JEE Main Mathematics Question (2026) — Solution

Step-by-step solution

Let f(x) = g(x) + h(x), where g(x) = \ 6x,\ 2 + 3x^2\ h(x) = |x - 1| | |x^2 - 1 4 | | We find the points of non-differentiability of each part in x (- , ). Non-differentiability of g(x): g(x) = \ 6x,\ 2 + 3x^2\ is non-differentiable where the two curves intersect with different slopes. 6x = 2 + 3x^2 3x^2 - 6x + 2 = 0 x = 6 36 - 24 6 = 6 2 3 6 = 1 1 3 Both values lie in (- , ). At these points, the derivatives of the two inner functions are 6 and 6x; since x 1, the slopes differ. So g(x) contributes 2 points. Non-differentiability of h(x): Since |y| = y: h(x) = |x - 1| | (x^2 - 1 4 ) | An expression of the form |u(x)| is non-differentiable at simple zeros of u(x). Case A: x - 1 = 0 x = 1 At x = 1: (1 - 1 4 ) = 3 4 0, so x = 1 is a point of non-differentiability. Contribution: 1 point. Case B: (x^2 - 1 4 ) = 0 x^2 - 1 4 = (2n + 1) 2 x^2 = 1 4 + (2n + 1) 2 Domain restriction: x (- , ) x^2 [0, ^2), and ^2 9.87. n = -1: x^2 = 0.25 - 2 n = 0: x^2 1.82 n = 1: x^2 4.96 n = 2: x^2 8.10 n = 3: x^2 11.25 > 9.87 (rejected) Contribution: 2 + 2 + 2 = 6 points. Checking for overlap: The points x = 1 1 3 are not equal to 1, and for these, x^2 - 1 4 = 13 12 2 3 , which are not odd multiples of 2 . So the non-differentiable points of g(x) and h(x) are disjoint. Total number of points of non-differentiability: 2 + 1 + 6 = 9 Hence, the answer is 9.