MHT CET202522 Apr 2025Evening ShiftMathematicsInverse Trigonometric FunctionsActual
The derivative of ⁻¹ ( 1+x^2 -1 x ) w.r.t. ⁻¹ ( 2 x 1-x^2 1-2 x^2 ) at x=0 is
Options
- A1 8
- B1 4
- C1 2
- D1
Correct answer
B. 1 4
Step-by-step solution
Let u = ⁻¹ ( 1+x^2 -1 x ) and v = ⁻¹ ( 2x 1-x^2 1-2x^2 ) . To find du dv at x=0 , first simplify both expressions using trigonometric substitutions. Substitute x = for u : u = ⁻¹ ( 1+ ^2 -1 ) = ⁻¹ ( -1 ) = ⁻¹ ( 1- ) . Using half-angle identities 1- = 2 ^2( /2) and = 2 ( /2) ( /2) : u = ⁻¹ ( 2 ^2( /2) 2 ( /2) ( /2) ) = ⁻¹ ( ( /2) ) = 2 . Since = ⁻¹x , we have u = 1 2 ⁻¹x , and du dx = 1 2 1 1+x^2 . Now substitute x = for v : v = ⁻¹ ( 2 1- ^2 1-2 ^2 ) = ⁻¹ ( 2 2 ) = ⁻¹ ( 2 2 ) = ⁻¹( 2 ) = 2 . Since = ⁻¹x , we have v