If the functions are defined as f(x) = √x and g(x) = √(1-x), then what is the common domain of the following functions: f + g, f - g, f/g, g/f, g - f where (f+g)(x) = f(x) + g(x), (f/g)(x) = f(x)/g(x)

Option 1 - 0 ≤ x < 1

Option 2 - 0 < x < 1

Option 3 - 0 < x ≤ 1

Option 4 - 0 ≤ x ≤ 1

2 Views|Posted 5 months ago

Asked by Shiksha User

1 Answer

Sort by:

Answered by

Vishal Baghel | Contributor-Level 10

5 months ago

Correct Option - 4

Detailed Solution:

f (x) + g (x) = √x + √1-x. The domain requires x ≥ 0 and 1-x ≥ 0, so x ≤ 1. Domain is [0,1].

f (x) - g (x) = √x - √1-x. Domain is [0,1].

f (x)/g (x) = √x / √1-x. Requires x ≥ 0 and 1-x > 0, so x < 1. Domain is [0,1).

g (x)/f (x) = √1-x / √x. Requires 1-x ≥ 0 and x > 0. Domain is (0,1].

The common domain for all these functi

Upvote

Similar Questions for you

$\underset{x \to \frac{π^{-}}{2}}{l i m} \frac{\int_{x^{3}}^{{(\frac{π}{2})}^{2}} c o s t^{1 / 3} d t}{{(x - \frac{π}{2})}^{2}}$

Alok Kumar Singh

$\underset{h \to 0}{l i m} \frac{\int_{(\frac{π}{2} - h)}^{{(\frac{π}{2})}^{3}} c o s (t^{1 / 3}) d t}{h^{2}}$

$= \underset{h \to 0}{l i m} \frac{0 + 3 {(\frac{π}{2} - h)}^{2} c o s (\frac{π}{2} - h)}{2 h}$

$= \underset{h \to 0}{l i m} \frac{3 {(\frac{π}{2} - h)}^{2} s i n h}{2 h}$

$= \frac{3 π^{2}}{8}$

$\int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{8 \sqrt[]{2} c o s x}{(1 + e^{s i n x}) (1 + s i n^{4} x)}$ dx = ap + b log $(3 + 2 \sqrt[]{2})$

Alok Kumar Singh

$\int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{8 \sqrt[]{2} c o s x}{(1 + e^{s i n x}) (1 + s i n^{4} x)}$ dx

$= \int_{0}^{\frac{π}{2}} {(\frac{8 \sqrt[]{2} c o s x}{(1 + e^{s i n x}) (1 + s i n^{4} x)} + \frac{8 \sqrt[]{2} c o s x}{(1 + e^{- s i n x}) (1 + s i n^{4} x)})} d x$

$= 8 \sqrt[]{2} \int_{0}^{\frac{π}{2}} \frac{c o s x}{1 + s i n^{4} x} d x$

Let sin x = t

$I = 8 \sqrt[]{2} \int_{0}^{1} \frac{d t}{1 + t^{4}}$

$= 4 \sqrt[]{2} \int_{0}^{1} \frac{(1 + \frac{1}{t^{2}}) - (1 - \frac{1}{t^{2}})}{t^{2} + \frac{1}{t^{2}}} d t$

$= 4 \sqrt[]{2} \int_{0}^{1} \frac{(1 + \frac{1}{t^{2}}) d t}{{(t - \frac{1}{t})}^{2} + 2} - 4 \sqrt[]{2} \int_{0}^{1} \frac{(1 - \frac{1}{t^{2}}) d t}{{(t + \frac{1}{t})}^{2} - 2}$

$= 4 \sqrt[]{2} \cdot \frac{1}{\sqrt[]{2}} {(t a n^{- 1} \frac{t - \frac{1}{t}}{\sqrt[]{2}})}_{0}^{1} - 4 \sqrt[]{2} \cdot \frac{1}{2 \sqrt[]{2}} {[l o g | \frac{t + \frac{1}{t} - \sqrt[]{2}}{t + \frac{1}{t} + \sqrt[]{2}} |]}_{0}^{1}$

$= 2 π - 2 l o g | \frac{2 - \sqrt[]{2}}{2 + \sqrt[]{2}} |$ &nbs

Vishal Baghel

$\int_{- a}^{a} (| x | + | x - 2 |) d x = 22, a > 2$

$\int_{- a}^{0} (- 2 x + 2) d x + \int_{0}^{2} (x - x + 2) d x + \int_{2}^{a} (2 x - 2) d x = 22$

$\Rightarrow 2 a^{2} + 2 = 20 \Rightarrow a^{2} = 9 \Rightarrow a = 3$

$∴ \int_{3}^{- 3} (x + [x]) d x = - \int_{- 3}^{3} (2 x - {x}) d x = - \int_{- 3}^{3} 2 x d x + 6 {\int_{0}^{1} x d x = 6 . \frac{x^{2}}{2} |}_{0}^{1} = 3$

Vishal Baghel

Let sin = t

$\int \frac{s i n θ (2 s i n θ . c o s θ) (s i n^{6} θ + s i n^{4} θ + s i n^{2} θ) \sqrt[]{2 s i n^{4} θ + 3 s i n^{2} θ + 6}}{2 s i n^{2} θ}$ d

sin = t

cos . d = dt

$\int \frac{u^{1 / 2}}{12} d u = \frac{u^{3 / 2}}{18} + C = \frac{{(2 t^{6} + 3 t^{4} + 6 t^{2})}^{3 / 2}}{18} + C = \frac{{(2 s i n^{6} θ + 3 s i n^{4} θ + 6 s i n^{2} θ)}^{3 / 2}}{18} + C$

The value of $\underset{x \to 1}{l i m} \frac{(x^{2} - 1) s i n^{2} (π x)}{x^{4} - 2 x^{3} + 2 x - 1}$ is equal to:

Vishal Baghel

$x^{4} - 2 x^{3} + 2 x - 1 = {(x - 1)}^{2} (x^{2} - 1)$

$s i n π x = s i n (π (1 - x)$

$= - s i n (s i n π (x - 1))$

$\underset{x \to 1}{l i m} \frac{(x^{2} - 1) \cdot s i n^{2} π x}{(x^{2} - 1) {(x - 1)}^{2}} = \underset{x \to 1}{l i m} \frac{s i n^{2} (π (x - 1))}{{(x - 1)}^{2}}$