Maths 3 Marks

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.

$\text{f}:\text{R}\rightarrow0,\infty;\ \text{g}:0,\infty\rightarrow\text{R}$

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(log_ex) = log_ee^x

= x

Computing gof: Clearly, the range of f is a subset of the domain of g.

⇒ fog : R → R

(gof)(x) = g(f(x))

= ge^x

= log_ee^x

= 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.

$\text{gof}:[-3,\infty)\rightarrow\text{R}$

$\Rightarrow\ \text{(gof)(x)}=\text{g(f(x)})$

$=\text{g}(\sqrt{\text{x}+3})$

$=(\sqrt{\text{x}+3})^2+1$

$=\text{x}+3+1$

$=\text{x}+4$

Range of f = R $\subset$ Domain of g = R ⇒ gof exists

Range 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)