99 Percentile Qs Bank for JEE MainMathematicsInverse Trigonometric Functions
If y(x)= ⁻¹ ( 1+a^2 x^2 -1 a x ) and (1+a^2 x^2 ) y^ +g(x) y^ =0 then, the sum of the roots of the equation 1+a^2 x^2+g(x)=0 is
Options
- A2 a
- B-2 a^2
- C2
- D-2
Correct answer
D. -2
Step-by-step solution
Given, y= ⁻¹ ( 1+a^2 x^2 -1 a x ) aligned & put a x= = ⁻¹(a x) & y= ⁻¹ ( -1 ) & y= ⁻¹ ( 1- )= ⁻¹ 2 = 2 & y= 1 2 ⁻¹(a x) y^ = 1 2 a 1+a^2 x^2 & (1+a^2 x^2 ) y^ = a 2 (1+a^2 x^2 ) y^ +2 a^2 x y^ =0 & g(x)=2 a^2 x 1+a^2 x^2+g(x)=0 & a^2 x^2+2 a^2 x+1=0 & Sum of roots = -2 a^2 a^2 =-2 aligned