Question:

If $f(x) = x^3 - \frac{1}{x^3}$, then $f(x) + f(\frac{1}{x})$ is equal

Updated On: Apr 16, 2024
  • $2 x^3$
  • $2 \frac{1}{x^3}$
  • 0
  • 1
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The Correct Option is C

Solution and Explanation

Since $f(x) = x^3 - \frac{1}{x^3}$ $f\left(\frac{1}{x}\right) = \frac{1}{x^{3}} - \frac{1}{\frac{1}{x^{3}}} = \frac{1}{x^{3}} - x^{3}$ Hence, $ f\left(x\right) + f\left(\frac{1}{x}\right) = x^{3} - \frac{1}{x^{3} }+ \frac{1}{x^{3}} - x^{3} = 0 $
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Concepts Used:

Relations and functions

A relation R from a non-empty set B is a subset of the cartesian product A × B. The subset is derived by describing a relationship between the first element and the second element of the ordered pairs in A × B.

A relation f from a set A to a set B is said to be a function if every element of set A has one and only one image in set B. In other words, no two distinct elements of B have the same pre-image.

Representation of Relation and Function

Relations and functions can be represented in different forms such as arrow representation, algebraic form, set-builder form, graphically, roster form, and tabular form. Define a function f: A = {1, 2, 3} → B = {1, 4, 9} such that f(1) = 1, f(2) = 4, f(3) = 9. Now, represent this function in different forms.

  1. Set-builder form - {(x, y): f(x) = y2, x ∈ A, y ∈ B}
  2. Roster form - {(1, 1), (2, 4), (3, 9)}
  3. Arrow Representation