Maths 2 Marks

$\text{x}^2-25\leq0$

$(\text{x}+5)(\text{x}-5)\leq0$

$\text{x}\in[-5,5]$

Now, $\text{z}\in\text{A}$ (co-domain of f) and f is a surjection.

So, z = f(y), where $\text{y}\in\text{A}$ (domain of f) .....(1)

Now, $\text{y}\in\text{A}$ (co-domain of g) and g is a surjection.

So, y = g(x), where $\text{x}\in\text{A}$ (domain of g) .....(2)

From (1) and (2),

z = f(y) = f(g(x)) = (fog)(x), where $\text{x}\in\text{A}$ (domain of fog)

So, fog is a surjection.

$\Rightarrow\ \text{f}^{-1}(-25)=\phi$ $[\because\ \sqrt{-25}\notin\text{R}]$

gof(x) = 0, for all x fog(-3) = 12

fog(5) = 0

gof(-2) = 0

⇒ f(y) = x

⇒ 3y - 4 = x

⇒ 3y = x + 4

$\Rightarrow\ \text{y}=\frac{\text{x}+4}{3}$

$\Rightarrow\ \text{f}^{-1}(\text{x})=\frac{\text{x}+4}{3}$ [from (1)]

Range of f = {-1, 1}