Aspire Faculty ID #18041 · Topic: JEE Main 2026 (24 January Morning Shift) · Just now
JEE Main 2026 (24 January Morning Shift)

If the domain of the function $f(x)=\log_{(10x^2-17x+7)}(18x^2-11x+1)$ is $(-\infty,a)\cup(b,c)\cup(d,\infty)-{e}$, then $90(a+b+c+d+e)$ equals:

Solution

For logarithm:

$18x^2 - 11x + 1 > 0$

$(2x-1)(9x-1)>0$

$x<\frac{1}{9}$ or $x>\frac{1}{2}$

Also base condition:

$10x^2 - 17x + 7 > 0$

$(x-1)(10x-7)>0$

$x<\frac{7}{10}$ or $x>1$

Also

$10x^2 -17x +7 \ne 1$

$\Rightarrow x \ne \frac{6}{5}$

Combine intervals

$x\in(-\infty,\frac{1}{9})\cup(\frac{1}{2},\frac{7}{10})\cup(1,\infty)\setminus\left{\frac{6}{5}\right}$

So

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

$90(a+b+c+d+e)$

$=90\left(\frac{1}{9}+\frac{1}{2}+\frac{7}{10}+1+\frac{6}{5}\right)$

$=10+45+63+90+108=316$

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