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f(x) = x + 1, g(x) = ex
Range of f = R
$\subset$ Domain of g = R ⇒ gof existRange of
$\text{g}=(0,\infty)\ \subset$ Domain of f = R ⇒ fog existNow,
gof(x) = g(f(x)) = g(x + 1) = ex+1
And,
fog(x) = f(g(x)) = f(ex) = ex + 1
f(x) = x2, g(x) = cosx
$\text{f}(\text{x})=\text{c},\text{c}\in \text{R},\text{g(x)}=\sin \text{x}^2$
Here, f : A → A is defined by
f(x) = x2
Clearly f in not injective,
$\because\ \text{f}(1)=\text{f}(-1)=1$
So, f is not bijective and hence not invertible.
Hence, f-1 does not exist.
f(x) = x2 + 2, $\text{g(x)}=1-\frac{1}{1-\text{x}}$
f(x) = |x|, g(x) = sinx
$\text{f}:\text{R}\rightarrow(0,\infty);\ \text{g}:\text{R}\rightarrow[-1,1]$
Computing fog: Clearly, the range of g is a subset of the domain of f.
⇒ fog : R → R
(fog)(x) = f(g(x))
= f(sinx)
= |sinx|
Computing gof: Clearly, the range of f is a subset of the domain of g.
⇒ fog : R → R
(gof)(x) = g(f(x))
= g(|x|)
= sin|x|
Injectivity: f(x) = 0 for all $\text{x}\in\text{Z}$
Therefore, f is not one-one.Surjectivity: Range of f = (0, 1) ≠ R.
Co-domain of f = R Both are not same. Therefore, f is not onto.Injection test: Let x and y be any two elements in the domain (A), such that f(x) = f(y).
f(x) = f(y) |x| = |y| $\text{x}=\pm\text{y}$ So, f is not one-one. Surjection test: For y = -1, there is no value of x in A. So, f is not onto. So, f is not bijective.$\Rightarrow\ \text{x}=\text{n}\pi+(-1)^{\text{n}}\text{y}$
$\Rightarrow\ \text{x}\neq\text{y}$
$\therefore$
f is not one-one.Surjective: Let
$\text{y}\in\text{R}$ be arbitrary such thatf(x) = y
⇒ sinx = y ⇒ x = sin-1y Now, for $\text{y}>1\times\notin\text{R}$ (domain)$\therefore$ f is not onto.
Therefore, the domains of all fog, and IR are the same.
(fog)(x) = f(g(x)) = f(x - 1) = x - 1 + 1 = x = IR(x) ..... (1)(gof)(x) = g(f(x)) = g(x + 1) = x + 1 -1 = x = IR(x) ..... (2)
From (1) and (2), (fog)(x) = (gof)(x) = IR(x), $\forall\text{ x}\in\text{R}$ Hence, fog = gof = IRh(x) = x2
$\text{x}=\pm\text{y}$
So, f is not one-one. Surjection test: For y = -1, there is no value of x in A. So, f is not onto. So, f is not bijective.= 5050
= g(x)
(fog)(x) = f(g(x)) = f|x| = |x|Injection test: Let x and y be any two elements in the domain (Z), such that f(x) = f(y).
f(x) = f(y) x - 5 = y - 5 x = y Therefore, f is an injection. Surjection test: Let y be any element in the co domain (Z), such that f(x) = y for some element x in Z (domain). f(x) = y x - 5 = y x = y + 5, which is in Z. Therefore, f is a surjection and f is a bijection.So,
f : R → R, fog(x) = f(g(x)) = f(2x) = sin2x Clearly, $\text{fog}\neq\text{gof}$⇒ h is onto.
⇒ h is bijection. ⇒ h has an inverse and it is given by, h - 1 = {(7, 2), (9, 3), (11, 4), (13, 5)}$\text{x}=\pm\text{y}$
So, f is not an injection. Surjection test: Let y be any element in the co-domain (Z), such that f(x) = y for some element x in Z (domain). f(x) = y x2 = y$\text{x}=\pm\sqrt{\text{y}}$
which may not be in Z.For example, if y = 3,
$\text{x}=\pm\sqrt{3}$ is not in Z.
