Question

The greatest value of $f\left(x\right)=\cos (x{e}^{\left[x\right]}+7x^2-3x),x\in [-1,\infty )$ is

• A. $-1$
• B. $1$
• C. $0$
• D. None of the above.

Answer 1

Answer: (B)

$1$

$f\left(x\right)=\cos (x{e}^{\left[x\right]}+7x^2-3x),xϵ(-1,\infty )$

At $x=0$

$f\left(0\right)=\cos (0+7\times {\left(0\right)}^2-3\times 0)$

$=\cos 0$

$=1$

Now, $-1\le \cos \theta \le 1$

Thus maximum value of $f\left(x\right)$ can be $1$

Therefore, maximum value of $f\left(x\right)=1$ at $x=0$