Question

Let $f\left(x\right)=x^2+\frac{x^2}{1+x^2}+\frac{x^2}{(1+x^2{)}^2}+\cdots$ and $y=f(1-x)$ is defined for $1\le x\le 2$ then which of the following(s) is(are) correct

• A. $y=x$ has exactly two real solutions in $[1,2]$
• B. $y=1-x$ has atleast one real solution in $[1,2]$
• C. $y=x+1$ has atleast one real solution in $[1,2]$
• D. $y=\sqrt{x}$ has atleast one real solution in $[1,2]$

Answer 1

Answer: (A), (D)

$y=\sqrt{x}$ has atleast one real solution in $[1,2]$

$f\left(x\right)=\frac{x^2}{1-\frac{1}{1+x^2}}=1+x^2$
$\therefore y=f(1-x)=x^2-2x+2$
when $x\in [1,2]$ Range of $y$ is $[1,2]$
So, from $I.V.T.:y=x$ and $y=\sqrt{x}$,
have atleast one solution in $[1,2]$.

Also, for $y=x\Rightarrow x^2-2x+2=x$
$\Rightarrow x^2-3x+2=0$
$\Rightarrow x=\frac{3\pm 1}{2}=1,2$
exactly two solutions in $[1,2]$
Now, for $y=1-x\Rightarrow x^2-x+1=0,D<0$(no solution)
and for $y=1+x\Rightarrow x^2-3x+1=0$
$\Rightarrow x=\frac{3\pm \sqrt{5}}{2}\notin [1,2]$
So, no solution in $[1,2]$