JEE Main Mathematics Question (2026) — Solution

Step-by-step solution

Let y = y_1 + y_2 y_1 = ^ -1 ( 3 x - 4 x 4 x + 3 x ) Dividing the numerator and denominator by 4 x: y_1 = ^ -1 ( 3 4 - x 1 + 3 4 x ) Using the formula ^ -1 ( A - B 1 + AB ) = ^ -1 A - ^ -1 B: y_1 = ^ -1 ( 3 4 ) - ^ -1 ( x) = ^ -1 ( 3 4 ) - x Differentiating with respect to x: dy_1 dx = -1 Now, y_2 = 2 ^ -1 ( x 1+ 1-x^2 ) Substituting x = : y_2 = 2 ^ -1 ( 1+ ) y_2 = 2 ^ -1 ( 2 ( /2) ( /2) 2 ^2( /2) ) y_2 = 2 ^ -1 ( 2 ) = 2 ( 2 ) = = ^ -1 x Differentiating with respect to x: dy_2 dx = 1 1-x^2 At x = 3 2 : dy_2 dx = 1 1 - ( 3 2 )^2 = 1 1 - 3 4 = 1 1 4 = 2 Therefore, dy dx = dy_1 dx + dy_2 dx = -1 + 2 = 1 Answer: 1