Let f : R -> R be a function satisfying (x - y) f(x + y) - (x + y) f(x - y) = 4xy (x^2 - y^2) for all x, y ∈ R. If f(1) = -2 (i) Find range of f(x) (ii) If f(x) = k has 3 distinct solutions find 'k'. (iii) If f(x) = k has exactly one solution find 'k' ---Here is the extraction of the content from the image: 1. Let f : R -> R be a function satisfying (x - y) f(x + y) - (x + y) f(x - y) = 4xy (x^2 - y^2) for all x, y ∈ R. If f(1) = -2 (i) Find range of f(x) (ii) If f(x) = k has 3 distinct solutions find 'k'. (iii) If f(x) = k has exactly one solution find 'k'.

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We have a functional equation problem. The equation is x minus y times f of x plus y, minus x plus y times f of x minus y, equals 4xy times x squared minus y squared, for all real x and y. We're given that f of 1 equals negative 2. We need to find the range of f, determine k for 3 distinct solutions, and k for exactly one solution. Let's start by solving the functional equation using substitution. Let u equal x plus y and v equal x minus y. Now let's substitute our variables. After substitution, we get v times f of u minus u times f of v equals u squared minus v squared times uv. This simplifies to u cubed v minus u v cubed. For non-zero u and v, we divide by uv to get f of u over u minus f of v over v equals u squared minus v squared. Let g of x equal f of x over x. Then g of u minus g of v equals u squared minus v squared. This means g of x minus x squared is constant. Therefore, f of x equals x cubed plus Cx. Now we use the condition f of 1 equals negative 2. Substituting into our general form, we get 1 plus C equals negative 2, so C equals negative 3. Therefore, f of x equals x cubed minus 3x. For part i, to find the range, we take the derivative: f prime of x equals 3x squared minus 3. Setting this to zero gives critical points at x equals plus or minus 1. We find f of negative 1 equals 2, which is a local maximum, and f of 1 equals negative 2, which is a local minimum. Since this is a cubic polynomial, the range is all real numbers. Now let's analyze the number of solutions. For part ii, we need f of x equals k to have exactly 3 distinct solutions. A cubic polynomial has 3 real solutions when the horizontal line y equals k intersects the curve at 3 points. This happens when k is strictly between the local maximum and minimum values. Since our local max is 2 and local min is negative 2, we need negative 2 less than k less than 2. For part iii, we need exactly one solution. This occurs when k is outside the interval from negative 2 to 2, so k is less than negative 2 or k is greater than 2. Let's summarize our complete solution. We started with a functional equation and used substitution to transform it into a simpler form. Through algebraic manipulation, we discovered that f of x equals x cubed minus 3x. For part i, the range is all real numbers since this is a cubic polynomial. For part ii, when we need exactly 3 distinct solutions, k must be in the open interval from negative 2 to 2. For part iii, when we need exactly one solution, k must be outside this interval. This functional equation problem demonstrates how substitution and analysis of critical points help us understand the behavior of polynomial functions.