JEE Main 2026 Jan 24 Shift 1, Mathematics Q6: Introduction Of Functions

Three conditions must hold for ${\log }_B(N)$:

(i) Argument positive: $18x^2-11x+1>0$. Factoring: $(2x-1)(9x-1)>0\Rightarrow x<\frac{1}{9}$ or $x>\frac{1}{2}$.

(ii) Base positive: $10x^2-17x+7>0$. Factoring: $(x-1)(10x-7)>0\Rightarrow x<\frac{7}{10}$ or $x>1$.

(iii) Base $\ne 1$: $10x^2-17x+6\ne 0$. Discriminant $=289-240=49$, roots $\frac{17\pm 7}{20}=\frac{1}{2},\frac{6}{5}$. So $x\ne \frac{1}{2}$ and $x\ne \frac{6}{5}$.

Intersecting (i) and (ii): $x\in \left(-\infty ,\frac{1}{9}\right)\cup \left(\frac{1}{2},\frac{7}{10}\right)\cup (1,\infty )$.

Removing $\frac{1}{2}$ (already excluded) and $\frac{6}{5}$ (lies in $(1,\infty )$): domain $=\left(-\infty ,\frac{1}{9}\right)\cup \left(\frac{1}{2},\frac{7}{10}\right)\cup (1,\infty )-\left\{\frac{6}{5}\right\}$.

So $a=\frac{1}{9}$, $b=\frac{1}{2}$, $c=\frac{7}{10}$, $d=1$, $e=\frac{6}{5}$.

$90(a+b+c+d+e)=90\left(\frac{1}{9}+\frac{1}{2}+\frac{7}{10}+1+\frac{6}{5}\right)=10+45+63+90+108=316$.