The domain for which the functions f(x)=2x2-1 and g(x)=1-3x are equal, is

Question

The domain for which the functions f(x)=2x2-1 and g(x)=1-3x are equal, is (A) R (B)  -2, (1/2)  (C) R+ (D)  2, (1/2)

Tardigrade Admin · Accepted Answer

For f(x) = g(x) ⇒ 2x2 - 1 = 1 - 3x ⇒ 2x2 + 3 x -2 = 0 ⇒ 2x2 + 4 x - x - 2 = 0 ⇒ (2x-1)(x+2)=0 Thus domain for which the function f(x)=g(x) is -2, (1/2) .

Q. The domain for which the functions $f (x) = 2 x^{2} - 1$ and $g (x) = 1 - 3 x$ are equal, is

$R$

${- 2, \frac{1}{2}}$

$R^{+}$

${2, \frac{1}{2}}$

For $f (x) = g (x)$
$\Rightarrow 2 x^{2} - 1 = 1 - 3 x$
$\Rightarrow 2 x^{2} + 3 x - 2 = 0$
$\Rightarrow 2 x^{2} + 4 x - x - 2 = 0$
$\Rightarrow (2 x - 1) (x + 2) = 0$
Thus domain for which the function $f (x) = g (x)$ is
${- 2, \frac{1}{2}}$ .

Q. The domain for which the functions f(x)=2x2−1 and g(x)=1−3x are equal, is

R

{−2,21​}

R+

{2,21​}

For f(x)=g(x) ⇒2x2−1=1−3x ⇒2x2+3x−2=0 ⇒2x2+4x−x−2=0 ⇒(2x−1)(x+2)=0 Thus domain for which the function f(x)=g(x) is {−2,21​}.

Q. The domain for which the functions $f (x) = 2 x^{2} - 1$ and $g (x) = 1 - 3 x$ are equal, is

$R$

${- 2, \frac{1}{2}}$

$R^{+}$

${2, \frac{1}{2}}$

For $f (x) = g (x)$
$\Rightarrow 2 x^{2} - 1 = 1 - 3 x$
$\Rightarrow 2 x^{2} + 3 x - 2 = 0$
$\Rightarrow 2 x^{2} + 4 x - x - 2 = 0$
$\Rightarrow (2 x - 1) (x + 2) = 0$
Thus domain for which the function $f (x) = g (x)$ is
${- 2, \frac{1}{2}}$ .