JEE Advanced Mathematics Question (2023) — Solution

Step-by-step solution

Given, f : ( 0 , 1 ) → ℝ f x = 4 x x - 1 4 2 x - 1 2 We know that,  4 x  will be discontinuous function at integer points, So,  critical point  = 1 4 , 1 2 , 3 4  as  x ∈ 0 , 1 And function will be discontinuous at x = 3 4  because for  1 2   &   1 4 ,  x - 1 2   &   x - 1 4 2  will make the function continuos Hence, the function is discontinuous exactly at one point in  0 , 1 , so option  A  is correct, Now, f x = 0 ,   0 < x < 1 4 1 · x - 1 4 2 x - 1 2 ,   1 4 ≤ x < 1 2 2 · x - 1 4 2 x - 1 2 ,   1 2 ≤ x < 3 4 3 · x - 1 4 2 x - 1 2 ,   3 4 ≤ x < 1 Now differentiating the above function we get, f ' x = 0 ,   0 < x < 1 4 x - 1 4 2 + 2 x - 1 4 x - 1 2 ,   1 4 ≤ x < 1 2 4 · x - 1 4 x - 1 2 + 2 x - 1 4 2 ,   1 2 ≤ x < 3 4 3 · x - 1 4 2 + 6 x - 1 4 x - 1 2 ,   3 4 ≤ x < 1 Now from above function we get, f ' 1 4 - = f ' 1 4 + = 0  so function is differentiable at  x = 1 4 And  f ' 1 2 - = 1 16 &   f ' 1 2 + = 2 × 1 16 = 1 8  so function is non-differentiable at  x = 1 2 , hence option  B  is correct, Now  f ' 3 4 - = 4 × 1 2 × 1 4 + 2 × 1 4 = 1   &   f ' 3 4 + = 3 × 1 4 + 6 × 1 2 × 1 4 = 3 2 Hence, the function is not differentiable at  x = 3 4 So, function non-differentiable at two points so option  C  is not correct. Now plotting the graph from the above value we get, Now  f x  will be minimum between  1 4 ≤ x < 1 2 So,  f x = x - 1 4 2 x - 1 2 ,   1 4 ≤ x < 1 2 ⇒ f ' x = x - 1 4 2 + 2 x - 1 4 x - 1 2 = x - 1 4 3 x - 5 4 Now critical points will be  f ' x = 0 ⇒ x = 1 4   or   x = 5 12 Now putting the value in first derivative we get, f ' 5 12 - < 0   &   f ' 5 12 + > 0 Hence,  x = 5 12  is point of minima, So,  f 5 12 m i n = 5 12 - 1 4 2 5 12 - 1 2 = - 1 432 Hence, option  D  is wrong.