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

Correct answer: C. 316 For f(x) = _ (10x^2-17x+7) (18x^2-11x+1) , we need: Base > 0 : 10x^2-17x+7 > 0 10(x-1)(x-7/10) > 0 x 1 Base ≠ 1 : 10x^2-17x+6 = 0 x = 1/2 or x = 6/5 Argument > 0 : 18x^2-11x+1 > 0 18(x-1/2)(x-1/9) > 0 x 1/2 Intersecting conditions 1 and 3: (-∞, 1/9) (1/2, 7/10) (1, ∞) Removing base = 1 points: x = 1/2 is already excluded; x = 6/5 ∈ (1,∞) must be removed. Domain: (-∞, 1/9) (1/2, 7/10) (1, ∞) - \ 6/5\ So a = 1/9,\; b = 1/2,\; c = 7/10,\; d = 1,\; e = 6/5 . 90(a+b+c+d+e) = 90((1)/(9)+(1)/(2)+(7)/(10)+1+(6)/(5)) = 10+45+63+90+108 = 316

JEE Main 202624 JanuaryMorning Shift

Exam: JEE Main
Year: 2026
Date: 24 January
Shift: Morning Shift
Subject: Mathematics
Chapter: Functions

JEE Main 2026 · 24 January · Morning Shift · Mathematics · Functions

If the domain of the function f(x)= _ (10 x^ 2 -17 x+7) (18 x^ 2...

If the domain of the function

f(x)=\log _{\left(10 x^{2}-17 x+7\right)}

\left(18 x^{2}-11 x+1\right)

(-\infty, a) \cup(b, c) \cup(d, \infty)-\{e\}

, then

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

equals:

Options

A170
B307
C316
D177

Correct answer

C. 316

Step-by-step solution

For

f(x) = \log_{(10x^2-17x+7)}(18x^2-11x+1)

, we need:
Base

> 0

10x^2-17x+7 > 0 \Rightarrow 10(x-1)(x-7/10) > 0 \Rightarrow x < 7/10

x > 1

Base

\neq 1

10x^2-17x+6 = 0 \Rightarrow x = 1/2

x = 6/5

Argument

> 0

18x^2-11x+1 > 0 \Rightarrow 18(x-1/2)(x-1/9) > 0 \Rightarrow x < 1/9

x > 1/2

Intersecting conditions 1 and 3:

(-\infty, 1/9) \cup (1/2, 7/10) \cup (1, \infty)

Removing base

= 1

points:

x = 1/2

is already excluded;

x = 6/5 \in (1,\infty)

must be removed.
Domain:

(-\infty, 1/9) \cup (1/2, 7/10) \cup (1, \infty) - \{6/5\}

a = 1/9,\; b = 1/2,\; c = 7/10,\; d = 1,\; e = 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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