JEE Main 2026 — Jan 24, Shift 1 — Mathematics solution

JEE Main2026Jan 24, Shift 1Mathematics

If the domain of the function

$f (x) = lo g_{(10 x^{2} - 17 x + 7)} (18 x^{2} - 11 x + 1)$

is $(- \infty, a) \cup (b, c) \cup (d, \infty) - {e}$ , then $90 (a + b + c + d + e)$ equals:

A
$170$
B
$177$
C
$307$
D
$316$

Solution

Three conditions must hold for $lo g_{B} (N)$ :

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

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

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

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

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

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

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

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