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The domain of sin^{1} [log_{3} (x/3)] is
Domain of sin^{1}x is [1,1]
Therefore, 1 ≤ log_{3}(x/3) ≤ 1
3^{1} ≤ x/3 ≤ 3
1 ≤ x ≤ 9
Therefore, the domain of sin^{1}[log_{3}(x/3)] is [1,9]
Domain of definiti on of the function
If f : R → R satisfies f (x + y) = f ( x) + f (y) , for all x,
A function f from the set of natural numbers to integers defined by
We have f: N →I
If x and y are two even natural numbers,
Again if x and y are two odd natural numbers then
Also each negative integer is an image of even natural number and each positive integer is an image of odd natural number.
∴ f is onto.
Hence f is one one and onto both.
The range of the function f (x) =^{ 7 x} P_{x 3} is
If f : R → S, defined by
f (x) = sin x  √3 cosx+ 1, is onto, then the interval of S is
f (x) is onto ∴ S = range of f (x)
The graph of the function y = f(x) is symmetrical about the line x = 2, then
Let us consider a graph symm. with respect to line x = 2 as shown in the figure.
From the figure
f (x_{1}) = f (x_{2}), where x_{1} = 2x and x_{2} = 2+x
∴ f (2  x) = f (2+x)
Taking common solution of (i) and (ii), we get 2 < x < 3 ∴ Domain = [2, 3)
L et f : (– 1, 1) → B , be a function defined by then f is both one  one and onto when B is the interval
Clearly, range of f (x)
For f to be onto, codomain = range
∴ Codomain of function
A function is matched below against an interval where it is supposed to be increasing. Which of the following pairs is incorrectly matched?
Clearly function f (x) = 3x^{2}  2x+1 is increasing when
f ' (x) = 6x – 2 > 0 ⇒ x ∈[1 / 3,∞)
∴ f (x) is incorrectly matched with
A real valued function f(x) satisfies the functional equation f(x – y) = f(x) f(y) – f(a – x) f(a + y) where a is a given constant and f(0) = 1, f(2a – x) is equal to
f(2a – x) = f (a – (x – a))
= f(a) f(x – a) – f(0) f(x) = f(a) f(x –a) – f(x)
= – f(x)
[∴ x = 0, y = 0, f(0) = f^{2}(0) f^{2}(a)
⇒ f^{2}(a) = 0
⇒ f(a) = 0]
⇒ f(2a  x) =  f(x)
The largest interval lying in for which the function, , is defined, is
Let f: N→Y be a function defined as f(x) = 4x + 3 where Y = {y ∈ N : y = 4x + 3 for some x ∈ N}.
Show that f is invertible and its inverse is
Clearly f is one one and onto, so invertible
Let f(x) = ( x + 1)^{2} – 1,x > –1
Statement 1 : The set {x : f(x) = f ^{–1}(x) = {0, –1}
Statement2 : f is a bijection.
Given that f (x) = (x + 1)^{2} –1, x > –1 Clearly D_{f} = [–, ∞) but codemain is not given.
Therefore f (x) need not be necessarily onto.
But if f (x) is onto then as f (x) is one one also, (x + 1) being something +ve, f^{–1}(x) will exist where (x + 1)^{2} –1 = y
∴ The statement1 is correct but statement2 is false.
For real x, let f (x) = x^{3} + 5x + 1, then
Given that f (x) = x^{3} + 5x + 1
⇒ f (x) is strictly increasing on R
⇒ f (x) is one one
∴ Being a polynomial f (x) is cont. and inc.
Hence f is onto also. So, f is one one and onto R.
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