Question

Solve :

${\log }_5(x^2-11x+43)<2$

• A. $x\in (2,9)$
• B. $x\in [2,9]$
• C. $x\in (-\infty ,2)\cup (9,\infty )$
• D. $x\in (1,9)$

Answer 1

The correct option is A $x\in (2,9)$

For the $\log$ to be defined $x^2-11x+43>0$

coefficient of $x^2$ is greater than $0$

so for $x^2-11x+43>0$

$\Rightarrow {11}^2-\left(4.43\right)<0$ $[\because forequationa{x}^2+bx+ctobe>0\Rightarrow b^2-4ac<0]$

$\Rightarrow 121-172<0$

$\Rightarrow -51<0$

$\Rightarrow x^2-11x+43$ is always then zero

${\log }_5(x^2-11x+43)<2$

$\Rightarrow x^2-11x+43<5^2$

$\Rightarrow x^2-11x+18<0$

$(x-2)(x-9)<0$

(Refer image)

$(\because (x-2)$ and $(x-9)$ has odd Power $\therefore$ sign change at $2,0$ and At $x<2(x-2)(x-9)>0)$

$\Rightarrow x\in (2,9)$