Question

Find the range of function $f:[0,1]\to R,f\left(x\right)=x^3-x^2+4x+2{\sin }^{-1}\left(x\right)$.

• A. $[-(\mathrm{\pi }+2),0]$
• B. $[0,(4+\mathrm{\pi }\left)\right]$
• C. $[2,3]$
• D. $[0,(2+\mathrm{\pi }\left)\right]$

Answer 1

The correct option is B

$[0,(4+\mathrm{\pi }\left)\right]$

Explanation for correct option.

Step:1 Simplify the function f(x).

$$\begin{gathered} f\left(x\right)=x^3-x^2+4x+2{\sin }^{-1}\left(x\right) \\ \Rightarrow f\left(x\right)=x^3+2{\sin }^{-1}\left(x\right)-x^2+4x+4-4 \\ \Rightarrow f\left(x\right)=x^3+2{\sin }^{-1}\left(x\right)-(x^2-4x+4)+4 \\ \Rightarrow f\left(x\right)=x^3+2{\sin }^{-1}\left(x\right)-{(x-2)}^2+4 \\ \Rightarrow f\left(x\right)=x^3+2{\sin }^{-1}\left(x\right)+4-{(x-2)}^2 \end{gathered}$$

Here, the part $x^3+2{\sin }^{-1}\left(x\right)+4$ of the function is increasing.

Step: Calculate the range of f(x)

Here, in

$$\begin{gathered} f\left(x\right),x\in (0,1) \\ \therefore (x-2)\in (-2,-1) \\ \Rightarrow {(x-2)}^2\in (4,1) \\ \Rightarrow {(x-2)}_{min}atx=1;{(x-2)}_{max}atx=0. \end{gathered}$$

$\begin{array}{rcl}f{\left(x\right)}_{max}atx& =& 1\\ \therefore f{\left(x\right)}_{x=1}& =& 4+1+\mathrm{\pi }-1\\ & =& 4+\mathrm{\pi }\\ \mathrm{f}{\left(\mathrm{x}\right)}_{\min }\mathrm{at}\mathrm{x}& =& 0;\\ \therefore \mathrm{f}{\left(\mathrm{x}\right)}_{\mathrm{x}=0}& =& 0+0+4-4\\ & =& 0\end{array}$

Range : $[0,(4+\mathrm{\pi }\left)\right]$.

Therefore, Option(B) is the correct answer.