Question

Find the greatest value of $f\left(x\right)=\cos (x{e}^{\left[x\right]}+7x^2-3x)$, $x\in [-1,\mathrm{\infty })$
A. -1
B. 1
C. 0
D. none of these

Answer 1

Hint: We first try to define the domain of the given trigonometric function. Then we state the range of the function for any values of t, $\cos t\in [-1,1]$. We also try to get one continuous domain of $2\pi$ distance. At the end we find one-point x for which it attains the maximum value.

Complete step by step answer:
The main function of the given $f\left(x\right)=\cos (x{e}^{\left[x\right]}+7x^2-3x)$ is $\cos$ function.
Now in a span of $2\pi$, it will take values of $[-1,1]$. Also, for any values of t, $\cos t\in [-1,1]$.
We just have to check that we can get a continuous domain of $2\pi$ distance.
Now we find a range of $x{e}^{\left[x\right]}+7x^2-3x$. Here $\left[x\right]$ is the greatest integer function which means the output is the greatest integer possible less than x.
As $x\in [-1,\mathrm{\infty })$, we can say $\left[x\right]\in [-1,\mathrm{\infty })$ with only integer values.
We also know for any value of x; the exponential function always gets only positive value.
So, $x\in [-1,\mathrm{\infty })$, we can say $e^{\left[x\right]}\in [{\displaystyle \frac{1}{e}},\mathrm{\infty })$.
So, we can see the function attains at least one continuous domain of $2\pi$ distance as it goes towards infinity.
So, we can say the greatest value of $f\left(x\right)=\cos (x{e}^{\left[x\right]}+7x^2-3x)$ is 1.
We can also prove it by just showing 1 value of x for which $f\left(x\right)=\cos (x{e}^{\left[x\right]}+7x^2-3x)=1$.
Let’s take $x=0$.
We find the value of $x{e}^{\left[x\right]}+7x^2-3x$. So, $x{e}^{\left[x\right]}+7x^2-3x=0$.
So, at $x=0$, $\cos 0=1$. We already got a point.

So, the correct answer is “Option B”.

Note: We need to always shoe at least one point which attains the maximum point. As the part $x{e}^{\left[x\right]}$ can have some fixed points due to the factor that $e^{\left[x\right]}$ attains only integer value. We have to show that at least one point of x satisfies the equation.