Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
BITSAT2024MathematicsFunctionsActual

The domain of the real valued function f(x)= 2 x^2-7 x+5 3 x^2-5 x-2 is

Options

  1. A(- ,- 1 3 ) [1,2) [ 5 2 , )
  2. B(- , 1) (2, )
  3. C(- 1 3 , 5 2 ]
  4. D(- , -1 3 ] [ 5 2 , )

Correct answer

A. (- ,- 1 3 ) [1,2) [ 5 2 , )

Step-by-step solution

Given function f(x)= 2 x^2-7 x+5 3 x^2-5 x-2 Here, f ( x ) should be greater than or equal to 0 . So, 2 x^2-7 x+5=0 2 x^2-5 x-2 x+5=0 (x-1)(2 x-5)=0 x=1, 5 2 3 x^2-5 x-2=0 3 x^2-6 x+x-2=03 x(x-2)+1(x-2)=0 (x-2)(3 x+1)=0 x=2, -1 3 When we include x= -1 3 , 2 then f(x) would give not define value so, we will exclude these values from the domain. When we take values between (- 1 3 , 1 ) and So, domain is (- , -1 3 ) [1,2) [ 5 2 , )

Practice Functions on Quantrex Academy →

More from Functions

Let [x] denote the greatest integer x . If f(x)=[x] and g(x)=|x| , then the value of f (g ( 8 5 ) )-g (f (- 8 5 ) ) is 2024Number of real solution of 5- ₂|x| =3- ₂|x| is equal to 2024f(x)= x [ 2 x ]+ 1 2 , where x is not an integral multiple of and [.] denotes the greatest integer function, is 2024The function f: R R defined by f(x)= x 1+x^2 is 2024If f: R R, g: R R are defined by f(x)=5 x-3, g(x)=x^2+3 , then g o f⁻¹(3) is equal to 2024If a function f : R - l R - m defined by f(x)= x+3 x-2 is a bijection, then 3 l+2 m= 2024f(x)= x+ x, g(x)=x^2-1 , then g(f(x)) is invertible if 2024Let the function g:(- , ) (- 2 , 2 ) be given by g(u)=2 ⁻¹ (e^u )- 2 . Then, g is 2024 Full Functions list All BITSAT PYQs