Question

Number of prime numbers satisfying the inequality ${\log }_3\frac{|x^2-4x|+3}{x^2+|x-5|}\ge 0$ is equal to

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

Answer: (A)

$1$

We have,

${\log }_3\frac{|x^2-4x|+3}{x^2+|x-5|}\ge 0$

$\frac{|x^2-4x|+3}{x^2+|x-5|}\ge 1$

$|x^2-4x|+3\ge x^2+|x-5|$

If $0<x<4$, then

$-x^2+4x+3\ge x^2-x+5$

$(2x-1)(x-2)\le 0$

$\frac{1}{2}<x<2⟹x=1$ is satisfying if x is prime

If $4<x<5$, then

$x^2-4x+3\ge x^2-x+5$

$3x+2\le 0⟹x\le \frac{-2}{3}$ which is not possible

If $x>5$, then

$x^2-4x+3\ge x^2+x-55x-8\le 0⟹x\le \frac{8}{5}$

Which is also not possible as $x\ge 5$

Therefore $x=1$ is the only satisfying prime value for $x$.