So, f is not a bijection.f : R → R, defined by f(x) = 1 + x2
Injection test: Let $\text{x, y}\in\text{R,}$ such that,
f(x) = f(y)
⇒ 1 + x2 = 1 + y2
⇒ x2 - y2 = 0
⇒ (x - y)(x + y) = 0
either x = y or x = -y or $\text{x}\neq\text{y}$
Therefore, f is not one-one.
Surjection: Let $\text{y}\in\text{R}$ be arbitrary, then
f(x) = y
⇒ 1 + x2 = y
⇒ x2 + 1 - y = 0
$\therefore\ \text{x}\pm\sqrt{\text{y}-1}\notin\text{R}$ or y < 1
$\therefore$ f is not onto.
f(x) = x + 1, g(x) = 2x + 3
f : R → R; g : R → R
Computing fog: Clearly, the range of g is a subset of the domain of f.
⇒ fog : R → R
(fog)(x) = f(g(x))
= f(2x + 3)
= 2x + 3 + 1
= 2x + 4
Computing gof: Clearly, the range of f is a subset of the domain of g.
⇒ fog : R → R
(gof)(x) = g(f(x))
= g(x + 1)
= 2(x + 1) + 3
= 2x + 5
f(x) = ex, g(x) = logex
$\text{f}:\text{R}\rightarrow0,\infty;\ \text{g}:0,\infty\rightarrow\text{R}$
Computing fog: Clearly, the range of g is a subset of the domain of f.
$\text{fog}(0,\infty)\rightarrow\text{R}$
fog(x) = f(g(x))
= f(logex) = logeex
= x
Computing gof: Clearly, the range of f is a subset of the domain of g.
⇒ fog : R → R
(gof)(x) = g(f(x))
= gex
= logeex
= x
For domain,
$\text{x}+3\geq0$
$\Rightarrow\ \text{x}\geq-3$
Domain of $\text{f}=[-3,\infty)$
Since f is a square root fuction, range of $\text{f}=[0,\infty)$
$\text{f}:[-3,\infty)\rightarrow[0,\infty)$
g(x) = x2 + 1 is a polynomial.
⇒ g : R → R
Computation of fog: Range of g is not a subset of the doamin of f.
and domain (fog) = {x : x $\in$ domain of g and g(x) $\in$ domain of f(x)}
⇒ Domain (fog) = {x : x $\in\text{R}$ and x2 + 1 $\in[-3,\infty)$}
⇒ Domain (fog) = {x : x $\in\text{R}$ and x2 + 1 $\geq-3$}
⇒ Domain (fog) = {x : x $\in\text{R}$ and x2 + 4 $\geq0$}
⇒ Domain (fog) = {x : x $\in\text{R}$ and $\text{x}\in\text{R}$}
⇒ Domain (fog) = R
fog : R → R
(fog)(x) = f(g(x))
= f(x2 + 1)
$=\sqrt{\text{x}^2+1+3}$
$=\sqrt{\text{x}^2+4}$
Computation of gof: Range of f is a subset of the doamin of g.
$\text{gof}:[-3,\infty)\rightarrow\text{R}$
$\Rightarrow\ \text{(gof)(x)}=\text{g(f(x)})$
$=\text{g}(\sqrt{\text{x}+3})$
$=(\sqrt{\text{x}+3})^2+1$
$=\text{x}+3+1$
$=\text{x}+4$
f(x) = x + 1, g(x) = sinx
Range of f = R
$\subset$ Domain of g = R ⇒ gof existsRange of g = [-1, 1] $\subset$ Domain of f = R ⇒ fog exists
Now,
fog(x) = f(g(x)) = f(sinx) = sinx + 1
And
gof(x) = g(f(x)) = g(x + 1) = sin(x + 1)