Show that the function f:R → R defined by f(x) = is neither one-one nor onto. Also,
if g:R → R is defined as g(x) = 2x – 1, find fog(x).
if g:R → R is defined as g(x) = 2x – 1, find fog(x).
![Question 2 Answer-Image](/wp-content/uploads/images/12_Maths/Relations And Functions_536_1.png)
![Question 3 Answer-Image](/wp-content/uploads/images/12_Maths/Relations And Functions_536_2.png)
![Question 4 Answer-Image](/wp-content/uploads/images/12_Maths/Relations And Functions_536_3.png)
![Question 5 Answer-Image](/wp-content/uploads/images/12_Maths/Relations And Functions_536_4.png)
Show that the function f : R {x ∈ R – 1 < x < 1} defined by f(x) = x ∈ R is one-one and onto function. Hence find f⁻¹(x).
If the function f : R → R is given by f(x) = and g : R → R is given by g(x) = 2x – 3, then find
(i) fog (ii) gof
Is f⁻¹ = g?
If the function f : R → R is given by f(x) = x² + 3x + 1 and g : R → R is given by g(x) = 2x – 3, then find
(i) fog (ii) gof
If the function f : R → R be defined by f(x) = 2x – 3 and g : R → R by g(x) = x³ + 5, then find fog and show that fog is invertible. Also, find (fog)⁻¹, hence find (fog)⁻¹(9).
Prove that the function f : [0, ∞) → R given by f(x) = 9x² + 6x – 5 is not invertible. Modify the Codomain of the function f to make it invertible, and hence find f⁻¹.
If the function f : R → R be given by f(x) = x² + 2 and g : R → R be given by g(x) = find fog and gof and hence find fog(2) and gof(– 3).
Show that the function f : R {x ∈ R – 1 < x < 1} defined by f(x) = x ∈ R is one-one and onto function. Hence find f⁻¹(x).
Find the equation of the tangent line to the curve y =x² -2x + 7 which is
(i) parallel to the line 2x-y +9 =0
(ii) perpendicular to the line 5y - 15x = 13.
LetA = {x ∈ Z:0≤ x ≤12}. Show that R={(a,b):a,b ∈ A, |a-b| is divisible by 4} is an equivalence relation. Find the set of all elements related to 1. Also write the equivalence class [2].
Find the angle of intersection of the curves y² = 4ax and x²= 4by.
Find the equation of the tangent to the curve which is parallel to the line 4x – 2y + 5 = 0.
Find the particular solution of the differential equation (x – sin y) dy + (tan y) dx = 0, given that y = 0 when x = 0.