JEE Main · 2026 · 24 Jan 2026 · Shift 1 · Sets Relations and Functions

The logarithm requires a positive argument, a positive base, and a base different from $1$. The argument $18x^2-11x+1=(2x-1)(9x-1)>0$ gives $x<\frac{1}{9}$ or $x>\frac{1}{2}$. The base $10x^2-17x+7=(x-1)(10x-7)>0$ gives $x<\frac{7}{10}$ or $x>1$. The base equals $1$ when $10x^2-17x+6=0$, i.e. $x=\frac{1}{2}$ or $x=\frac{6}{5}$, and $x=\frac{1}{2}$ is already excluded. Intersecting the two inequalities and removing $x=\frac{6}{5}$ gives the domain $\left(-\infty ,\frac{1}{9}\right)\cup \left(\frac{1}{2},\frac{7}{10}\right)\cup (1,\infty )-\left\{\frac{6}{5}\right\}$. Thus $a=\frac{1}{9},b=\frac{1}{2},c=\frac{7}{10},d=1,e=\frac{6}{5}$, and $90(a+b+c+d+e)=10+45+63+90+108=316$.