$\therefore$ One root in $(\frac{a}{2},\frac{b}{2})$ and another root in $(\frac{c}{2},\frac{d}{2}).$