**Even and Odd Functions Worksheet : **

Worksheet given in this section will be much useful for the students who would like to practice problems on even and odd functions.

Let f(x) be a function.

To find whether f(x) is even or odd, we have to replace "x" by "-x" in f(x). We have to conclude f(x) as even or odd function from the result of f(-x) as shown below.

**1. If f(-x) = f(x), then f(x) is even function**

**2. If f(-x) = - f(x), then f(x) is odd function**

If f(-x) is neither equal to f(x) nor -f(x), we have to conclude f(x) is neither even nor odd.

**Problem 1 : **

Let f(x) = x^{3}, is f(x) odd or even function ?

**Problem 2 : **

Let f(x) = x^{2} + 2, is f(x) odd or even function ?

**Problem 3 : **

Let f(x) = x^{3} - 2x, is f(x) odd or even function ?

**Problem 4 : **

Let f(x) = 5x^{3} + x^{2 }- 1, is f(x) odd or even function ?

**Problem 5 : **

Let f(x) = x^{4} + 2x^{2} + 5, is f(x) odd or even function ?

**Problem 6 : **

Is sinx odd or even function ?

**Problem 7 : **

Is cscx odd or even function ?

**Problem 8 : **

Is secx odd or even function ?

**Problem 9 :**** **

Is cosx odd or even function ?

**Problem 10 :**** **

Is tanx odd or even function ?

**Problem 11 :**** **

Let f(x) = sinx + tanx, is f(x) odd or even function ?

**Problem 12 :**** **

Let f(x) = secx + cosx, is f(x) odd or even function ?

**Problem 1 : **

Let f(x) = x^{3}, is f(x) odd or even function ?

**Solution : **

To know f(x) is odd or even function, let us plug x = -x in f(x).

Then, we have

f(-x) = (-x)^{3}

f(-x) = -x^{3}

f(-x) = - f(x)

So, f(x) is odd function.

**Problem 2 : **

Let f(x) = x^{2} + 2, is f(x) odd or even function ?

**Solution : **

To know f(x) is odd or even function, substitute -x for x in f(x).

Then, we have

f(-x) = (-x)^{2} + 2

f(-x) = x^{2} + 2

f(-x) = f(x)

So, f(x) is even function.

**Problem 3 : **

Let f(x) = x^{3} - 2x, is f(x) odd or even function ?

**Solution : **

To know f(x) is odd or even function, substitute -x for x in f(x).

Then, we have

f(-x) = (-x)^{3} - 2(-x)

f(-x) = -x^{3} + 2x

f(-x) = -(x^{3} - 2x)

f(-x) = -f(x)

So, f(x) is odd function.

**Problem 4 : **

Let f(x) = 5x^{3} + x^{2 }- 1, is f(x) odd or even function ?

**Solution : **

To know f(x) is odd or even function, substitute -x for x in f(x).

Then, we have

f(-x) = 5(-x)^{3} + (-x)^{2 }- 1

f(-x) = 5(-x^{3}) - x^{2 }- 1

f(-x) = -5x^{3} - x^{2 }- 1

f(-x) = -(5x^{3} + x^{2 }+ 1)

f(-x) can not be expressed as either as f(x) or -f(x).

So, f(x) is neither even nor odd function.

**Problem 5 : **

Let f(x) = x^{4} + 2x^{2} + 5, is f(x) odd or even function ?

**Solution : **

To know f(x) is odd or even function, substitute -x for x in f(x).

Then, we have

f(-x) = (-x)^{4 }+ 2(-x)^{2} + 2

f(-x) = x^{4 }+ 2x^{2} + 2

f(-x) = f(x)

So, f(x) is even function.

**Important Note : **

In trigonometric ratios, if we have negative angle, we have to understand that the angle will fall in the IV^{th }quadrant.

In IV^{th} quadrant, the trigonometric ratios "cos" and "sec" are positive and all other trigonometric ratios are negative

**Problem 6 : **

Is sinx odd or even function ?

**Solution : **

Let f(x) = sinx

To know f(x) is odd or even function, substitute -x for x in f(x).

Then, we have

f(-x) = sin(-x)

Because the angle is negative, it falls in the IV^{th} quadrant. In IV^{th} quadrant "sin" is negative.

So, we have

f(-x) = - sinx

f(-x) = - f(x)

f(x) is odd function

So, sinx is odd function.

**Problem 7 : **

Is cscx odd or even function ?

**Solution : **

Let f(x) = cscx

To know f(x) is odd or even function, substitute -x for x in f(x).

Then, we have

f(-x) = csc(-x)

Because the angle is negative, it falls in the IV^{th} quadrant. In IV^{th} quadrant "csc" is negative.

So, we have

f(-x) = - cscx

f(-x) = - f(x)

f(x) is odd function

So, cscx is odd function.

**Problem 8 : **

Is secx odd or even function ?

**Solution : **

Let f(x) = secx

To know f(x) is odd or even function, substitute -x for x in f(x).

Then, we have

f(-x) = sec(-x)

Because the angle is negative, it falls in the IV^{th} quadrant. In IV^{th} quadrant "sec" is positive.

So, we have

f(-x) = secx

f(-x) = f(x)

f(x) is even function

So, secx is even function.

**Problem 9 :**** **

Is cosx odd or even function ?

**Solution : **

To know f(x) is odd or even function, substitute -x for x in f(x).

Then, we have

f(-x) = cos(-x)

Because the angle is negative, it falls in the IV^{th} quadrant. In IV^{th} quadrant "cos" is positive.

So, we have

f(-x) = cosx

f(-x) = f(x)

f(x) is even function

So, cosx is even function.

**Problem 10 :**** **

Is tanx odd or even function ?

**Solution : **

Let f(x) = tanx

To know f(x) is odd or even function, substitute -x for x in f(x).

Then, we have

f(-x) = tan(-x)

Because the angle is negative, it falls in the IV^{th} quadrant. In IV^{th} quadrant "tan" is negative.

So, we have

f(-x) = - tanx

f(-x) = - f(x)

f(x) is odd function

So, tanx is odd function.

**Problem 11 :**** **

Let f(x) = sinx + tanx, is f(x) odd or even function ?

**Solution : **

To know f(x) is odd or even function, substitute -x for x in f(x).

Then, we have

f(-x) = sin (-x) + tan(-x)

Because the angle is negative, it falls in the IV^{th} quadrant. In IV^{th} quadrant both "sin" and "tan" are negative.

So, we have

f(-x) = - sinx - tanx

f(-x) = - (sinx + tanx)

f(-x) = - f(x)

So, f(x) is odd function.

**Note : **The sum or difference of two odd functions is always odd.

**Problem 12 :**** **

Let f(x) = secx + cosx, is f(x) odd or even function ?

**Solution : **

To know f(x) is odd or even function, substitute -x for x in f(x).

Then, we have

f(-x) = sec(-x) + cos(-x)

Because the angle is negative, it falls in the IV^{th} quadrant. In IV^{th} quadrant both "sec" and "cos" are positive.

So, we have

f(-x) = secx + cosx

f(-x) = f(x)

So, f(x) is even function.

**Note : **The sum or difference of two even functions is always even.

After having gone through the stuff given above, we hope that the students would have understood even and odd functions.

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