Advertisements
Advertisements
Question
Answer the following:
Find the range of the following function.
f(x) = `1/(1 + sqrt(x))`
Solution
f(x) = `1/(1 + sqrt(x))` = y, (say)
∴ `sqrt(x) y + y` = 1
∴ `sqrt(x) = (1 - y)/y ≥ 0`
∴ `(y - 1)/y ≤ 0`
∴ 0 < y ≤ 1
Range = (0, 1]
APPEARS IN
RELATED QUESTIONS
Let A = {−2, −1, 0, 1, 2} and f : A → Z be a function defined by f(x) = x2 − 2x − 3. Find:
(b) pre-images of 6, −3 and 5.
et A = (12, 13, 14, 15, 16, 17) and f : A → Z be a function given by
f(x) = highest prime factor of x.
Find range of f.
If f(x) = (x − a)2 (x − b)2, find f(a + b).
If f, g and h are real functions defined by
Write the range of the function f(x) = cos [x], where \[\frac{- \pi}{2} < x < \frac{\pi}{2}\] .
Let f and g be two real functions given by
f = {(0, 1), (2, 0), (3, −4), (4, 2), (5, 1)} and g = {(1, 0), (2, 2), (3, −1), (4, 4), (5, 3)}
Find the domain of fg.
If \[f\left( x \right) = \frac{2^x + 2^{- x}}{2}\] , then f(x + y) f(x − y) is equal to
If f : R → R and g : R → R are defined by f(x) = 2x + 3 and g(x) = x2 + 7, then the values of x such that g(f(x)) = 8 are
If f : [−2, 2] → R is defined by \[f\left( x \right) = \begin{cases}- 1, & \text{ for } - 2 \leq x \leq 0 \\ x - 1, & \text{ for } 0 \leq x \leq 2\end{cases}\] , then
{x ∈ [−2, 2] : x ≤ 0 and f (|x|) = x} =
If \[f\left( x \right) = 64 x^3 + \frac{1}{x^3}\] and α, β are the roots of \[4x + \frac{1}{x} = 3\] . Then,
The domain of the function \[f\left( x \right) = \sqrt{\frac{\left( x + 1 \right) \left( x - 3 \right)}{x - 2}}\] is
The domain of the function \[f\left( x \right) = \sqrt{5 \left| x \right| - x^2 - 6}\] is
A function f is defined as follows: f(x) = 5 − x for 0 ≤ x ≤ 4. Find the value of x such that f(x) = 3
Check if the relation given by the equation represents y as function of x:
3x − 6 = 21
If f(m) = m2 − 3m + 1, find `f(1/2)`
If f(m) = m2 − 3m + 1, find f(x + 1)
Find x, if g(x) = 0 where g(x) = x3 − 2x2 − 5x + 6
Find x, if f(x) = g(x) where f(x) = x4 + 2x2, g(x) = 11x2
Find the domain and range of the following function.
f(x) = `sqrt((x - 3)/(7 - x))`
lf f(x) = 3(4x+1), find f(– 3)
Express the following exponential equation in logarithmic form
231 = 23
Express the following exponential equation in logarithmic form
10−2 = 0.01
Express the following logarithmic equation in exponential form
ln 1 = 0
Given that log 2 = a and log 3 = b, write `log sqrt(96)` in terms of a and b
If x = loga bc, y = logb ca, z = logc ab then prove that `1/(1 + x) + 1/(1 + y) + 1/(1 + z)` = 1
Select the correct answer from given alternative.
The domain and range of f(x) = 2 − |x − 5| is
Answer the following:
Show that, `log |sqrt(x^2 + 1) + x | + log | sqrt(x^2 + 1) - x|` = 0
Answer the following:
Simplify `log_10 28/45 - log_10 35/324 + log_10 325/432 - log_10 13/15`
Answer the following:
If `log (("a" + "b")/2) = 1/2(log"a" + log"b")`, then show that a = b
A function f is defined by f(x) = 3 – 2x. Find x such that f(x2) = (f(x))2
A plane is flying at a speed of 500 km per hour. Express the distance ‘d’ travelled by the plane as function of time t in hour
The data in the adjacent table depicts the length of a person's forehand and their corresponding height. Based on this data, a student finds a relationship between the height (y) and the forehand length (x) as y = ax + b, where a, b are constant.
| Length ‘x’ of forehand (in cm) |
Height 'y' (in inches) |
| 35 | 56 |
| 45 | 65 |
| 50 | 69.5 |
| 55 | 74 |
Find a and b
The range of the function f(x) = `(x^2 - 3x + 2)/(x^3 - 4x^2 + 5x - 2)` is ______
If f(x) = `{{:(x^2",", x ≥ 0),(x^3",", x < 0):}`, then f(x) is ______.
If f(x) = `(x - 1)/(x + 1)`, then show that `f(1/x)` = – f(x)
The domain of the function f(x) = `sin^-1((|x| + 5)/(x^2 + 1))` is (–∞, –a] ≈ [a, ∞). Then a is equal to ______.
The expression \[\begin{array}{cc}\log_p\log_p\sqrt[p]{\sqrt[p]{\sqrt[p]{\text{...........}\sqrt[p]{p}}}}\\
\phantom{...........}\ce{\underset{n radical signs}{\underline{\uparrow\phantom{........}\uparrow}}}
\end{array}\]where p ≥ 2, p ∈ N; ∈ N when simplified is ______.
The domain of the function f(x) = `1/sqrt(|x| - x)` is ______.
