Consider given by f(x) = Show that f is bijective. Find the inverse of f and hence find f⁻¹(0) and x such that f⁻¹(x) = 2.
![Question 2 Answer-Image](/wp-content/uploads/images/12_Maths/Relations And Functions_534_1.png)
![Question 3 Answer-Image](/wp-content/uploads/images/12_Maths/Relations And Functions_534_2.png)
![Question 4 Answer-Image](/wp-content/uploads/images/12_Maths/Relations And Functions_534_3.png)
![Question 5 Answer-Image](/wp-content/uploads/images/12_Maths/Relations And Functions_534_4.png)
![Question 6 Answer-Image](/wp-content/uploads/images/12_Maths/Relations And Functions_534_5.png)
![Question 7 Answer-Image](/wp-content/uploads/images/12_Maths/Relations And Functions_534_6.png)
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).
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?
Let A = R – {3}, B = R – {1}. Let f : A → B be defined by f(x) = Show that f is bijective. Also, find
(i) x, if = 4
(ii)
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).
Let f : W → W be defined as show that f is invertible. Find the inverse of f,where W is the set of all whole numbers.
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).
Find the particular solution of the differential equation given that y = 0 when x = 1.
Find the value of p for which the curves x² = 9p(9 – y) and x² = p(y + 1) cut each other at right angles.
Find the equation of the tangent to the curve which is parallel to the line 4x – 2y + 5 = 0.
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].
Show that the normal at any point θ to the curve x = a cos θ + aθ sin θ, y = a sin θ – aθ cos θ is at a constant distance from the origin.