Question

The equation ${|x+1|}^{{\log }_{(x+1)}(3+2x-x^2)}=(x-3)\left|x\right|$ has

• A. Unique
• B. Two solution
• C. No solution
• D. More that two

Answer 1

Answer: (C)

No solution

For the domain of the function
$3+2x-x^2>0$
Or
$x^2-2x-3<0$
$(x-1{)}^2-4<0$
$(x-1{)}^2<4$
$-2<x-1<2$
$-1<x<3$
And
$x+1>0$
Or
$x>-1$
Hence from the above the domain of the function is $(-1,3)$.
Now for $xϵ(-1,0)$
$RHS$
$=(x-3)(-x)$
$=(3-x)\left(x\right)$
Now RHS is negative but LHS is positive (exponential function).
Hence no solution in $(-1,0)$.
Now for $0<x<3$
$RHS$
$=x(x-3)$
Which is again negative, while RHS is positive.
Hence again no solution in $(0,3)$.
Hence there exists no real solution for the above equation.