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JEE Mathematics 2026 Question with Solution

If the domain of the function f(x)=log⁡(10x2−17x+7)(18x2−11x+1)f(x)=\log\left(10x^2-17x+7\right)\left(18x^2-11x+1\right) is (−∞,a)∪(b,c)∪(d,∞)−{e}(-\infty,a)\cup(b,c)\cup(d,\infty)-\{e\}, then 90(a+b+c+d+e)90(a+b+c+d+e) equals

  • A

    177177

  • B

    170170

  • C

    307307

  • D

    316316

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: f(x)=log⁡(10x2−17x+7)(18x2−11x+1)f(x)=\log\left(10x^2-17x+7\right)\left(18x^2-11x+1\right)

Find: 90(a+b+c+d+e)90(a+b+c+d+e) from the domain (−∞,a)∪(b,c)∪(d,∞)−{e}(-\infty,a)\cup(b,c)\cup(d,\infty)-\{e\}.

For the logarithm to be defined, its argument must be positive.

(10x2−17x+7)(18x2−11x+1)>0(10x^2-17x+7)(18x^2-11x+1)>0

Now find the roots of each quadratic.

10x2−17x+7=0⇒x=1,  71010x^2-17x+7=0 \Rightarrow x=1,\;\frac{7}{10} 18x2−11x+1=0⇒x=19,  1218x^2-11x+1=0 \Rightarrow x=\frac{1}{9},\;\frac{1}{2}

Arrange the critical points in increasing order:

19<12<710<1\frac{1}{9}<\frac{1}{2}<\frac{7}{10}<1

Using sign analysis, the product is positive in

(−∞,19)∪(12,710)∪(1,∞)(-\infty,\tfrac{1}{9})\cup(\tfrac{1}{2},\tfrac{7}{10})\cup(1,\infty)

Therefore,

a=19,b=12,c=710,d=1a=\frac{1}{9},\quad b=\frac{1}{2},\quad c=\frac{7}{10},\quad d=1

and the excluded value is

e=12e=\frac{1}{2}

Now compute:

a+b+c+d+e=19+12+710+1+12=5930a+b+c+d+e=\frac{1}{9}+\frac{1}{2}+\frac{7}{10}+1+\frac{1}{2}=\frac{59}{30}

Hence,

90(a+b+c+d+e)=90×5930=17790(a+b+c+d+e)=90\times\frac{59}{30}=177

Therefore, the correct option is A.

Sign Chart View

Given: The argument of the logarithm is (10x2−17x+7)(18x2−11x+1)(10x^2-17x+7)(18x^2-11x+1).

Find: The interval endpoints and then the value of 90(a+b+c+d+e)90(a+b+c+d+e).

The zeros of the factors are 19,12,710,1\frac{1}{9},\frac{1}{2},\frac{7}{10},1. These points divide the real line into five intervals.

For a logarithmic function, values where the argument is zero are excluded, and only intervals where the argument is strictly positive belong to the domain. From the sign chart, the valid intervals are

(−∞,19)∪(12,710)∪(1,∞)(-\infty,\tfrac{1}{9})\cup(\tfrac{1}{2},\tfrac{7}{10})\cup(1,\infty)

Matching with (−∞,a)∪(b,c)∪(d,∞)−{e}(-\infty,a)\cup(b,c)\cup(d,\infty)-\{e\} gives

a=19,b=12,c=710,d=1,e=12a=\frac{1}{9},\quad b=\frac{1}{2},\quad c=\frac{7}{10},\quad d=1,\quad e=\frac{1}{2}

So,

90(a+b+c+d+e)=90(19+12+710+1+12)=17790(a+b+c+d+e)=90\left(\frac{1}{9}+\frac{1}{2}+\frac{7}{10}+1+\frac{1}{2}\right)=177

Thus, the correct option is A.

Common mistakes

  • Students often treat the domain condition as (10x2−17x+7)(18x2−11x+1)≥0(10x^2-17x+7)(18x^2-11x+1)\ge 0. That is incorrect because the argument of a logarithm must be strictly positive. Values making the argument zero must be excluded.

  • A common error is to solve each quadratic correctly but arrange the roots in the wrong order. The sign chart depends on the ordered critical points 19<12<710<1\frac{1}{9}<\frac{1}{2}<\frac{7}{10}<1, so misordering them gives wrong intervals.

  • Some students forget to match the obtained domain carefully with (−∞,a)∪(b,c)∪(d,∞)−{e}(-\infty,a)\cup(b,c)\cup(d,\infty)-\{e\}. The interval endpoints give a,b,c,da,b,c,d, while ee is the excluded point written separately in the expression.

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