Question

If $f(x) = \dfrac{1}{2} - \tan \left( {\dfrac{{\pi x}}{2}} \right), - 1 < x < 1$ and $g(x) = \sqrt 3 + 4x - 4{x^2}$ , then find the domain of $(f + g)$ .
A.$\left[ {\dfrac{1}{2},1} \right]$
B. $[\dfrac{1}{2}, - 1)$
C. $[ - \dfrac{1}{2},1)$
D. $\left[ { - \dfrac{1}{2}, - 1} \right]$

Answer 1

Hint: The domain of $f(x)$ is given, set $\sqrt 3 + 4x - 4{x^2}$ to zero and obtain the values of x. Then obtain the common points between $ - 1 < x < 1$ and the obtained values of x to obtain the required answer.

Formula Used:
The quadratic formula of a quadratic equation $a{x^2} + bx + c = 0$ is $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ .
Domain of $(f + g)$=domain of $f(x) \cap $ domain of $g(x)$.

Complete step by step solution:
The given equation is $g(x) = \sqrt 3 + 4x - 4{x^2}$.
So, set $\sqrt 3 + 4x - 4{x^2}$to zero and obtain the values of x
$\sqrt 3 + 4x - 4{x^2} = 0$
$4{x^2} - 4x - \sqrt 3 = 0$
Apply the quadratic formula to obtain the values of x.
$x = \dfrac{{4 \pm \sqrt {{{( - 4)}^2} - 4.4.\left( { - \sqrt 3 } \right)} }}{{2.4}}$
$ = \dfrac{{4 \pm \sqrt {16 + 16\sqrt 3 } }}{8}$
$ = \dfrac{{4 \pm 4\sqrt {1 + \sqrt 3 } }}{8}$
$ = \dfrac{{1 \pm \sqrt {1 + \sqrt 3 } }}{2}$
Therefore, $x \approx - \dfrac{1}{2},\dfrac{3}{2}$
Now, let’s draw a number line to obtain the intersection of domain to given functions.

From the number line we have the intersection as $x \in [ - \dfrac{1}{2},1)$.

Option ‘C’ is correct

Note: Students are sometimes unable to factor the given quadratic equation because the equation lacks integer roots. So, to find the values of x, use the quadratic formula $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ and then find the intersection of two domains.