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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

  1. A1 8
  2. B1 4
  3. C1 2
  4. 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

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