Mathematics - Functions Question with Solution | TestHub

(a) The function f(x) is defined if (i) $x^2-5x+10>0$, (ii) $1-{\log }_{10}\left(x^2-5x+10\right)>0$

Now, (ii) $\Rightarrow {\log }_{10}\left(x^2-5x+10\right)<1\Rightarrow x^2-5x+10<10$ $\Rightarrow x^2-5x<0\Rightarrow x(x-5)<0\Rightarrow 0<x<5$..........(A)

Again, $x^2-5x+10>0$ for all x , ............(B)

Since the discriminant of the corresponding equation $x^2-5x+10=0$ is negative, so that the roots of the equation are imaginary. Combining $(A)$ and $(B)$, we find that the domain of $f(x)$ is $(0,5)$. \item[(b)] The function$\mathrm{g}(\mathrm{x})$is defined if

(iii)]${\log }_3\left\{{\log }_4(x-4{)}^2\right\}>0$

(i) is true for all x . ..........(A)

(ii) is true if$(x-4{)}^2>1\Rightarrow x^2-8x+15>0\Rightarrow (x-3)(x-5)>0\Rightarrow x<3$ or $x>5$.....(B)

[(iii)] is true if${\log }_4(x-4{)}^2>1\Rightarrow (x-4{)}^2>4\Rightarrow x^2-8x+12>0$ $\Rightarrow (x-2)(x-6)>0\Rightarrow x<2\text{ or }x>6$...........(C)

Hence combining $(A),(B)$ and $(C)$, we find that the domain of $g(x)$ is $(-\infty ,2)\cup (6,\infty )$.