JEE Main Mathematics Question (2026) — Solution

Step-by-step solution

For the function f(x) = _ 0.6 ( | 2x-5 x^2-4 | ) to be defined, two conditions must be satisfied: 1. The argument of the logarithm must be strictly positive: | 2x-5 x^2-4 | > 0 2x - 5 0 x 5 2 Also, the denominator cannot be zero: x^2 - 4 0 x 2 2. The expression inside the square root must be non-negative: _ 0.6 ( | 2x-5 x^2-4 | ) 0 Since the base of the logarithm is 0.6 | 2x-5 x^2-4 | (0.6)^0 | 2x-5 x^2-4 | 1 Since both sides are non-negative, we can square both sides: (2x-5)^2 (x^2-4)^2 (x^2-4)^2 - (2x-5)^2 0 Using the identity A^2 - B^2 = (A-B)(A+B): (x^2 - 4 - 2x + 5)(x^2 - 4 + 2x - 5) 0 (x^2 - 2x + 1)(x^2 + 2x - 9) 0 (x - 1)^2 (x^2 + 2x - 9) 0 Since (x - 1)^2 0 for all real x, the inequality holds if: x - 1 = 0 x = 1 or x^2 + 2x - 9 0 Finding the roots of x^2 + 2x - 9 = 0 using the quadratic formula: x = -2 4 - 4(1)(-9) 2 = -2 40 2 = -1 10 Thus, x^2 + 2x - 9 0 gives: x (- , -1 - 10 ] [-1 + 10 , ) Combining this with x = 1, the solution to the inequality is: x (- , -1 - 10 ] \ 1\ [-1 + 10 , ) Now, we must exclude the restricted values x = 2 and x = 5 2 . Note that -1 - 10 -4.16 and -1 + 10 2.16. The values -2 and 2 do not fall in the above intervals, so they are already excluded. However, 5 2 = 2.5, which lies in the interval [-1 + 10 , ). We must exclude it by splitting the interval: [-1 + 10 , 5 2 ) ( 5 2 , ) The final domain of the function is: (- , -1 - 10 ] \ 1\ [-1 + 10 , 5 2 ) ( 5 2 , ) Comparing this with the given domain (- , a] \ b\ [c, d) (e, ), we get: a = -1 - 10 b = 1 c = -1 + 10 d = 5 2 e = 5 2 We need to find the value of a + b + c + d + e: a + b + c + d + e = (-1 - 10 ) + 1 + (-1 + 10 ) + 5 2 + 5 2 = -1 - 10 + 1 - 1 + 10 + 5 = 4 Answer: 4