Question

The Domain for which the functions $f(x) = 2{x^2} - 1$ and $g(x) = 1 - 3x$ are equal i.e., $f(x) = g(x)$, is
A. $\left\{ {0,2} \right\}$
B. $\left\{ {\dfrac{1}{2}, - 2} \right\}$
C. $\left\{ { - \dfrac{1}{2},2} \right\}$
D. $\left\{ {\dfrac{1}{2},2} \right\}$

Answer 1

Hint: Equate both the equations as given in the question. Solve the equality and form the quadratic equation. Find the roots of the required quadratic equation. As, the domain of the function is the value of its variable.

Complete step by step solution: 
Given that,
$f(x) = g(x)$
$ \Rightarrow 2{x^2} - 1 = 1 - 3x$
$2{x^2} - 1 - 1 = 1 - 3x - 1$
$2{x^2} - 2 = - 3x$
$2{x^2} - 2 + 3x = - 3x + 3x$
$2{x^2} + 3x - 2 = 0$
Finding roots of above quadratic equation
$2{x^2} + 4x - x - 2 = 0$
$2x(x + 2) - 1(x + 2) = 0$
$(2x - 1)(x + 2) = 0$
$2x - 1 = 0,x + 2 = 0$
$x = \dfrac{1}{2},x = - 2$
Roots of the quadratic equations are the domain of any equation
$x = \left\{ {\dfrac{1}{2}, - 2} \right\}$
Hence, the domain of the function $f(x) = g(x)$ is $\left\{ {\dfrac{1}{2}, - 2} \right\}$.
Therefore Option (2) is the correct answer i.e., $\left\{ {\dfrac{1}{2}, - 2} \right\}$.


Note: (General quadratic equation $a{x^2} + bx + c$)In such questions, while finding the roots of quadratic equation multiply the value of $a,c$ and then try to open that multiple in such form where you will get the value of $b$. For example – Let, the quadratic equation be $2{x^2} + 3x - 2$. Here, the multiple of $a,c$ is $ - 4$ and it can be written as $ + 4 - 1 = 3 = b$.