MHT CET202526 Apr 2025Evening ShiftMathematicsInverse Trigonometric FunctionsActual
If y= ⁻¹ ( 1 1+x+x^2 )+ ⁻¹ ( 1 x^2+3 x+3 )+ ⁻¹ ( 1 x^2+5 x+7 ) then the value of y ^ (0) is
Options
- A9 10
- B1 10
- C- 9 10
- D- 1 10
Correct answer
C. - 9 10
Step-by-step solution
Given: y = ⁻¹ ( 1 1+x+x^2 ) + ⁻¹ ( 1 x^2+3x+3 ) + ⁻¹ ( 1 x^2+5x+7 ) Each term simplifies using the identity ⁻¹(A) - ⁻¹(B) = ⁻¹ ( A-B 1+AB ) : ⁻¹ ( 1 1+x+x^2 ) = ⁻¹(x+1) - ⁻¹(x) ⁻¹ ( 1 x^2+3x+3 ) = ⁻¹(x+2) - ⁻¹(x+1) ⁻¹ ( 1 x^2+5x+7 ) = ⁻¹(x+3) - ⁻¹(x+2) Substituting simplifies y to a telescoping sum: y = ⁻¹(x+3) - ⁻¹(x) Differentiate: y'(x) = 1 1+(x+3)^2 - 1 1+x^2 Evaluate at x = 0 : y'(0) = 1 10 - 1 = - 9 10 Result: y'(0) = - 9 10 The correct choice is C