Chapter 8, Problem 16CT

a.

To determine

To find:The value of a, b and c so that function $f\left(x\right)=a{x}^2+bx+c$ is an even function.

Answer to Problem 16CT

The function $f\left(x\right)=a{x}^2+bx+c$ is an even function when $b=0$ and a, c are any real numbers.

Explanation of Solution

Given:

The function $f\left(x\right)=a{x}^2+bx+c$ is an even function.

Concept used:

Afunction $f\left(x\right)$ is an even function $f\left(-x\right)=f\left(x\right)$ .

Calculations:

The given function is $f\left(x\right)=a{x}^2+bx+c$ .

  $\Rightarrow f\left(-x\right)=a{\left(-x\right)}^2+b\left(-x\right)+c$

  $\Rightarrow f\left(-x\right)=a{x}^2-bx+c$

But $f\left(-x\right)=f\left(x\right)$

  $\Rightarrow a{x}^2-bx+c=a{x}^2+bx+c$

  $\Rightarrow -2bx=0$

  $\Rightarrow b=0$

Hence, the function is an even function when $b=0$ and a, c are any real numbers.

Conclusion:

The function $f\left(x\right)=a{x}^2+bx+c$ is an even function when $b=0$ and a, c are any real numbers.

b.

To determine

To find:The value of a, b and c so that function $f\left(x\right)=a{x}^2+bx+c$ is an odd function.

Answer to Problem 16CT

The function $f\left(x\right)=a{x}^2+bx+c$ is an odd function when $a=0$ , $c=0$andb is any real number.

Explanation of Solution

Given:

The function $f\left(x\right)=a{x}^2+bx+c$ is an odd function.

Concept used:

Afunction $f\left(x\right)$ is an even function $f\left(-x\right)=-f\left(x\right)$ .

Calculations:

The given function is $f\left(x\right)=a{x}^2+bx+c$ .

  $\Rightarrow f\left(-x\right)=a{\left(-x\right)}^2+b\left(-x\right)+c$

  $\Rightarrow f\left(-x\right)=a{x}^2-bx+c$

But $f\left(-x\right)=-f\left(x\right)$

  $\Rightarrow a{x}^2-bx+c=-\left(a{x}^2+bx+c\right)$

  $\Rightarrow a{x}^2-bx+c=-a{x}^2-bx-c$

  $\Rightarrow 2a{x}^2+2c=0$

  $\Rightarrow 2a{x}^2+2c=0x^2+0$

  $\Rightarrow a=0,c=0$

Hence, the function $f\left(x\right)=a{x}^2+bx+c$ is an odd function when $a=0$ , $c=0$andb is any real number.

Conclusion:

The function $f\left(x\right)=a{x}^2+bx+c$ is an odd function when $a=0$ , $c=0$andb is any real number.

c.

To determine

To find: The value of a, b and c so that function $f\left(x\right)=a{x}^2+bx+c$ is neither even nor odd function.

Answer to Problem 16CT

The function $f\left(x\right)=a{x}^2+bx+c$ is neither an even function nor an odd function when $a=0$ , $b=0$ , and c is any real number.

Explanation of Solution

Given:

The function $f\left(x\right)=a{x}^2+bx+c$ is an even function.

Concept used:

A constantfunction is neither an even function nor an odd.

Calculations:

The given function is $f\left(x\right)=a{x}^2+bx+c$ .

The function is a constant function when coefficients of x and $x^2$ are zero.

Hence, the function $f\left(x\right)=a{x}^2+bx+c$ is neither an even function nor an odd function when $a=0$ , $b=0$ , and c is any real number.

Conclusion:

The function $f\left(x\right)=a{x}^2+bx+c$ is neither an even function nor an odd function when $a=0$ , $b=0$ , and c is any real number.