The function $f:[0,3] \rightarrow [1,29]$,defined by $f(x)=2x^3-15x^2+36x+1$,is

Let $f:[0,10] \rightarrow [1,20]$ be a function defined as $f(x) = \begin{cases} \frac{60-5x}{3}, & 0 \leq x \leq 6 \\ 10, & 6 \leq x \leq 7 \\ 31-3x, & 7 \leq x \leq 10 \end{cases}$. The function $f$ is:

If $f: R \rightarrow R$ is defined by $f(x) = \begin{cases} x+4 & \text{for } x < -4 \\ 3x+2 & \text{for } -4 \leq x < 4 \\ x-4 & \text{for } x \geq 4 \end{cases}$ then the correct matching of List-$I$ from List-$II$ is:

List-$I$	List-$II$
$(A)$ $f(-5) + f(-4)$	$(i)$ $14$
$(B)$ $f(|f(-8)|)$	$(ii)$ $4$
$(C)$ $f(f(-7) + f(3))$	$(iii)$ $-11$
$(D)$ $f(f(f(f(0)))) + 1$	$(iv)$ $-1$
	$(v)$ $1$
	$(vi)$ $0$

If $f: R \rightarrow R$ is defined by $f(x) = x + 2|x + 1| + 2|x - 1|$, then the element in the co-domain, which has a unique pre-image in the domain is

If $f: R \rightarrow R$ is defined by $f(x) = x^2 + 3x + 4$,then the function $f$ is . . . . . . .

Show that the function $f: R \rightarrow R$ defined as $f(x) = x^{2}$ is neither one-one nor onto.