Advertisements
Advertisements
प्रश्न
The function f is defined by \[f\left( x \right) = \begin{cases}x^2 , & 0 \leq x \leq 3 \\ 3x, & 3 \leq x \leq 10\end{cases}\]
The relation g is defined by \[g\left( x \right) = \begin{cases}x^2 , & 0 \leq x \leq 2 \\ 3x, & 2 \leq x \leq 10\end{cases}\]
Show that f is a function and g is not a function.
उत्तर
The function f is defined by
\[f\left( x \right) = \begin{cases}x^2 & 0 \leqslant x \leqslant 3 \\ 3x & 3 \leqslant x \leqslant 10\end{cases}\]
It is observed that for 0 ≤ x < 3, f (x) = x2.
3 < x ≤ 10, f (x) = 3x
Also, at x = 3, f(x) = 32 = 9. And
f (x) = 3 × 3 = 9.
That is, at x = 3, f (x) = 9.
Therefore, for 0 ≤ x ≤ 10, the images of f (x) are unique.
Thus, the given relation is a function.
Again,
the relation g is defined as
g(x) = 3 × 2 = 6.
Hence, 2 in the domain of the relation g corresponds to two different images, i.e. 4 and 6.
Hence, this relation is not a function.
APPEARS IN
संबंधित प्रश्न
Find the domain of the function f(x) = `(x^2 + 2x + 1)/(x^2 - 8x + 12)`
Define a function as a correspondence between two sets.
If \[f\left( x \right) = \frac{1}{1 - x}\] , show that f[f[f(x)]] = x.
Write the domain and range of the function \[f\left( x \right) = \frac{x - 2}{2 - x}\] .
Write the domain and range of \[f\left( x \right) = \sqrt{x - \left[ x \right]}\] .
Let f and g be two functions given by
f = {(2, 4), (5, 6), (8, −1), (10, −3)} and g = {(2, 5), (7, 1), (8, 4), (10, 13), (11, −5)}.
Find the domain of f + g
If f : Q → Q is defined as f(x) = x2, then f−1 (9) is equal to
If 2f (x) − \[3f\left( \frac{1}{x} \right) = x^2\] (x ≠ 0), then f(2) is equal to
If \[e^{f\left( x \right)} = \frac{10 + x}{10 - x}\] , x ∈ (−10, 10) and \[f\left( x \right) = kf\left( \frac{200 x}{100 + x^2} \right)\] , then k =
If f(x) = sin [π2] x + sin [−π]2 x, where [x] denotes the greatest integer less than or equal to x, 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 definition of \[f\left( x \right) = \sqrt{4x - x^2}\] is
The domain of the function \[f\left( x \right) = \sqrt{5 \left| x \right| - x^2 - 6}\] is
If f(x) = 3x + a and f(1) = 7 find a and f(4).
Which sets of ordered pairs represent functions from A = {1, 2, 3, 4} to B = {−1, 0, 1, 2, 3}? Justify.
{(1, 1), (2, 1), (3, 1), (4, 1)}
Check if the relation given by the equation represents y as function of x:
2x + 3y = 12
Check if the relation given by the equation represents y as function of x:
2y + 10 = 0
Find x, if f(x) = g(x) where f(x) = x4 + 2x2, g(x) = 11x2
Check the injectivity and surjectivity of the following function.
f : Z → Z given by f(x) = x2
Find the domain of f(x) = log10 (x2 − 5x + 6)
Write the following expression as a single logarithm.
5 log x + 7 log y − log z
Solve for x.
log2 x + log4 x + log16 x = `21/4`
If `log((x + y)/3) = 1/2 log x + 1/2 logy`, show that `x/y + y/x` = 7
If `log(( x - y)/4) = logsqrt(x) + log sqrt(y)`, show that (x + y)2 = 20xy
Select the correct answer from given alternatives.
If f : R → R is defined by f(x) = x3 then f–1 (8) is equal to :
A function f is defined as : f(x) = 5 – x for 0 ≤ x ≤ 4. Find the value of x such that f(x) = 3
Answer the following:
If a2 + b2 = 7ab, show that, `log(("a" + "b")/3) = 1/2 log "a" + 1/2 log "b"`
Answer the following:
If a2 = b3 = c4 = d5, show that loga bcd = `47/30`
Answer the following:
Find the range of the following function.
f(x) = `1/(1 + sqrt(x))`
Answer the following:
Find the range of the following function.
f(x) = 1 + 2x + 4x
Given the function f: x → x2 – 5x + 6, evaluate f(2)
A function f is defined by f(x) = 2x – 3 find x such that f(x) = f(1 – x)
If f(x) = `(x - 1)/(x + 1), x ≠ - 1` Show that f(f(x)) = `- 1/x`, Provided x ≠ 0
The range of 7, 11, 16, 27, 31, 33, 42, 49 is ______.
Find the domain of the following functions given by f(x) = x|x|
Let f(x) = `sqrt(x)` and g(x) = x be two functions defined in the domain R+ ∪ {0}. Find `(f/g)(x)`
If f(x) = x3 – 1 and domain of f = {0, 1, 2, 3}, then domain of f–1 is ______.
The range of the function f(x) = x2 + 2x+ 2 is ______.
