99 Percentile Qs Bank for JEE MainMathematicsInverse Trigonometric Functions
The set of values of such that f: R [0, 2 ) defined by f(x)= ⁻¹ (x^2+x+ ^2 ) is onto is
Options
- A( -1 2 , 1 2 )
- B( -1 4 , 1 4 )
- C(- , -1 2 ) ( 1 2 , )
- D(- , -1 4 ) ( 1 4 , )
Correct answer
C. (- , -1 2 ) ( 1 2 , )
Step-by-step solution
Let A= x: 0 x < 2 Since f: R A is an onto function, therefore, Range of f=A 0 f(x) 2 for all x R 0 ⁻¹ (x^2+x+ ^2 ) 2 for all x R 0 x^2+x+ ^2 for all x R x^2+x+ ^2 0 for all x R 1-4 ^2 0 ^2 1 4 (- ,- 1 2 ) ( 1 2 , )