Question

If f(x) = $x^3+x^2$ for 0 $\le x\le 2$ then the odd extension of f(x) would be -

x +2 for 2 < x $\le$ 4

• A. x³ + x² for -2 ≤ x ≤ 0 f(x) = x +2 for -4 ≤ x < -2
• B. -x³ + x² for -2 ≤ x ≤ 0 f(x) = -x +2 for -4 ≤ x < -2
• C. x³ - x² for -2 ≤ x ≤ 0 f(x) = x -2 for -4 ≤ x < -2
• D. None of these

Answer 1

The correct option is C

x³ - x² for -2 ≤ x ≤ 0

f(x) =

x -2 for -4 ≤ x < -2

Given,

$x^3+x^2$ for 0 $\le$ x $\le$ 2

f(x) =

x +2 for 2 < x $\le$ 4

The function is defined only on positive side of X- axis. Now in order to extend the function we'll have to extend the domain on negative side as well.

To make it an odd function we'll replace x by –x and multiply the whole expression by (-1). So we'll have

(-1) $(-x^3+x^2)$ for -2 $\le$ x $\le$ 0

f(x) =

(-1) (-x +2) for -4 $\le$ x < -2

Or the complete function defined would be

$x^3+x^2$ for 0 $\le$ x $\le$ 2

f(x) =

x +2 for 2 < x $\le$ 4

$x^3-x^2$ for -2 $\le$ x $\le$ 0

f(x) =

x -2 for -4 $\le$ x < -2