Question

The number of solutions of the equation ${|x+1|}^{{\log }_{x+1}(3+2x-x^2)}=(x-3)\left|x\right|$ is/are

• A. only one
• B. two
• C. no solution
• D. more than two

Answer 1

Answer: (C)

no solution

By the definition of ${\log }_{a}b;a,b>0$

Thus, $x+1>0$

$\therefore |x+1|=x+1$ ...(1)
Also, $3+2x-x^2>0$

$\Rightarrow x^2-2x-3<0$

or $(x-3)(x+1)<0$

$\therefore -1<x<3$

or $-1<x<0$ and $0<x<3$ ...(2)

Hence, the given equation by (1) reduces to

$3+2x-x^2=(x-3)\left|x\right|$

[$\because a^{{\log }_{a}x}=x$]

Now, $\left|x\right|=-x$ if $x<0$, $\left|x\right|=x$ when $x>0$

When $-1<x<0$

then $3+2x-x^2=(x-3)(-x)$

or $3+2x-x^2=-x^2+3x$

$\therefore x=3\notin [-1,0]$
When $0<x<3,$ then by (2)

$3+2x-x^2=(x-3)x=x^2-3x$

or $2x^2-5x-3=0\therefore x=-\frac{1}{2},3$

These values do not lie in $0<x<3.$

Hence, the equation has no solution.
Ans: C